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LETTER FROM THE EDITORS This issue sees a change in the Editorship of CRUX MATHEMATICORUM. After many successful years under the accomplished leadership of Bill Sands and Robert Woodrow, the editorial oce has moved from the University of Calgary to Memorial University of Newfoundland. It is our intention to maintain the high standard that has led to CRUX MATHEMATICORUM having an excellent international reputation as one of the world's top problem solving journals. This high standard is a result of the submissions from the readership, and we encourage you all to continue to provide this. Please write to the new editor if you have suggestions for improvements. Readers will be delighted to know that Robert Woodrow has agreed to continue as editor of the Olympiad Corner. However, he will soon be relinquishing the Skoliad Corner. Contributions intended for that section should be sent directly to the editor. Bill Sands will remain \in the background", and the new editor is indeed delighted to be able to call on Bill's sage advice whenever necessary. There are a few changes in the works. We shall continue with the present format, which the readership appears to enjoy, but we shall change from 10 issues of 36 pages (360 pages per volume) to 8 issues of 48 pages (384 pages per volume). Although this has two fewer issues, the bonus is that there are 24 extra pages per year. This allows increased eciency of printing and helps to control mailing costs. The new schedule will have issues for February, March, April, May, September, October, November and December. We feel that this ts in with the teaching year of the majority of the readership. Since postal rates for mailing outside of Canada are signi cantly higher than within Canada, and as is customary for many publications, the Society has reluctantly adopted the policy that all subscription and other rates for subscribers with nonCanadian addresses must be paid in US funds. Although this policy was instituted in 1995, the Society has, until the present time, accepted payment in Canadian funds from any subscriber. However, eective January 1996, the Society will require payment in US funds where applicable. We trust that our nonCanadian subscribers will understand the necessity of this change in payment policy. We are also considering how to make CRUX available electronically to its subscribers. One possibility would be to send postscript les to subscribers. Please communicate your thoughts to the EditorinChief or to the Managing Editor (email: gpwright@acadvm1.uottawa.ca). We are now encouraging electronic submission of material. Please email submissions to cruxeditor@cms.math.ca. We use LATEX2e. Those of you who would like some of the technical details of the style use are asked to request this by email. Bruce Shawyer, EditorinChief Graham Wright, Managing Editor
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LETTRE DES REDACTEURS Le present numero marque l'entree en fonction d'une nouvelle e quipe a la direction de CRUX MATHEMATICORUM. Apres de nombreuses annees sous la direction experte de Bill Sands et de Robert Woodrow, le conseil de redaction quitte l'Universite de Calgary pour s'installer a l'Universite Memorial de TerreNeuve. Nous avons l'intention de maintenir le niveau de qualite qui a fait de CRUX MATHEMATICORUM une publication reconnue mondialement comme l'une des meilleures en resolution de problemes. L'excellence de notre periodique tient a votre apport, chers lecteurs, et nous vous encourageons tous a poursuivre dans la m^eme voie. Si vous avez des ameliorations a proposer, veuillez en faire part au nouveau redacteur en chef. Vous serez enchante d'apprendre que Robert Woodrow a accepte de conserver la direction de la chronique \Olympiade"; il delaissera toutefois la chronique \Skoliad". Si vous desirez contribuer a cette derniere, veuillez vous adresser au redacteur en chef. Bill Sands, pour sa part, demeurera \en veilleuse"; fort heureusement, nous pourrons compter sur ses judicieux conseils au besoin. Quelques changements sont a venir. Nous conserverons la presentation actuelle de la revue puisqu'elle semble plaire a nos lecteurs, mais nous produirons desormais huit numeros de 48 pages (env. 384 pages par volume) au lieu de dix numeros de 36 pages (env. 360 pages par volume). Vous recevrez donc deux numeros de moins par annee, mais vous aurez droit en prime a 24 pages de plus. Cette formule s'avere plus avantageuse du point de vue de l'impression et des frais d'expedition. Le nouveau calendrier de publication prevoit la parution d'un numero en fevrier, mars, avril, mai, septembre, octobre, novembre et decembre. Nous pensons qu'il correspond au calendrier scolaire de la majorite de nos lecteurs. Comme il lui en co^ute beaucoup plus cher d'expedier ses publications a l'etranger qu'au Canada, la Societe, a l'instar de nombreux autres e diteurs de revues, a e te contrainte d'adopter une nouvelle politique; dorenavant, tous les abonnes dont l'adresse postale n'est pas au Canada devront regler leurs frais d'abonnement et autres en devises americaines. M^eme si cette politique avait e te introduite en 1995, la Societe a jusqu'a present accepte les paiements en dollars canadiens qu'elle avait recus de ses abonnes. Mais a partir de janvier 1996, elle exigera des paiements en dollars americains, le cas e cheant. Nos abonnes de l'exterieur du Canada comprendront qu'il nous e tait devenu necessaire de modi er ainsi notre grille tarifaire. Par ailleurs, nous songeons a distribuer CRUX par voie e lectronique. L'une des options envisageesserait l'envoi de chiers postscript aux abonnes. Veuillez transmettre vos commentaires au redacteur en chef ou au redactiongerant (courrier e lectronique : gpwright@acadvm1.uottawa.ca).
3 Nous encourageons desormais nos lecteurs a soumettre leurs contributions par courrier e lectronique a cruxeditor@cms.math.ca. Nous utilisons LATEX2e. Si vous desirez obtenir certains details techniques quant au style, faitesen la demande par courrier e lectronique. Bruce Shawyer, Redacteur en chef Graham P. Wright, Redacteurgerant
CONGRATULATIONS
CRUX would like to extend its collective congratulations to Professor Ron Dunkley, founder of the Canadian Mathematics Competitions, on his appointment as a Member of the Order of Canada.
GUIDELINES FOR ARTICLES
Articles for this section of CRUX should satisfy the following: have length of two to four pages, ideally (we have allowed up to six pages in exceptional circumstances); be of interest to advanced high school and rst or second year university students; contain some new material that leads to further interesting questions (for this level); be well referenced as to origin of the problem and related material; not contain long involved formulas or expressions, that is, we like more elegant mathematics as opposed to that which involves tedious calculations and attention to detail. We really want to emphasize ideas and avoid many, many formulas. CRUX does not want to be too technical in its appeal, rather we want to have wide general interest.  Denis Hanson
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UNITARY DIVISOR PROBLEMS K.R.S. Sastry Suppose d is a (positive integral) divisor of a (natural) number n. Then d is a unitary divisor of n if and only if d and n=d are relatively prime, that is (d; n=d) = 1. For example, 4 is a unitary divisor of 28 because (4; 28=4) = (4; 7) = 1. However, 2 is not a unitary divisor of 28 because (2; 28=2) = (2; 14) = 2
6= 1. Our aim is to consider
(i) the unitary analogue of a number theory result of Gauss, and (ii) the unitary extension of super abundant numbers.
THE NUMBER d (n) AND THE SUM (n) OF UNITARY DIVISORS of n.
To familiarize ourselves with the novel concept of unitary divisibility, the following table, adapted from the one in [2], is presented. It consists of n; 1 n 12; n's unitary divisors; d (n), the number of unitary divisors of n; and (n), the sum of unitary divisors of n. TABLE
n
1 2 3 4 5 6 7 8 9 10 11 12
unitary divisors of n
1 1, 2 1, 3 1, 4 1, 5 1, 2, 3, 6 1, 7 1, 8 1, 9 1, 2, 5, 10 1, 11 1, 3, 4, 12
d(n) 1 2 2 2 2 4 2 2 2 4 2 4
(n) 1 3 4 5 6 12 8 9 10 18 12 20
From the above table it is clear that if n = p1 1 is the prime decomposition of n, then its unitary divisors are 1 and p1 1 . So, d (n) = 2 and (n) = 1+ p1 1 . If n1= p121 p221 isthe prime decomposition of n, then its unitary divisors are 1; p1 ; p2 ; p1 p2 2 . So d (n) = 4 = 22 and (n) = (1+ p1 1 )(1+ p2 2 ).
5 Now it is a simple matter to establish the following results: Let n = p1 1 p2 2 pk k denote the prime decomposition of n. Then
d(n) = 2k; (n) = (1 + p1 1 )(1 + p2 2 ) (1 + pk ); If (m; n) = 1; then (mn) = (m) (n):
(1)
k
(2) (3)
THE EULER FUNCTION AND A RESULT OF GAUSS. The Euler function counts the number (n) of positive integers that are less than and relatively prime to n. Also, (1) = 1 by de nition. For example, (6) = 2 because 1 and 5 are the only positive integers that are less than and relatively prime to 6. The following results are known [1]: then (n) = (p1 , 1)p1 1 ,1 :
If n = p1 1 ; If n =
k Y i=1
k Y
pi ; then (n) = i
If (m; n) = 1;
i=1
(4)
(pi ) : i
(5)
then (mn) = (m)(n):
(6)
P
Let D = fd : d is a divisor of ng. Then Gauss showedP that (d) = n. For example, if n = 12, then D = f1; 2; 3; 4; 6; 12g and (d) = (1) + (2)+ (3)+ (4)+ (6)+ (12) = 1+1+2+2+2+4P= 12. Analogously, if D = fd : d is a unitary divisor of ng, then what is (d )? The answer is given by Theorem 1. k Y
Theorem 1 Let n = pi i denote the prime decomposition of n and D = i=1 fd : d is a unitary divisor of ng. Then
X
(d) =
k Y
[1 + (pi )] : i
i=1
Proof: The unitary divisors of n are the 2k elements, see (1), in the set
D = f1; p1 1 ; p2 2 ; ; pk ; p1 1 p2 2 ; ; pk,,11 pk ; ; k
k
k
k Y i=1
pi g: i
6 Hence
! k X Y j (d) = (1) + (pi i ) + pi i pj + + pi i : i=1 i<j i=1
X
k X
On repeated applications of (5) and (6) we nd that
X
(d) = 1 +
k X i=1
(pi ) + i
X i<j
Yk (pi ) pj + + (pi ) : i
j
i
i=1
But the righthand side expression in the above equation is precisely the k Y expansion of [1 + (pi i )]. i=1
a numerical illustration, let n = 108. Then D = f1; 4; 27; 108g, P (dFor ) = (1)+ (4)+ (27)+ (108) = 1 + 2 + 18 + 36 = 57 by actual count. AlsoPn = 108 = 22 33 . , Therefore (d ) = [1 + (22)][1 + 33 ] = [1 + 2][1 + 18] = 57, from
Theorem 1. It is the converse problem that is more challenging to P solve: Given a positive integer m, nd the set N such that N = fn : (d ) = mg. k Y There is a solution n if and only if m has the form [1 + (pi i )]. To see i=1 P this, let m = 4. Then there is no solution n such that (d ) = m. This follows because if n = p , then 1 + (p) = 4 yields (p , 1)p,1 = 3. If p = 2, then 2,,11 = 3 has no solution for a positive integer . If p,is1odd, then (p , 1)p is even and hence there is no solution of (p , 1)p = 3 for a positive integer . We leave it as an exercise to show that N = if m = 2 for > 1. It is easy to verify that if m = p is an odd prime, then N = fp; 2pg. As another exercise, the reader may nd N when m = 1995. It is an open problem to determine the integers m for which N = . In the next section, we consider an extension of the concept of superabundant numbers in the context of unitary divisors.
UNITARY SUPER ABUNDANT NUMBERS. We call a natural number n unitary abundant if (n) is greater than 2n. For example, 150 is unitary abundant because (150) = 312 > 2(150).
To extend the work of Erdos and Alaoglu [3], we call a natural number
n unitary super abundant if n(n) m(m) for all natural numbers m n. Theorem 2 shows that the product of the rst k primes, k = 1; 2; is a unitary super abundant number.
7 Theorem 2 Let pk denote the kth prime, k = 1; 2; : Then n = p1 ; p2 pk is a unitary super abundant number. Proof: Consider the natural numbers of m n. Then m belongs to one of the three groups described below. I. m is composed of powers of primes pj
pk. pk and powers of some
II. m is composed of powers of some primes pj primes p` > pk . III. m is composed of powers of primes p` pk .
First of all we note that the total number of primes composing m in any of the three groups does not exceed k. We now show that m(m) n(n) in all the above three cases. Case I. Let m = p1 1 p2 2 pj j where some 's, except j , may be zero. From (2) we see that
(m) = Yj 1 + pi m pi i=1 6=0 j Y = 1 + p1 i i=1 6=0 Yj 1 1+ p i i=1 6=0 k Y 1 + p1 i i=1 = n(n) : i
i
i
i
i
i
(7)
k+1 Case II. In this case m = p1 1 p2 2 pj j pk+1 p` ` : Here too (7) holds. Some 's, except j and ` , may be zero. Furthermore,
` > k ) 1 + p1 < 1 + p1 < 1 + p 1 < < 32 : `
k
k,1
(8)
8 Again, from (2) and (3)
,k 1 + p + ! (m) = Yj 1 + pi `Y k+i + m p p i k+i i=1 6=0 i=1 6=0 ! `Y ,k Yj 1 1 = 1 + p 1 + p + i k+i i =1 i =1 6=0 6=0 Yk 1 1 + p i i=1 k Y 1 1+ p i i=1 = n(n) ; k i
i
k i
i
i
i
k i
i
i
i
i
on using (7) and (8). Case III. In this case m = pk k p` ` . Here too (7) holds. Some 's, except ` , may be zero. As in earlier cases
! ,k (m) = `Y 1 1 + p + m k+i i =0 6=0 `Y ,k 1 + p1 k+i i=0 k Y 1 1+ p i i=1 = n(n) on using (8). We now give an example of a number n to show that there are values of m less than n that can belong to any of the three groups described in k i
i
Theorem 2. 96 Let n = 210 = 2:3:5:7. Then n(n) = 35 . I. m = 120 = 23 :3:5 is less than n and belongs to Group I.
(m) = 9 (n) holds. m 5 n II. m = 165 = 3:5:11 is less than n and belongs to Group II. (m) = 96 (n) holds. Here m 55 n Here
9 III. m = 143 = 11:13 is less than n and belongs to Group III. (m) = 168 (n) holds. Here too,
m 143 n If we write nk = p1 p2 pk ; k = 1; 2; , then we observe that (nk+1) = (nk) 1 + 1 nk+1 nk pk+1 > n(nk) ; k = 1; 2; : k
This above observation, coupled with the argument used in the proof of Theorem 2, implies Theorem 3. Theorem 3 The only unitary super abundant numbers are
nk; k = 1; 2; 3; : That is
2; 6; 30; 210; 2310; :
Acknowledgement.
The author thanks the referees for their suggestions. References
[1] L.E. Dickson, History of the Theory of Numbers, Vol. I, Chelsea Publications (1971), 113{114. [2] R.T. Hanson and L.G. Swanson, Unitary Divisors, Mathematics Magazine, 52 (September 1979), 217{222. [3] R. Honsberg, Mathematical Gems I, MAA, 112{113. 2943 Yelepet Dodballapur 561203 Bangalore District Karnataka, India
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THE SKOLIAD CORNER No. 11 R.E. Woodrow
This number marks the rst anniversary of the Skoliad Corner. While I have not received a lot of correspondence about the column per se, what I have received has been positive. A good source of more entry level problems seems needed, and it would be nice to nd a way of recognizing eorts by students at this level to solve the problems. As a problem set this issue, we give the Sharp U.K. Intermediate Mathematical Challenge which was written by about 115,000 students February 2, 1995. Entrants had to be in the equivalent of School Year 11 or below for England and Wales. Questions 1{15 were worth 5 marks each and Questions 16{25 were worth 6 marks each. The allotted time was one hour. The use of calculators, rulers, and measuring instruments was forbidden. The contest was organized by the U.K. Mathematics Foundation with additional help from The University of Birmingham. Many thanks go to Tony Gardiner, University of Birmingham, for sending me this contest and others sponsored by the Foundation.
SHARP U.K. INTERMEDIATE MATHEMATICAL CHALLENGE February 2, 1995 Time: 1 hour
1. Which of these divisions has a whole number answer? A. 1234 5 B. 12345 6 C. 123456 7 D. 1234567 8
E. 12345678 9
2. The word \thirty" has six letters, and 6 is a factor of 30. How many of the numbers from one up to twenty have this curious property? A. 14 B. 2 C. 3 D. 4 E. 5 3. Rene Descartes (1596{1650) published his most famous book Discourse on Method in 1637. To illustrate his \method" he included an appendix on Geometry, in which he introduced the idea of cartesian coordinates and showed how their use allowed one to solve many problems which had previously been unsolved. Descartes is best known for a sentence on which
11 he based part of his philosophy. The famous sentence (in Latin) is \Cogito, ergo sum". This is usually translated into the ve English words given below (in dictionary order). When they are written in the correct order to make Descartes' saying, which word is in the middle? A. am B. I C. I D. therefore E. think 4. What is the sum of the rst nine prime numbers? A. 45 B. 78 C. 81 D. 91 E. 100 5. Three hedgehogs Roland, Spike and Percival are having a race against two tortoises Esiotrot and Orinoco. Spike (S) is 10 m behind Orinoco (O), who is 25 m ahead of Roland (R). Roland is 5 m behind Esiotrol (E), who is 25 m behind Percival (P). Which of the following words gives the order ( rst to
fth) at this point in the race? A. POSER B. SPORE C. ROPES D. PORES E. PROSE 6. Nabil was told to add 4 to a certain number and then divide the answer by 5. Instead he rst added 5 and then divided by 4. He came up with the \answer" 54. What should his answer have been? A. 34 B. 43 C. 45 D. 54 E. 211 7. Find the value of the angle BAD, where AB = AC , AD = BD, and angle DAC = 39 . A
A. 39
,@D , , DD @@ , DD @@ , B D C
B. 45 C. 47 D. 51 E. 60 8. A maths teacher who lives \opposite" (or is it \adjacent to"?) our local hospital claims they recently hung up a banner saying: \Selly Oak Hospital Can Always Handle the Odd Accident". In fact, between March 1993 and March 1994 the hospital handled 53; 744 accident and emergency cases. Approximately how many cases was this per day? A. 90 B. 150 C. 250 D. 500 E. 1000 9. The diagonal of a square has length 4 cm. What is its area (in cm2 )? p A. 2 B. 4 C. 4 2 D. 8 E. 16 10. If x = 3, which expression has a dierent value from the other four? A. 2x2 B. x2 + 9x C. 12x D. x2 (x , 1)2 E. 2x2 (x , 1)
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11. The triangle formed by joining the points with coordinates (,2; 1),
(2; ,1) and (1; 2) is
A. scalene B. rightangled but not isosceles C. equilateral D. isosceles but not rightangled E. rightangled and isosceles
12. The number of students who sat the 1992 U.K. S.M.C. was 80; 000. The number who sat the 1993 U.K. S.M.C. was 105; 000. Which calculation gives the right percentage increase? 80;000 ;000 105; 000 ;000 100 A. 104 B. 80100 C. 105 ;000 100 80;000 ,80;000) 100 D. (105;000 80;000
000,80;000) E. (105;105 100 ;000
13. Professor Hardsum was very absentminded. He kept forgetting his fourdigit \PIN number"; without it he could not use his bank card to get cash out of the bank cash machines. Then one day he noticed that none of the digits is zero, the rst two digits form a power of ve, the last two digits form a power of two, and the sum of all the digits is odd. Now he no longer needs to remember the number because he can always work it out each time he needs it! What is the product of the digits of his PIN number? A. 60 B. 120 C. 240 D. 480 E. 960 14. A stopped clock may be useless, but it does at least show the correct time twice a day. A \good" clock, which gains just one second each day, shows the correct time far less often. Roughly how often?
'$ t &% 6 
A. once every 60 days B. once every 72 days C. once every 360 days D. once every 12 years E. once every 120 years
15. What is the value of the fraction 1 + 1 +2 3
1+4
when written as a decimal? A. 1:5 B. 2:25
C. 2:5
D. 2:6
E. 3:5
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16. Each year around 750; 000 bottles of water are prepared for the runners in the London Marathon. Each bottle holds 200 ml. George (who
nished last year in 3hr 39min) reckoned that of each bottle used \one quarter was more or less drunk, one half was sloshed all over the runner, and one quarter went straight into the gutter". If four fths of the bottles prepared were actually used by the runners, roughly how many litres were \more or less drunk"? A. 30; 000 B. 36; 000 C. 120; 000 D. 150; 000 E. 190; 000 17. Augustus Gloop eats x bars of chocolate every y days. How many bars does he get through each week? 1 A. 7yx B. 7xy C. 7xy D. 7xy E. 7xy 18. How many squares can be formed by
qq qq qq qq qq qq
joining four dots in the diagram? A. 4 B. 5 C. 9 D. 11 E. 13 19. Timmy Riddle (no relation) never gives a straightforward answer. When I asked him how old he was, he replied: \If I was twice as old as I was eight years ago, I would be the same age as I will be in four years time". How old is Timmy? A. 12 B. 16 C. 20 D. 24 E. 28 20. Ivor Grasscutter's lawn, which is circular with radius 20 m, needs retur ng. Ivor buys the turf in 40 cm wide strips. What is the approximate total length he needs? A. 300 m B. 600 m C. 1500 m D. 3000 m E. 6000 m 21. Last year's carnival procession was 1 12 km long. The last oat set o, and nished, three quarters of an hour after the rst oat. Just as the
rst oat reached us, young Gill escaped. She trotted o to the other end of the procession and back in the time it took for half the procession to pass us. Assuming Gill trotted at a constant speed, how fast did she go? A. 3 km/h B. 4 km/h C. 5 km/h D. 6 km/h E. 7 km/h 22. In this 4 by 4 square you have to get from X to Y moving only along black lines. How many dierent shortest routes are there from X to Y ? Y
A. 18
B. 26
r
X C. 28
r
D. 32
E. 34
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23. When we throw two ordinary dice, the possible totals include all the numbers from 2 to 12. Suppose we have two dice with blank faces. If we mark the six faces of one dice 1; 2; 2; 3; 3; 4 how should we mark the faces of the second dice so that all totals from 2 to 12 are possible, and each total has exactly the same probability of occurring as with two ordinary dice? A. B. C. D. E. 1; 2; 2; 7; 7; 8 1; 3; 4; 5; 6; 8 1; 4; 4; 5; 5; 8 1; 3; 4; 5; 7; 8 1; 2; 3; 6; 7; 8
24. A regular pentagon is inscribed in a circle of radius p cm, and the vertices are joined up to form a pentagram. If AB has length 1 cm, nd the area inside the pentagram (in cm2 ). B ZZ BB Z ZZBB A BB
A. p2
B. 5p=2 C. 5p2=2 D. 5p E. 5p2 25. The six faces of a madeup cube are labelled F; H;I; N; X; Z . Three views of the labelled cube are shown. The cube is then opened up to form the net shown here (the net has been turned so that the F is upright). What should be drawn (upright) in the shaded square?
X XXX(X(XX XXX( XXXX XXX XX(X(( X X X X XX X X X XXXX X XXX
A. H
B. I
C. N
D. X
E. Z
Last month we gave the problems of the Saskatchewan Senior Mathematics Contest 1994. This month we give the \ocial" solutions. My thanks go to Gareth Grith, University of Saskatchewan, longtime organizer of the contest, for supplying the problems and answers.
15
Saskatchewan Senior Mathematics Contest 1994
1. Solve the equation
1 + 68x,4 = 21x,2 :
Solution. The equation 1+68x,4 = 21x,2 is equivalent to u2 , 21u + 68 = 0 wherepu = x2 . Since (u , 4)(u , 17) = 0, x2 = 4 or 17. Therefore x = 2 or 17. 2. Find the number of divisors of 16 128 (including 1 and 16 128). Solution. To begin with, consider a simpler example: Find the number of divisors of 12. The divisors are:
20 30; 20 31;
21 30; 21 31;
22 30 ; 22 31 :
There are 6 divisors. We notice that we arrive at the number 6 as (a+1)(b+1) where 12 = 2a 3b . This results is true in general
16128 = 28 32 7: The number of divisors is (8 + 1)(2 + 1)(1 + 1) = 54.
3. In the gure, lines ABC and ADE intersect at A. The points BCDE are chosen such that angles CBE and CDE are equal. Prove that the rectangle whose sides have length AB and AC and the rectangle whose sides have length AD and AE are equal in area.
q qq
@@ ,, QQ @A, BQ Q,Q@D@ , , QQ@ C,, Q@Q @Q@QE , @QQ@Q ,, @
q
q
Solution. Since angle CBE = angle CDE , the points BCDE are concyclic. (Converse of the theorem \angles in the same segment are equal".) (This is known as the theorem of Thales.) ABC and ADE are secants of this circle. Therefore AB AC = AD AE .
16
4. (a) State the domain and range of the functions
f (x) = tan x and g(x) = loga x; where a = 54 : (b) Determine the smallest value of x for which tan x = log5=4 x. Solution. (a) The domain of tan x is all the real numbers except for odd multiplies of =2. The range of tan x is all real numbers. The domain of loga x is all positive real numbers. The range is the real numbers. This is true for all a > 0, a = 6 1, and thus is true for a = 54 . (b) Consider the graphs of y = tan x, y = log(5)=4 x for 0 < x < 32 .
y = log5=4 (x)
,=2
0
1
=2
3
=2
log(5)=2 x is not de ned for x 0.
The graphs indicate that the smallest value for x at which the graphs intersect lies between and 32 . But tan 54 = 1 and log(5)=4 54 = 1. Therefore this point is x = 54 . Note: for students who have studied calculus: tan x 6= log(5)=4 x for 0 < x < ;
2
since (i) tan 1 > tan = 1, log(5)=4 1 = 0 and and
4
(ii) sec2 x > 1 >
1 x ln 54 for 1 < x < 2 .
5 > 3 > e; 4
17
5. (a) Prove that the system of equations x +y =1 x23 + y23 = 2 x +y =3
has no solution. (b) Determine all values of k such that the system
x +y =1 x32 + y32 = 2 x +y =k
has at least one solution. x + y = 1 Solution. (a) x2 + y 2 = 2 . Therefore 1 = (x + y )2 = x2 + 2xy + y 2 = 2 + 2xy . Therefore xy = , 12 . If x3 + y 3 = 3 then
(x + y)(x2 , xy + y 2) = (1)(2 , xy ) = 3: Therefore xy = ,1. This is a contradiction and so the system has no solution.
(b) In order that the system admits at least one solution,
(x + y)(x2 , xy + y 2) = (1)(2 , xy) = k: Therefore xy = 2 , k and xy = , 12 so that 2 , k = , 21 ; k = 52 .
6. (Contributed by Murray Bremner, University of Saskatchewan.) This problem shows how we may nd all solutions to the equation X 2 + Y 2 = Z 2 where X , Y and Z are positive integers. Such a solution (X;Y; Z ) is called a Pythagorean triple. If (X;Y; Z ) have no common factor (other than 1), we call (X; Y; Z ) a primitive Pythagorean triple. Part I Remarks. Part I shows that if a, b are positive integers with no common factor and a > b then X = a2 , b2, Y = 2ab, Z = a2 + b2 is a primitive Pythagorean triple. (a) Let a, b be two positive integers with a > b. Show that X = a2 , b2, Y = 2ab, Z = a2 + b2 is a Pythagorean triple. Solution. Part I (a). X 2 + Y 2 = (a2 , b2)2 + 4a2b2 = a4 , 2a2b2 + b4 + 4a2 b2 = a4 + 2a2b2 + b2 = Z 2;
18 (b) Now assume that a, b have no common factor and not both are odd. Show that (X; Y; Z ) in (a) is a primitive Pythagorean triple. (Hint: Suppose that X , Y , Z have a common factor, p = some prime number. Then p divides Z + X and Z , X . Note that Z + X = 2a2 and Z , X = 2b2. So what is p? But Z must be odd (why?) so p can't be 2). Solution. Part I (b). Since a, b have no common factor, they cannot both be even. Since a, b are not both odd (given), one is odd and the other even. Therefore Z = a2 + b2 is odd. (The square of an odd number is odd; the square of an even number is even.) By the hint, if we suppose that there exists a prime p which divides X , Y and Z , p divides Z + X = 2a2 and Z , X = 2b2. But p cannot divide a and b (given) and so p divides 2. Therefore p = 2. But Z is odd. This contradiction implies that there does not exist such a prime p. Therefore, (X;Y; Z ) is a primitive Pythagorean triple. Part II Remarks. Part II shows that every primitive Pythagorean triple arises this way (for suitable choice of a, b). (a) Let (X;Y; Z ) be any Pythagorean triple. Show that the point ( XZ ; YZ ) lies on the unit circle x2 + y 2 = 1. Solution. Part II (a). The point ( XZ ; YZ ) lies on the circle x2 + y 2 = 1 if and only if ( XZ )2 + ( YZ )2 = 1. This is true since (X; Y; Z ) is a Pythagorean triple. (b) Let the slope of the line l which joins (,1; 0) to ( XZ ; YZ ) be b=a where a, b are positive integers with no common factor and a > b. Find the points of intersection of the line l and the unit circle in terms of a, b to show that 2 2
X = a ,b ; Z a2 + b2
Y = 2ab : Z a2 + b2 Solution. Part II (b). The line l, through (,1; 0) and with slope ab has equation y = ab (x + 1). l intersects the circle at those points for which 2 x2 + ab (x + 1) = 1 or (a2 + b2)x2 + 2b2x + b2 , a2 = 0 2b2 b2 , a2 or x2 + 2 2 x + 2 2 = 0 since a2 + b2 6= 0: a +b a +b b2 , a2 (x + 1) x + a2 + b2 = 0: 2 2 2 2 Therefore x = h,1 or x = aa2i,+bb2 . So XZ = aa2,+bb2 . 2 2 2 2 Further y = ab a ,ab 2++ab2 +b = a22+abb2 and so YZ = a22+abb2 as required.
19 (c) If (X;Y; Z ) is a primitive Pythagorean triple and if a, b are not both odd, show that X = a2 , b2 , Y = 2ab, Z = a2 + b2. Solution. Part II (c). Since (X; Y; Z ) is a Pythagorean triple, it follows from parts (a) and (b) there exist a, b such that
X = a2 , b2 ; Z a2 + b2
Y = 2ab : Z a2 + b2 Further, since a, b are not both odd, a2 + b2 and a2 , b2 are both odd. If a2 , b2 and a2 + b2 have a common factor then that factor would be a factor of their sum and dierence, namely 2a2 , 2b2. Note that 2 is not a common factor and therefore since (X;Y; Z ) is a primitive triple a, b have no common factor. Thus
X = a2 , b2; Y = 2ab; Z = a2 + b2: (d) If (X; Y; Z ) is a primitive Pythagorean triple and if a, b are both odd, a > b, we let r = 12 (a + b), s = 21 (a , b). (i) Prove that r, s are positive integers, that r > s, that r, s have no common factor (other than 1) and that r, s are not both odd. (ii) Let X 0 = 2(2rs), Y 0 = 2(r2 , s2 ), Z 0 = 2(r2 + s2 ). Show that X2 = X2 0 =2, Y = Y 0 =2, Z = Z 0=2. (Thus X = 2rs, Y = r2 , s2, Z = r + s ). Solution. Part II (d). (i) r = 12 (a + b), s = 12 (a , b) and a, b are both odd. The sum and dierence of two odd numbers is even and one half of an even number is an integer. Since a > b > 0 (given), r, s are positive integers. r , s = b > 0 and therefore r > s: Suppose that r, s have a common factor, p > 1. Then p divides r + s = a and r , s = b. But a, b have no common factor. Therefore neither do r, s. Suppose that r, s are both odd. Then r + s is even. But r + s = a which is odd. Therefore, r, s are not both odd. (ii) Note that
X 0 = 2(2rs) = (a + b )( a , b) = a2 , b2; 1 0 2 2 Y = 2(r , s ) = 2 14 (a2 + 2ab + b2 , a2 + 2ab , b2) = 2ab; Z 0 = 2(r2 + s2) = 2 4 (a2 + 2ab + b2 + a2 , 2ab + b2) = a2 + b2: Therefore (X 0 ; Y 0 ; Z 0 ) is a Pythagorean triple, but it0 is not primitive since 0 0 X Y Z 0 0 0 2 divides X , Y and Z . However X = 2 , Y = 2 , Z = 2 is primitive (given) and thus X = 2rs, Y = r2 , s2 , Z = r2 + s2 . For the sake of interest, we list the Pythagorean triples that result from small values of a, b (a > b):
20
a b a2 , b2 2ab a2 + b2 2 1
3
4
5
3 3 4 4
2 1 3 2
5 8 7 12
12 6 24 16
13 10 25 20
4 5 5 5 5
1 4 3 2 1
15 9 16 21 24
8 40 30 20 10
17 41 34 29 26
Remarks Surely, the most famous Pythagorean triple Both a; b are odd; see case d(ii) Since a; b are both even, the triple cannot be primitive. The corresponding primitive triple is the case a = 2, b = 1. Both a; b are odd; see case d(ii) Case d(ii) again
That completes the Skoliad Corner for this month. Please send me your preOlympiad contests and suggestions for future directions for this feature of Crux.
Historical Titbit Taken from a 1894 Ontario Public School textbook. Show how to dissect a rhombus, and reassemble the pieces to make a rectangle. Now this is quite easy, so we shall restate the problem: Given a rhombus with diagonals of length a and b, show how to dissect the rhombus, and reassemble the pieces to make a rectangle of maximum possible area. Call this maximum area . Let 0 < < . Show how to dissect the rhombus, and reassemble the pieces to make a rectangle of area . In each case, the dissection should be with the minimum possible number of pieces.
21
THE OLYMPIAD CORNER No. 171 R.E. Woodrow
All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. Another year has passed, and we begin the 1996 volume of Crux Mathematicorum with a change in the number of issues as well as in the Editorship. This is the rst number of the Corner which I will submit that does not fall under the critical gaze of Bill Sands who stepped down as EditorinChief at the end of December. I want to express my particular gratitude to Bill for his many hours of hard work and devotion to the publication over ten years. It is also time to welcome the new EditorinChief, with whom I look forward to working over the next years. Before launching into new material, let us pause to thank those who contributed to the Corner in 1995. My particular thanks go to Joanne Longworth whose layout skills with LATEX enhance the readability of the copy. It is also time to thank those who have contributed problem sets, solutions, comments, corrections, and criticisms over the past year. Among our contributors we have: Miquel Amenguel Covas Federico Ardila M. S efket Arslanagic SeungJin Bang Gerd Baron Bruce Bauslaugh Francisco Bellot Rosado Christopher Bradley Himadri Choudhury Tim Cross George Evagelopoulos Tony Gardiner Murray Grant Gareth Grith Georg Gunther Walther Janous Georey Kandall Derek Kisman Murray Klamkin Joseph Ling Andy Liu Sam Maltby Beatriz Margolis Stewart Metchette John Morvay Waldemar Pompe Bob Prielipp Toshio Seimiya Michael Selby Bruce Shawyer D.J. Smeenk Daryl Tingley Jim Totten Panos E. Tsaoussoglou Stan Wagon Edward T.H. Wang Martin White Chris Wildhagen C.S. Yogananda Thank you all (and anyone I've left out by accident). As an Olympiad set this number, we give the problems of the 29th Spanish Mathematical Olympiad (National Round). My thanks go to Georg
22 Gunther, Sir Wilfred Grenfell College and Canadian Team leader at the 34th I.M.O. at Istanbul, Turkey; and to Francisco Bellot Rosada, Valladolid, Spain, for collecting and sending me a copy of the contest.
29th SPANISH MATHEMATICAL OLYMPIAD (National Round) Madrid, February 26{27, 1992 FIRST DAY (Time: 4.5 hours)
1. At a party there are 201 people of 5 dierent nationalities. In each group of six, at least two people have the same age. Show that there are at least 5 people of the same country, of the same age and of the same sex. 2. Given the number triangle 0
1
1 2 3 4 : : : : : : 1991 1992 1993 3 5 7 :::::: 3983 3985 4 8 12 :::::: 7968
in which each number equals the sum of the two above it, show that the last number is a multiple of 1993. 3. Show that in any triangle, the diameter of the incircle is not bigger than the circumradius. SECOND DAY (Time: 4.5 hours)
4. Show that each prime number p (dierent from 2 and from 5) has an in nity of multiples which can be written as 1 1 1 1 : : : 1. 5. Given 16 points as in the gure
bb br
bb bb
bb bb
rb bb
D (lattice points)
A in which the points A and D are marked, determine, for all the possible manners, two other points B , C given that the six distances between these four points should be dierent. In this set of 4 points, (a) How many gures of four points are there (with the condition above)? (b) How many of them are noncongruent?
23 (c) If each point is represented by a pair of integers (Xi; Yi), show that the sum jXi , Xj j + jYi , Yj j, extended to the six pairs AB , AC , AD, BC , BD, CD, is constant. 6. A gamemachine has a screen in which the gure below is showed. At the beginning of the game, the ball is in the point S .
C A B D S G HHH HH
bb bb b
With each impulse from the player, the ball moves up to one of the neighbouring circles, with the same probability for each. The game is over when one of the following events occurs: (1) The ball goes back to S , and the player loses. (2) The ball reaches G, and the player wins. Determine: (a) The probability for the player to win the game. (b) The mean time for each game. Next I'd like to give six Klamkin Quickies. Many thanks to Murray Klamkin, University of Alberta, for creating them for us. The answers will appear in the next number.
SIX KLAMKIN QUICKIES
1. Which is larger p
( 3 2 , 1)1=3
or
2. Prove that
p3
p
p
1=9 , 3 2=9 + 3 4=9?
3 min ab + bc + ac ; ab + bc + ac (a + b + c) a1 + 1b + 1c
where a, b, c are sides of a triangle.
24
3. Let ! = ei= . Express ,! as a polynomial in ! with integral coecients. 4. Determine all integral solutions of the simultaneous Diophantine equations x + y + z = 2w and x + y + z = 2w . 5. Prove that if the line joining the incentre to the centroid of a triangle 13
2
2
1
2
2
1
4
4
4
4
is parallel to one of the sides of the triangle, then the sides are in arithmetic progression and, conversely, if the sides of a triangle are in arithmetic progression then the line joining the incentre to the centroid is parallel to one of the sides of the triangle. 6. Determine integral solutions of the Diophantine equation
x,y + y,z + z ,w + w,x = 0 x+y y+z z +w w+x
(joint problem with Emeric Deutsch, Polytechnic University of Brooklyn). To nish this number of the Corner, we turn to readers' solutions to problems from the May 1994 number of the Corner and the Final Round of the 43rd Mathematical Olympiad (1991{92) in Poland [1994: 129{130]. 1. Segments AC and BD intersect in point P so that PA = PD, PB = PC . Let O be the circumcentre of triangle PAB. Prove that lines OP and CD are perpendicular. Solutions by Joseph Ling, University of Calgary; by Toshio Seimiya, Kawasaki, Japan; and by Chris Wildhagen, Rotterdam, The Netherlands. We give Seimiya's solution and comment.
A
O B
D
P
E C
Because PA = PD, PB = PC , and \APB = \DPC , we get PAB PDC , so that \BAP = \CDP:
(1)
At least one of \PAB and \PBA is acute, so we may assume without loss of generality that \PAB is acute. Since O is the circumcentre of PAB we get OB = OP and \BOP = 2\BAP , so that
1 2
\OPB = 90 , \BOP = 90 , \BAP:
Let E be the intersection of OP with CD. Then \EPD = \OPB:
(2)
(3)
25 From (1), (2) and (3) we have
\EPD = 90 , \CDP: Thus \EPD + \EDP = \EPD + \CDP = 90 . Therefore OP ? CD. Comment: Generally if A, B , C , D are concyclic, we have OP ? CD and this theorem is an extension of Brahmagupta's theorem. 2. Determine all functions f de ned on the set of positive rational numbers, taking values in the same set, which satisfy for every positive rational number x the conditions
f (x + 1) = f (x) + 1 and f (x3) = (f (x))3:
Solution by Edward T.H. Wang and by Siming Zhan, Wilfrid Laurier University, Waterloo, Ontario. Let N and Q+ denote the set of positive integers. and the set of positive rational numbers, respectively. We show that f (x) = x, for all x 2 Q+, is the only function satisfying the given conditions. First of all, by the rst condition and an easy induction we see that f (x + n) = f (x) + n, for all x 2 Q+, and for all n 2 N. Now for arbitrary pq 2 Q+, where p; q 2 N, we have
3! p3 p 2 f q =q = f q 3 + 3p2 + 3pq 3 + q 6 p 3! =f q + 3p2 + 3pq 3 + q 6:
On the other hand
(1)
p 3! p 3 p 3 2 2 f q +q =f q + q = f q + q2 3 2 2 = f qp +3 f pq q 2 +3 f pq q 4 + q 6: (2) 3 Letting t = f ( qp ) and comparing (1) and (2), we get, since f ( pq3 ) = (f ( qp ))3, p(p+q3) = q2t23+q4t or q2t2+q4t,p(p+qp 3) = 0 orp (qt,pp)(qt+p+q3) = 0. Since qt + p + q > 0, we must have t = q , i.e. f ( q ) = q , and we are done. 3. Prove that the inequality r X r am an ! X 0 n=1 m=1 m + n holds for any real numbers a1 ; a2 ; : : : ; ar . Find conditions for equality.
26 Solutions by SeungJin Bang, Seoul, Korea; by Joseph Ling, University of Calgary; and by Chris Wildhagen, Rotterdam, The Netherlands. We give Ling's solution. Consider the polynomial
p(x) = Then
r X r X
n=1 m=1
amanxm+n,1
!
:
! X ! r xp(x) = am anxm+n = am xm anxn n=1 m=1 m=1 !2 n=1 r X = aixi 0; r X r X
r X
i=1
for all x 2 R. In particular p(x) 0 for all x 0. Hence
1! r X r am an X m +n 0 p(x) dx = x 0 0 n=1 m=1 m + n r X r am an X = n=1 m=1 m + n The inequality is strict unless xp(x) 0, that is a1 = a2 = = ar = 0. Z1
[Editor's Remark. All three solutions involved the integral. Can you furnish a nice solution avoiding the calculus?] 4. De ne the sequence of functions f0; f1; f2; : : : by f0(x) = 8 for all x 2 R; p fn+1(x) = x2 + 6fn(x) for n = 0; 1; 2; : : : and for all x 2 R: For every positive integer n, solve the equation fn(x) = 2x. Solutions by SeungJin Bang, Seoul, Korea; and by Chris Wildhagen, Rotterdam, The Netherlands. We use Bang's comment and solution. This problem is the same as problem 2 of the 19th Annual U.S.A. Mathematical Olympiad (which appeared in Math. Magazine 64, 3 (1991), pp. 211{ 213.) Since fn (x) is positive, fn(x) = 2x has only positive solutions. We show for each n, fn (x) = 2x has a solution x = 4. Since = p42 +f16(fx)(4) px2 +that, 48 , x = 4 is a solution of f ( x ) = 2 x . Now f (4) = 2 n+1 n = p42 + 6 8 = 8 = 2 4, which completes the inductive step. Next, induction on n gives us that for each n, fnx(x) decreases as x increases in (0; 1). It follows that fn(x) = 2x has the unique solution x = 4.
27
(k!)
6. Prove that, for every natural k, the number (k3)! is divisible by
k2+k+1
. Solution by Chris Wildhagen, Rotterdam, The Netherlands. Applying the well known fact that (ab)! is divisible by (a!)b b! yields 2 3 2 k (k )! = (k k )! is divisible by (k!) (k2)! and (k2)! = (k k)! is divisible by (k!)k+1 from which the required result follows immediately. The astute reader will notice we did not have a solution on le from the readership to problem 5. There's your challenge! That completes the Olympiad Corner for this issue. The Olympiad season is fast approaching. Please send me your national and regional contests.
Introducing the new EditorinChief This issue of CRUX marks the change of the Editorin Chief from Bill Sands and Robert Woodrow to Bruce Shawyer. For those of you who do not know Bruce, here is a short pro le: Born: Kirkcaldy, Scotland 1 Educated Kirkcaldy High School University of St. Andrews, Scotland Employment University of Nottingham, England University of Western Ontario, Canada Memorial University of Newfoundland, Canada (Head of Department, 1985{91)
Visiting Positions
Service Research Home Page
Universit at Ulm, Germany University of St. Andrews Canadian Mathematical Society
Team Leader, Team Canada at IMO 1987 and 1988
IMO95, Principal Organiser Summability of Series and Integrals Approximation Theory LATEX macros for the picture environment http://www.math.mun.ca/~bshawyer
Kirkcaldy, an industrial town north of Edinburgh, is the birthplace of Sir Sandford Fleming, who invented time zones, and of Adam Smith, the Father of economics. 1
28
THE ACADEMY CORNER No. 1
Bruce Shawyer All communications about this column should be sent to Professor Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7 With this issue, we start a new corner, which, for want of a better name, is being called \the Academy Corner". It will be concerned, in particular, with problem solving at the undergraduate level. This can take the form of a competition, a \problem of the week", a course in problem solving, etc. Your submissions and comments are welcome! Many Universities hold their own undergraduate Mathematics Competitions. We invite subscribers to send us information about them for publication here. We also solicit nice solutions to these problems. Please send them to the EditorinChief. We shall begin with the competition with which the editor is most familiar: the Undergraduate Mathematics Competition at Memorial University of Newfoundland. This competition is used as primary information for choosing Memorial's team for the Atlantic Provinces Council on the Sciences Mathematics Competition (more on that in a later issue).
Memorial University Undergraduate Mathematics Competition 1995 Time allowed  3 hours 1. Find all integer solutions of the equation x4 = y 2 + 71. 2. (a) Show that x2 + y 2 2xy for all real numbers x, y . (b) Show that a2 + b2 + c2 ab + bc + ca for all real numbers a, b, c. 3. Find the sum of the series
1 + 2 + 3 + 4 + : : : + 99 : 2! 3! 4! 5! 100!
29 4. If a, b, c, d are positive integers such that ad = bc, prove that a2 + b2 + c2 + d2 is never a prime number. 5. Determine all functions f : R ! R which satisfy
(x , y)f (x + y) , (x + y )f (x , y ) = 4xy(x2 , y 2)
for all real numbers x, y . 6. Assume that when a snooker ball strikes a cushion, the angle of incidence equals the angle of re ection. For any position of a ball A, a point P on the cushion is determined as shown. Prove that if the ball A is shot at point P , it will go into the pocket B .
i
B
ir
i
rw ri r r i A
i
P
This concludes the rst Academy Corner.
Mathematical Literacy 1. Who thought that the binary system would convince the Emperor of China to abandon Buddhism in favour of Christianity? 2. Who asked which king for one grain of wheat for the rst square of a chess board, two grains for the second square, four grains for the third square, and so on? 3. In which well known painting, by whom, does a Magic Square appear? 4. Where was bread cut into \Cones, Cylinders, Parallelograms, and several other Mathematical Figures"? 5. Which mathematician said: \Philosophers count about twohundred and eighty eight views of the sovereign good"? Answers will be given in a subsequent issue.
30
BOOK REVIEWS Edited by ANDY LIU The Monkey and the Calculator (Le singe et la calculatrice), softcover, ISBN 2909737071, 128 pages, 48 French francs (plus mailing). Aladdin's Sword (Le sabre d'Aladin), softcover, ISBN 290973708X, 128 pages, 48 French francs (plus mailing). Both published in 1995 by Production et Organisation du Loisir Educatif (POLE), 31 avenue des Gobelins, 75013 Paris, France. Reviewed by Claude La amme, University of Calgary. These two delightful little pocket books each contain a selection of almost 100 puzzles from the eighth International Championship of Mathematical Puzzles. This is an annual competition in four stages and in seven categories from elementary to advanced, open to anyone and administered by FFJM, the French Federation of Mathematical Games. These books are numbers 14 and 15 respectively in a series taken from these competitions and published by POLE, and most of these books are still in print. The rst book is from the Junior High level and the second from the Grand Public competition, slightlymore sophisticated. Nevertheless, no specialized knowledge is required for the competitions and appropriately the statements of these puzzles are all very elementary; but they puzzle your puzzler to the point that once you have read one of the problems, you cannot leave the book until you gure out a solution. Now this is a serious competition and a mere solution is usually not enough, the most elegant and complete one is the favourite! These puzzles should be of interest to anyone looking for some \fun brain gymnastics" (gymnastique intellectuelle), or \neurobics" as I have seen somewhere in the books. I can see here a great resource for teachers trying to interest students with some challenging but elementary and accessible problems. Here is a sample from the second book, called \A Numismatic Coincidence": By multiplying the number of pieces in her coin collection by 1994, the clever Miss Maths arrives at a product the sum of the digits of which is exactly equal to the number of pieces in her coin collection. How many pieces does Miss Maths have in her coin collection? All the problems are translated into English, a job very well done even if this means that some problems require a new answer, such as the problem asking us to ll in the blank with a number (written out in words) making the following sentence true (a hyphen doesn't count as a letter):
31 \This sentence has
letters."
There are two answers in English, and only one (and it is dierent) for the French version. The problem solutions are in French only but you will not be puzzled here, even a small diagram or a few numbers and the solution jumps at you. Apart from the puzzles themselves, you will nd a few advertisements for other mathematics magazines, even a puzzling one on music and mathematics, as well as for camps and Summer Schools. You will also nd details regarding the competition itself. By the way, the titles refer to some of the more puzzling puzzles in the books! Tournament of the Towns, 1980{1984, Questions and Solutions; published in 1993, softcover, 117+ pages, Australian $18. Tournament of the Towns, 1989{1993, Questions and Solutions; published in 1994, softcover, 209+ pages, Australian $22. Both edited by Peter J. Taylor, Australian Mathematics Trust, University of Canberra, P. O. Box 1, Belconnen, A.C.T. 2616, Australia. Reviewed by Murray S. Klamkin, University of Alberta. In my previous review [1992: 172] of the rst book of this series, which are the problems and solutions for 1984{1989 (and which was published rst), I had given a short description of the competition and references for further information on the competition. More complete information on this is given in the prefaces of the books. This competition, as I said before, is one of the premier mathematics competitions in the world for secondary school students and contains many wonderful challenging problems (even to professional mathematicians). In view of my previous review, I just give another sampling of these problems.
Junior Questions 1. Construct a quadrilateral given its side lengths and the length joining the midpoints of its diagonal. [1983] 2. (a) A regular 4kgon is cut into parallelograms. Prove that among these there are at least k rectangles. (b) Find the total area of the rectangles in (a) if the lengths of the sides of the 4kgon equal a. [1983] 3. Prove that in any set of 17 distinct natural numbers one can either nd
ve numbers so that four of them are divisible into the other or ve numbers none of which is divisible into any other. [1983]
32 4. Given the continued fractions
a= 2+ 1 1 3+
prove that ja , bj <
1 + 991
and
b = 2+
1 : 99! 100!
1
3+
1 1 + 99+1 1 100
; [1990]
5. The numerical sequence fxn g satis es the condition
xn+1 = jxn j , xn,1
for all n > 1. Prove that the sequence is periodic with period 9, i.e., for any n 1, we have xn = xn+9 . [1990] 6. There are 16 boxers in a tournament. Each boxer can ght no more often than once a day. It is known that the boxers are of dierent strengths, and the stronger man always wins. Prove that a 10 day tournament can be organized so as to determine their classi cation (put them in order of strength). The schedule of ghts for each day is xed on the evening before and cannot be changed during the day. [1991]
Senior Questions 1. We are given 30 nonzero vectors in 3 dimensional space. Prove that among these there are two such that the angle between them is less than 45. [1980] 2. Prove that every real positive number may be represented as a sum of nine numbers whose decimal representation consists of the digits 0 and 7. [1981] 3. A polynomial P (x) has unity as coecient of its highest power and has the property that with natural number arguments, it can take all values of form 2m, where m is a natural number. Prove that the polynomial is of degree 1. [1982] 4. A square is subdivided into K 2 equal small squares. We are given a broken line which passes through the centres of all the smaller squares (such a broken line may intersect itself). Find the minimum number of links in this broken line. [1982] 5. Do there exist 1; 000; 000 distinct positive numbers such that the sum of any collection of these numbers is never an exact square? [1989] 6. There are 20 points in the plane and no three of them are collinear. Of these points 10 are red while the other 10 are blue. Prove that there exists a straight line such that there are 5 red points and 5 blue points on either side of this line. [1990]
33
PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 21 " 11" or A4 sheets of paper. These may be typewritten or neatly handwritten, and should be mailed to the EditorinChief, to arrive no later that 1 September 1996. They may also be sent by email to cruxeditor@cms.math.ca. (It would be appreciated if email proposals and solutions were written in LATEX, preferably in LATEX2e). Solutions received after the above date will also be considered if there is sucient time before the date of publication. 2101. Proposed by Ji Chen, Ningbo University, China. Let a; b; c be the sides and A; B; C the angles of a triangle. Prove that for any k 1,
X ak 3 X k A a; where the sums are cyclic. [The case k = 1 is known; see item 4.11, page 170
of Mitrinovic, Pecaric, Volenec, Recent Advances in Geometric Inequalities.] 2102. Proposed by Toshio Seimiya, Kawasaki, Japan. ABC is a triangle with incentre I . Let P and Q be the feet of the perpendiculars from A to BI and CI respectively. Prove that
AP + AQ = cot A : BI CI 2
2103. Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a triangle. Let D be the point on side BC produced beyond B such that BD = BA, and let M be the midpoint of AC . The bisector of \ABC meets DM at P . Prove that \BAP = \ACB .
34
2104. Proposed by K. R. S. Sastry, Dodballapur, India. In how many ways can 111 be written as a sum of three integers in geometric progression? U. K.
2105. Proposed by Christopher J. Bradley, Clifton College, Bristol, Find all values of for which the inequality
2(x3 + y3 + z3 ) + 3(1 + 3)xyz (1 + )(x + y + z )(yz + zx + xy)
holds for all positive real numbers x; y;z .
2106. Proposed by Yang Kechang, Yueyang University, Hunan, China. A quadrilateral has sides a; b;c; d (in that order) and area F . Prove that
2a2 + 5b2 + 8c2 , d2 4F:
When does equality hold?
2107. Proposed by D. J. Smeenk, Zaltbommel, The Netherlands.
Triangle ABC is not isosceles nor equilateral, and has sides a; b; c. D1 and E1 are points of BA and CA or their productions so that BD1 = CE1 = a. D2 and E2 are points of CB and AB or their productions so that CD2 = AE2 = b. Show that D1 E1 k D2 E2.
2108. Proposed by Vedula N. Murty, Dover, Pennsylvania. Prove that
s a + b + c 1 3 (b + c)2(c + a)2(a + b)2 ; 3 4 abc where a; b; c > 0. Equality holds if a = b = c.
2109. Proposed by Victor Oxman, Haifa, Israel.
In the plane are given a triangle and a circle passing through two of the vertices of the triangle and also through the incentre of the triangle. (The incentre and the centre of the circle are not given.) Construct, using only an unmarked ruler, the incentre.
35
2110. Proposed by Jordi Dou, Barcelona, Spain.
Let S be the curved Reuleaux triangle whose sides AB , BC and CA are arcs of unit circles centred at C , A and B respectively. Choose at random (and uniformly) a point M in the interior and let C (M ) be a chord of S for which M is the midpoint. Find the length l such that the probability that C (M ) > l is 1=2.
2111. Proposed by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore. Does there exist a function f : N ,! N (where N is the set of positive integers) satisfying the three conditions: (i) f (1996) = 1; (ii) for all primes p, every prime occurs in the sequence f (p), f (2p), f (3p); : : : ; f (kp); : : : in nitely often; and (iii) f (f (n)) = 1 for all n 2 N ? 2112. Proposed by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario. Find a fourdigit baseten number abcd (with a 6= 0) which is equal to aa + bb + cc + dd. 2113. Proposed by Marcin E. Kuczma, Warszawa, Poland. Prove the inequality
! X n ! X n ! n n ai bi ! X X ai bi (ai + bi) i=1 i=1 i=1 i=1 ai + bi for any positive numbers a1 ; : : : ; an ; b1; : : : ; bn .
36
SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems. 1827. [1993: 78; 1994: 57; 1995: 54] Proposed by Sefket Arslanagic, Trebinje, Yugoslavia, and D. M. Milosevic, Pranjani, Yugoslavia. Let a; b; c be the sides, A; B; C the angles (measured in radians), and s the semiperimeter of a triangle. (i) Prove that
X
bc 12s ; A(s , a)
where the sums here and below are cyclic. (ii) It follows easily from the proof of Crux 1611 (see [1992: 62] and the correction on [1993: 79]) that also
X b + c 12s A :
Do the two summations above compare in general? IV. Comment by Waldemar Pompe, student, University of Warsaw, Poland. Here we give a short demonstration of the equality
X bc (4R + r)2 ; = s + s,a s
which has appeared on [1995: 55]. Since
X
X a = 2s; bc = s2 + r2 + 4Rr; abc = 4sRr ; we get X X (s , b)(s , c) = 3s2 , 4s2 + bc = r2 + 4Rr : Using the above equality we obtain
X 1 r2 + 4Rr r2 + 4Rr = r + 4R : = = s , a (s , a)(s , b)(s , c) sr2 sr
Therefore
X bc X 1 X 1 1 = abc = 4 Rr + s,a a) a s,a s2a(+s , 2 r + 4Rr + r + 4R = 4Rr 4sRr sr 2 + 4Rr 2 r 4 Rr + 16 R (4R + r)2 : = s+ s + = s + s s
(1)
37 A (not quite as) short demonstration of (1) has also been sent in by Toshio Seimiya, Kawasaki, Japan.
2006. [1995: 20] Proposed by John Duncan, University of Arkansas, Fayetteville; Dan Velleman, Amherst College, Amherst, Massachusetts; and Stan Wagon, Macalester College, St. Paul, Minnesota. Suppose we are given n 3 disks, of radii a1 a2 an . We wish to place them in some order around an interior disk so that each given disk touches the interior disk and its two immediate neighbours. If the given disks are of widely dierent sizes (such as 100, 100, 100, 100, 1), we allow a disk to overlap other given disks that are not immediate neighbours. In what order should the given disks be arranged so as to maximize the radius of the interior disk? [Editor's note. Readers may assume that for any ordering of the given disks the con guration of the problem exists and that the radius of the interior disk is unique, though, as the proposers point out, this requires a proof (which they supply).] Solution by the proposers. Let r be the radius of the central disk. First look at a single tangent con guration made up of the central disk, and two disks of radii x and y . The three centres form a triangle with sides r + x, r + y , and x + y ; let = r (x; y) be the angle at the centre of the disk with radius r. Applying the law of cosines to this angle and simplifying gives r (x; y) = arccos 1 , r2 + ry2+xyrx + xy :
A routine calculation shows that the mixed partial derivative 12 is given by 2 12 = 2pxy(r2 +rrx + ry)3=2 ;
therefore 12 > 0 (for x; y > 0). Integrating from a to a + s and b to b + t (where s; t > 0) yields:
(a + s; b + t) + (a; b) > (a + s; b) + (a; b + t): (1) Note that equality occurs if and only if s = 0 or t = 0. We may assume n 4 and a1 Pa2 an ; let Di denote the disk with radius ai . And let S (r) denote ni=1 r (ai; ai+1), the total angle made by the con guration; when S < 2 , then the disk of radius r is too large and
the ring is not yet closed up. THEOREM. The largest inner radius r occurs by placing D2 and D3 on either side of D1 , then D4 alongside D2 , D5 alongside D3 , and so on around the ring.
38 Proof. By induction on n. We actually prove the stronger assertion that for any radius r, S (r) for the con guration of the statement of the theorem is not less than S (r) for any other con guration. This suces, for if r is such that S (r) = 2 , then S (r) for any other con guration is not greater than 2 . For n = 4 there are only three arrangements: D1 D1 D1
D3
'$ '$ '$ &% &% &% D2
D4 Arrangement I
D4
D2
D3 Arrangement II
D4
D3
D2 Arrangement III
for which (suppressing the subscripts r in the 's)
SI (r) = (a1 ; a2) + (a2 ; a4) + (a3; a4 ) + (a1 ; a3); SII (r) = (a1; a2) + (a2 ; a3) + (a3; a4) + (a1; a4); SIII (r) = (a1; a3) + (a2; a3) + (a2 ; a4) + (a1; a4): By (1), SI (r) SII (r) SIII (r), which proves our assertion for this case. For the general case, suppose E2 ; : : : ; En is an arrangement of D2 ; : : : ; Dn, with bi denoting the radius of Ei. Then a2 b2 and we may also assume a3 b3 (otherwise ip and relabel). Now it is sucient, by the induction hypothesis, to show that
(a1; a2 ) + (a1 ; a3) , (a2; a3) (a1; b2) + (a1; b3) , (b2; b3): But (1) implies that
(a1; a2 ) + (a1 ; a3) + (b2; b3) (a1; a3) + (a1 ; b2) + (a2 ; b3); and this means it is sucient to prove:
(a1; a3 ) + (a2 ; b3) (a1; b3) + (a2; a3): But this too is a consequence of (1). Note. A similar argument shows that the smallest inner radius occurs for the con guration: : : : D5 Dn,3 D3Dn,1 D1Dn D2 Dn,2 D4 : : : . There were no other solutions sent in. The problem was motivated by the special case of three pennies and two nickels, which was in (the late) Joe Konhauser's collection of problems.
39 2007. [1995: 20] Proposed by Pieter Moree, Macquarie University, Sydney, Australia. Find two primes p and q such that, for all suciently large positive real numbers r, the interval [r; 16r=13] contains an integer of the form 2n ; 2n p; 2n q; or 2n pq for some nonnegative integer n. Solution by Peter Dukes, student, University of Victoria, B.C. The prime numbers p = 3 and q = 13 solve the problem. To prove this, it suces to nd an increasing sequence of positive integers s1 ; s2 ; : : : such that each si is of one of the forms stated in the problem, and si+1 =si 16=13 for all i = 1; 2; : : : . Then, given any real number r s1 , the integer sk (where k = minfi 2 Z+ : r sig), is no more than 16r=13. For if sk 2= [r; 16r=13], then sk =sk,1 16=13, contrary to construction. Of course if k = 1 then r = s1 ; so s1 2 [r; 16r=13]. Now, consider the sequence
fsig : 24; 26; 32; 39; 48; 52; 64; 78; : : : : : : ; 3 2m+3 ; 13 2m+1; 2m+5; 3 13 2m; : : : : It is a simple matter to verify that the ratio of consecutive terms si+1 =si does not exceed 16=13 for any i 2 Z+. Thus, for all r 24, the interval [r; 16r=13] contains an integer of the form 2n , 2n p, 2n q , or 2n pq .
Also solved by FEDERICO ARDILA, student, MassachusettsInstitute of Technology, Cambridge; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; RICHARD I. HESS, Rancho Palos Verdes, California; PETER HURTHIG, Columbia College, Burnaby, British Columbia; ROBERT B. ISRAEL, University of British Columbia; DAVID E. MANES, State University of New York, Oneonta; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; CHRIS WILDHAGEN, Rotterdam, The Netherlands; and the proposer. One other solution contained a correct pair of primes, but the editor could not decipher the proof that they worked. Most solvers simply found a single pair of primes, some of which allow the constant 16=13 to be replaced by a slightly smaller constant. As Engelhaupt, Hess and the proposer point out, the best you can do inp4this direction is that 16=13 can be replaced by any real number greater than 2. Hess says how: nd primes \near" 2k1+1=4 and 2k2 +1=2 for integers k1 and k2 (and p he gives the example p = 4871, q = 181, which gets you within :00013 of 4 2). More precisely, if p = 2k1+1=4 t1 and q = 2k2+1=2 t2 (1) where k1 ; k2 are positive integers and t1 ; t2 are real numbers close to 1, then 2k1+k2 < p 2k2 < q 2k1 < pq , and the ratios of consecutive terms are
p4 2 t ; p4 2 t2 ; p4 2 t ; p4 2 1 ; 1 1 t1 t1t2
40
p
all of which can be made arbitrarily close to 4 2. The reason (1) is possible is because (by the Prime Number Theorem) there is always a prime between n and nt for any given t > 1, if n is big enough. By increasing the number of primes allowed, with an analogous change in the kind of integers you want the interval to contain, the proposer can get the interval down to [r; r] for any given > 1. Subsequent to submitting the problem the proposer was able to generalize it. He writes: For given > 1 we claim that there exists squarefree D such that the consecutive integers n1 ; n2 ; n3; : : : of the form 2 d, with djD satisfy ni+1=ni for all i suciently large. It suces to nd an in nite subsequence fmi g1 i=1 such that mi+1=mi for all i suciently large. Let 3 = p1 < p2 < p3 < : : : be the sequence of odd consecutive primes. For n 1 let pin denote the smallest prime exceeding q n and let pjn denote n+1. Using the Prime Number Theorem the greatest prime less than q it fol= 1 , so there exists n such that p =p lows that limi!1 pip+1 i i+1 for i n i = inQ; s: : : ; jn + 1. Put q0 = 2 , qr1 = pin ; q2 = p1+in ; : : : ; qs = pjn and D = i=1 qi. Put ma+r(s+1) = 2 qa, for 0 a s, r 0. Notice that 1 < mi+1=mi for every i 1. Thus the claim holds with D = Qsi=1 qi. Remark. A similar result holds with 2 replaced by an arbitrary prime.
2008. [1995: 20] Proposed by Junhua Huang, The Middle School Attached To Hunan Normal University, Changsha, China. Let I be the incentre of triangle ABC , and suppose there is a circle with centre I which is tangent to each of the excircles of ABC . Prove that ABC is equilateral. Solution by Waldemar Pompe, student, University of Warsaw, Poland. We solve the problem assuming that the circle with centre I is tangent either externally or internally to all the excircles. Without this assumption the problem is not true; below we give a counterexample. (The triangle ABC in the gure has sides 2; 9; 9, and of course is not equilateral.) Assume rst there is a circle C with centre I tangent externally to all the excircles [i.e., C does not contain the excircles  Ed.]. Then according to Feuerbach's theorem the circle C is the ninepoint circle of ABC . Therefore C is also tangent to the incircle of ABC , but since I is the centre of C, the circle C and the incircle of ABC must coincide. It follows that the triangle ABC is equilateral. Now let C be the circle with centre I and tangent internally to all the excircles. Let DE and FG be the chords of C containing the segments AB and AC respectively. Since C and the incircle of ABC are concentric, the lines DE and FG are symmetric to each other with respect to the line AI . Therefore, since the excircles lying opposite to B and C are uniquely determined by the chords DE , FG and the circle C , they also have to be symmetric
41
A
B I
q
C
to each other with respect to the line AI . Thus rB = rC . Analogously we show that rC = rA , which implies that ABC has to be equilateral, as we wished to show. Also solved by FEDERICO ARDILA, student, Massachusetts Institute of Technology, Cambridge; FRANCISCO BELLOT ROSADO, I.B. Emilio Fer LOPEZ rari, and MARIA ASCENSION CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, U. K.; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; TOSHIO SEIMIYA, Kawasaki, Japan; ASHISH KR. SINGH, Kanpur, India; D. J. SMEENK, Zaltbommel, The Netherlands; HOE TECK WEE, student, Hwa Chong Junior College, Singapore; CHRIS WILDHAGEN, Rotterdam, The Netherlands; and the proposer. There was one incorrect solution received. Pompe was the only solver to nd the counterexample. Everyone else either ignored the possibility, or (in a couple of cases at least) believed that it could not occur! Half the solvers considered both cases of tangency, that is, where the circle is tangent externally to all the excircles or internally to all of them, and the others considered only one.
2009. [1995: 20] Proposed by Bill Sands, University of Calgary.
Sarah got a good grade at school, so I gave her N two{dollar bills. Then, since Tim got a better grade, I gave him just enough ve{dollar bills so that he got more money than Sarah. Finally, since Ursula got the best grade, I gave her just enough ten{dollar bills so that she got more money than Tim. What is the maximum amount of money that Ursula could have received? (This is a variation of problem 11 on the 1994 Alberta High School Mathematics Contest, First Part; see the January 1995 Skoliad Corner [1995: 6].)
42 Solution by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario. The most Ursula could receive is 2N + 14 dollars. Sarah received 2N dollars and Tim received 2N + 1, 2N + 2, 2N + 3, 2N + 4 or 2N + 5 dollars depending on what residue class N belongs to modulo 5. But since Tim gets $5 bills his amount is divisible by 5. Ursula will then receive either $5 more than Tim (if Tim's amount is not divisible by 10) or $10 more than Tim (if Tim's amount is divisible by 10). Clearly 2N +5 is odd, thus not divisible by 10. So the maximum occurs when Tim receives 2N + 4 dollars, which means 2N 1 mod 5, i.e. N 3 mod 5, and that maximum is 2N + 14. Also solved by HAYO AHLBURG, Benidorm, Spain; FEDERICO ARDILA, student, Massachusetts Institute of Technology, Cambridge; CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, U. K.; PAUL COLUCCI, student, University of Illinois; PETER DUKES, student, University of Victoria, B.C.; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; J.K. FLOYD, Newnan, Georgia; TOBY GEE, student, The John of Gaunt School, Trowbridge, England; RICHARD K. GUY, University of Calgary; DAVID HANKIN, John Dewey High School, Brooklyn, New York; RICHARD I. HESS, Rancho Palos Verdes, California; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; J. A. MCCALLUM, Medicine Hat, Alberta; ROBERT P. SEALY, Mount Allison University, Sackville, New Brunswick; SEIFFERT, Berlin, Germany; and the proposer. Three other HEINZJURGEN readers sent in solutions which the editor judges are not precise enough. Here Ursula gets at most $14 more than Sarah, which is 14=15 of the obvious maximum dierence $5+$10 = $15. How small can this ratio be, if we replace the denominations $2, $5, $10 by three other positive integers? (Choosing N; 2N , 1; 2N for large N gets it down arbitrarily close to 3=4.)
2010. [1995: 20] Proposed by Marcin E. Kuczma, Warszawa, Poland.
In triangle ABC with \C = 2\A, line CD is the internal angle bisector (with D on AB ). Let S be the centre of the circle tangent to line CA (produced beyond A) and externally to the circumcircles of triangles ACD and BCD. Prove that CS ? AB . Composite of solutions by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore and Roland H. Eddy, Memorial University, St. John's, Newfoundland. First, note that \BCD = \CAD, so that the line BC is tangent to the circumcircle of ACD. Next, perform an inversion with respect to C . The line BC inverts into the line B 0 C , so that the circumcircle of ACD inverts into the line A0 D0 , which is parallel to B 0 C . The circumcircle of BCD inverts into the line B 0 D0 . The line AD inverts into the circumcircle of the quadrilateral CA0 D0 B 0 . Finally, the circle tangent to the
43 line CA, and externally tangent to the circumcircles of ACD and BCD inverts into a circle ,, which is tangent to the line segments A0 C , A0 D0 and B0 D0 as shown in the diagram.
A0
C
D0
B0
Now, \BCD = \ACD, so that \B 0 CD0 = \A0 CD0 . Thus B 0 D0 = 0 0 A D . Since A0D0 and B0C are parallel, we have A0 0D0 = B0D0 = A0C . 0 0 Hence, the circumcentre of the quadrilateral A D B C lies on the angle bisectors of \CA0 D0 and \A0 D0 B 0 . These two angle bisectors intersect at the centre of ,. Thus, quadrilateral A0 D0 B 0 C and , are concentric. Let their
common centre be O. Thus, C , O and S 0 (the inverse of S ) are collinear, and so CS is perpendicular to AB . Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, U. K.; KEEWAI LAU, Hong Kong; D. J. SMEENK, Zaltbommel, The Netherlands; and the proposer.
2012. [1995: 52] Proposed by K. R. S. Sastry, Dodballapur, India.
Prove that the number of primitive Pythagorean triangles (integersided right triangles with relatively prime sides) with xed inradius is always a power of 2. Solution by Carl Bosley, student, Topeka, Kansas The formula for the inradius of a right triangle, r = (a + b , c)=2, where a, b are the legs and c is the hypotenuse, together with the formula a = 2mn, b = m2 , n2, c = m2 + n2, for the sides of the triangle, where m and n are relatively prime and not both odd, gives r = n(m , n). Let p be a prime which divides r. If p = 2, p can divide n but not m , n, since then m and n would either be both even or both odd. If p is not 2, then p must divide n or m , n, but cannot divide both, since m and n would have a common factor. Since each prime p other than 2 which divides r divides n or m , n, but not both, and each combination of choices produces a pair m, n which generates a right triangle, so if r has k distinct prime factors greater than 2, there are 2k primitive Pythagorean triangles with inradius r. Also solved by CLAUDIO ARCONCHER, Jundia, Brazil; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, U. K.; DAVID DOSTER, Choate Rose
44 mary Hall, Wallingford, Connecticut; H. ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; TOBY GEE, student, The John of Gaunt School, Trowbridge, England; RICHARD I. HESS, Rancho Palos Verdes, California; PETER HURTHIG, Columbia College, Burnaby, British Columbia; JAMSHID Y, Ferris State University, Big KHOLDI, New York, N. Y.; VACLAV KONECN SEIFFERT, Berlin, Germany; LAWRENCE Rapids, Michigan; HEINZ{JURGEN SOMER, Catholic University of America, Washington, D. C.; CHRIS WILDHAGEN, Rotterdam, The Netherlands: and the proposer. There were seven incorrect solutions received, most of which did not handle the case p = 2 correctly. Kholdi points out that the problem is solved on page 43 of Sierpinski's Pythagorean Triangles, published by the Graduate School of Science, Yeshiva University, 1962.
2013. [1995: 52] Proposed by Waldemar Pompe, student, University of Warsaw, Poland. Given is a convex ngon A1 A2 : : : An (n 3) and a point P in its plane. Assume that the feet of the perpendiculars from P to the lines A1 A2 ; A2A3 ; : : : ; AnA1 all lie on a circle with centre O. (a) Prove that if P belongs to the interior of the ngon, so does O. (b) Is the converse to (a) true? (c) Is (a) still valid for nonconvex ngons? Solution by Jerzy Bednarczuk, Warszawa, Poland. (a) We show that (a) is true for a polygon that is not necessarily convex (so we will have treated (a) and (c) at the same time). Let us call the given circle C . We rst see that when P is inside the ngon, then it must also be inside C . The halfline on OP starting at P and going away from O intersects some side of the ngon, say Ai Ai+1 , at a point S . If P were exterior to C , then the circle with diameter PS would lie in the exterior of C . Since the latter circle is the locus of points B with \PBS = 90 , the foot of the perpendicular from P to AiAi+1 would then lie outside of C , contrary to the de nition of C . Also by de nition, P cannot lie on C so we conclude that P must lie inside C as claimed. By way of contradiction we now assume that O lies outside of the given ngon. Since P is inside, the segment OP would intersect some side of the ngon, say Aj Aj +1 , at some point T . Since P is also inside C , the circle whose diameter is PT would be contained inside C , which would force the foot of the perpendicular from P to Aj Aj+1 to lie inside C contrary to its de nition. We conclude that O lies inside the given ngon as desired. (b) No. We can nd a counterexample for any n > 2 ! First draw a circle C and then choose any point P lying OUTSIDE of this circle. Next take n points B1; B2; : : : ; Bn on the circle C and through each Bi draw the line perpendicular to the line PBi. Those lines will form an ngon which will
45 contain O if you choose each Bi so that the halfplane determined by the line perpendicular to PBi will contain all other Bj together with O. Also solved by the proposer.
2014. [1995: 52] Proposed by Murray S. Klamkin, University of Alberta. (a) Show that the polynomial ,
,
2 x7 + y7 + z 7 , 7xyz x4 + y 4 + z 4
has x + y + z as a factor. (b) Is the remaining factor irreducible (over the complex numbers)? I. Solution to ,(a) by Jayabrata India. Das, Calcutta, , Let f (x;y;z ) = 2 x7 + y 7 + z 7 , 7xyz x4 + y 4 + z 4 . If we can show that f (x;y;z ) = z when x + y + z = 0, we are done. We know, for x + y + z = 0, that x3 + y 3 + z 3 = 3xyz . Thus
x7 + y,7 + z7 + x3y4 +, x3 z4 + y3z4 + y3x4 + z3y4 + z3x4 = x3 + y 3 + z3 x4 + y 4 + z 4 = 3xyz ,x4 + y 4 + z4
so that
x7 + y7 + z7 = 3xyz ,x4 + y4 + z4 ,x3y4 , x3z4 , y3z4 , y3x4 , z3y4 , z3x4
Therefore
7 f (x;y; z) = 2 ,x7 + y + z 7 , 7xyz ,x4 + y 4 + z 4 = 6xyz ,x4 + y 4 + z 4 ,2 ,x3y4 ,+ x3z4 + y3z4 + y3x4 + z3y4 + z3x4 ,7,xyz x4 + y4+ z4 = ,xyz x4 + y 4 + z 4 ,2x,3y3(x + y) , 2y3z3(y + z) , 2z3x3(z + x) = ,xyz ,x4 + y 4 + z 4 + 2x3 y 3z + 2xy 3z 2 + 2x3yz 3 = ,xyz x4 + y 4 + z 4 , 2x2y 2 , 2y2z 2 , 2z 2x2 , , = ,xyz x2 + y 2 + z 2 2 , 4 x2 y2 + y 2z 2 + z 2 x2 : Since x2 + y 2 + z 2 = ,2(xy + yz + zx), we now have that f (x; y;z) = ,xyz 4 (xy + yz + zx)2 , 4 ,x2y2 + y2z2 + z2x2 , 2 = ,4xyz 2x2 yz + 2xy2z + 2 xyz = ,8xyz xyz (x + y + z ) = 0 :
46 II. Solutionto (a) by Cyrus Hsia, student, University of Toronto, Toronto, Ontario. Consider the sequence an = xn + y n + z n . The characteristic equation with roots x, y , z , is
a3 , Aa2 + Ba , C = 0; where A = x + y + z , B = xy + yz + zx and C = xyz . The sequence fan g follows the recurrence relation: an+3 = A an+2 , B an+1 + C an : Now, we have
a0 = x0 + y0 + z0 = 3 ; a1 = x1 + y1 + z1 = A ; a2 = x2 + y2 + z2 = (x + y + z)2 , 2(xy + yz + zx) = A2 , 2B
From the recurrence relation, we see:
a3 = A a2 , B a1 + C a0 = A3 , 2AB , AB + 3C = A k3 + 3C; where k3 is some term in x, y and z
Similarly
a4 a5 a6 a7
Thus,
= = = =
A k4 + 2B2; where k4 is some term in x, y and z, A k5 , 5BC; where k5 is some term in x, y and z, A k6 , 2B3 + 3C 2; where k6 is some term in x, y and z, A k7 + 7B2C; where k7 is some term in x, y and z.
,
,
2 x7 + y7 + z7 , 7xyz x4 + y4 + z 4 = 2a,7 , 7C a4 , = a A k7 + 7B 2C , 7C A k4 + 2B 2 = Ak
where k is some term, in x, y and z; that is, , x + y + z divides 2 x7 + y7 + z7 , 7xyz x4 + y4 + z4. Berlin, Germany; Part (a) was also solved by SEFKET ARSLANAGIC, CHRISTOPHER J. BRADLEY, Clifton College, Bristol, U. K.; MIGUEL ANGEL OCHOA, Logro~no, Spain; ADRIAN CHAN, Grade 8 student, UpCABEZON per Canada College, Toronto, Ontario; TIM CROSS, Wolverley High School, Kidderminster, U. K.; RICHARD I. HESS, Rancho Palos Verdes, Califor nia; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; VACLAV Y, Ferris State University, Big Rapids, Michigan; J. A. MCCALLUM, KONECN
47 Medicine Hat, Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; CHRIS WILDHAGEN, Rotterdam, The Netherlands; and the proposer. One incorrect solution to part (b) was received. Other solvers of part (a) made use of computer algebra, properties of roots, the substitution x = ,y , z , or direct division. Some solvers pointed out that computer algebra failed to factorize the remaining factor. The editor (Shawyer) tried using DERIVE on a MSDOS 486 66MHZ computer. The factorisation stopped with no factors after 155 seconds. MAPLE also failed to nd any factors. McCallum commented \An asterisk on a question of Klamkin's is equivalent to a DO NOT ENTER sign!"
2015. [1995: 53 and 129 (Corrected)] Proposed by ShiChang Chi and Ji Chen, Ningbo University, China. Prove that 1 1 1 27p3 sin(A) + sin(B ) + sin(C ) A + B + C 2 ; where A, B , C are the angles (in radians) of a triangle.
I. Solution by Douglass L. Grant, University College of Cape Breton, Sydney, Nova Scotia, Canada. If A = B = C = =3, equality obtains. It then suces to show that each factor has an absolute minimum at the point. Note that C = , (A + B ). Let S = f(A; B ) : A > 0; B > 0 A + B < g. 1 1 1 Let f (A;B ) = + + A B , (A + B) . Then f is unbounded on S. So, if there is a unique critical point for f on S , it must be an absolute minimum. 1 1 Now, FA (A; B ) = , 2 + 2 = 0 implies that , (A +
A , (A + B) B) = A. Similarly, FB (A; B) = 0 implies that , (A + B) = B, and so that A = B . Hence A = B = , (A + B ) = 3 . Let g (A;B ) = sin(A) + sin(B ) + sin , (A + B ) = sin(A) + sin(B)+sin(A+B ). We now obtain that 0 = gA(A; B ) = cos(B )+cos(A + B), 0 = gB (A;B) = cos(A)+ cos(A + B), so that cos(A) = cos(B). Since no two distinct angles in (0; ) have equal cosines, we have that A = B. 2 Then 0 = cos(A)+cos(2A) = 2 cos (A)+cos(A),1= 2 cos(A) , 1 cos(A) + 1 : Since cos(A) cannot have the value ,1, it must then have value 12 , and so we have A = B = C = 3 .
48 II. Solution by the proposers. Let y (x) = x,1=3 cos(x) for 0 < x 2 . Dierentiating twice yields
x) + 4 , x2 x7=3 sec(x):y00 (x) = 2x tan( 3 x3 9 4 2 x > 3 x + 3 + 9 , x2 2 23 2 3 2 = 9 x , 4 + 72 > 0 :
By the AM{GM inequality and the Jensen inequality, we have
X
sin(A)
Y X1 X1 = 4 cos( A= 2) A A Y 6 cos(A=2) Y 1=3 A 0 2 13 p3 BB cos 6 CC 27 6 B@ 1=3 CA = 2 : 6
Berlin, Germany; FRANCISCO Also solved by SEFKET ARSLANAGIC, BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; and BOB PRIELIPP, University of Wisconsin{Oshkosh. Their solutions were based on known geometric inequalities. Also some readers wrote in after the initial publication of the problem, pointing out that the original result could not be true, and suggesting possible corrections. Thank you.
D
(
Xh
h(
B
E
49
On The Generalized Ptolemy Theorem Shailesh Shirali
Rishi Valley School, Rishi Valley 517 352, Chittoor Dt., A.P, INDIA Introduction. The following note describes a few uses of a relatively less known result in plane geometry, the Generalized Ptolemy Theorem (GPT, for short), also known as Casey's Theorem (see Johnson[1]). Featured will be two proofs of the problem proposed by India for the 33rd IMO in Moscow [1993: 255; 1995: 86]. Theorem 1. Circles 1 and 2 are externally tangent at a point I , and both are enclosed by and tangent to a third circle . One common tangent to 1 and 2 meets in B and C , while the common tangent at I meets in A on the same side of BC as I . Then I is the incentre of triangle ABC .
A r
1
B
I
2
r
C
r
r
Figure 1. Proof 1. Notation: Let tij refer to the length of the external common tangent to circles i and j (thus the two circles lie on the same side of the tangent). We use the GPT, which we state in the following manner. (The GPT) Let circles , , , all touch the circle ,, the contacts being all internal or all external and in the cyclical order , , , . Then:
t t + t t = t t :
50 Moreover, a converse also holds: if circles , , , are located such that
t t t t t t = 0 for some combination of +, , signs, then there exists a circle
that touches all four circles, the contacts being all internal or all external. For a proof of the GPT and its converse, please refer to [1].
A r
c z b
1
B
I r
X r
D r
2
r
Y r
BC = a BX = x CY = y DX = DI = DY =
C
r
Figure 2. Consider the con guration shown in Figure 2, where x and y are, respectively, the lengths of the tangents from B and C to 1 and 2 ; D is AI \ BC ; z = jAI j; u = jIDj; and a, b, c are the sides of 4ABC . We apply the GPT to the two 4tuples of circles (A; 1 ; B; C ) and (A; 2 ; C; B). We obtain: az + bx = c(2u + y) (1) az + cy = b(2u + x) : (2) Subtracting (2) from (1) yields bx , cy = u(c , b), so (x + u)=(y + u) = c=b, that is, BD=DC = AB=AC , which implies that AI bisects \BAC and that BD = ac=(b + c). Adding (1) and (2) yields az = u(b + c), so z=u = (b + c)=a, that is, AI=ID = AB=BD, which implies that BI bisects \ABC . This proves the result. Proof 2. Lemma. Let BC be a chord of a circle ,, and let S1 , S2 be the two arcs of , cut o by BC . Let M be the midpoint of S2, and consider all possible circles
that touch S1 and BC . Then the length tM of the tangent from M to
is constant for all such . (See Figure 3.)
51
R r
B
S
r
S1
r
r
C
tM
S2
,
M r
Figure 3. Proof of Lemma. Let \ , = R, \ BC = S . Applying the GPT to the 4tuple (B; ; C; M ), we nd: BS CM + CS BM = tM BC . Since BM = CM , we obtain: tM = BM , a constant. Proof of Theorem 1. (See gure 4.) Let S1 , S2 be the two arcs of , cut o by chord BC , S1 being the one containing A, and let M denote the midpoint of S2. Using the above lemma,
tM 1 = MB = MI = MC = tM 2 :
Therefore M has equal powers with respect to 1 and 2 and lies on their radical axis, namely AI . It follows that AI bisects \A of 4ABC .
A r
1
B
I
2
r
r
C
r
M r
Figure 4.
52 Next, 4IBM is isosceles, so \IBM = =2 , C=2. Also, \CBM = A=2, so \IBC = =2 , C=2 , A=2 = B=2, that is, IB bisects \B of 4ABC . It follows that I is the incentre of 4ABC .
}
}
}
}
}
For nonbelievers, here are two more illustrations of the power and economy of the GPT. Theorem 2. Let 4ABC have circumcircle ,, and let be a circle lying within , and tangent to it and to the sides AB (at P ) and AC (at Q). Then the midpoint of PQ is the incentre of 4ABC . (See Figure 5.)
A
,
r
R
S
r
r
P B
I r
r
Q
r
r
D
r
C r
Figure 5.
Proof. Let the GPT be applied to the 4tuple of circles (A; B; ; C ). Let AP = x = AQ. Then:
tAB = c; tA = AP = x; tAC = b ; tB = BP = c , x; tBC = a; t C = CQ = b , x : The GPT now gives: c(b , x) + (c , x)b = ax, so x = bc=s, where s = (a + b + c)=2 is the semiperimeter of 4ABC . Let I denote the midpoint of PQ; then IP = x sin A=2 = (bc=s) sin A=2, and the, perpendicular distance , cos A= = from I to AB is IP cos A= 2 , which equals ( bc=s ) sin A= 2 ,(1=2)bc sin A=s. But this is just the radius of the incircle of 4ABC . 2Since I is equidistant from AB and AC , it follows that I is the incentre of the triangle. The next illustration concerns one of the most celebrated discoveries in elementary geometry made during the last two centuries.
53 Theorem 3. (Feuerbach's Theorem) The incircle and ninepoint circle of a triangle are tangent to one another.
Proof. Let the sides BC , CA, AB of 4ABC have midpoints D, E , F respectively, and let be the incircle of the triangle. Let a, b, c be the sides of 4ABC , and let s be its semiperimeter. We now consider the 4tuple of circles (D; E; F; ). Here is what we nd:
tDE = 2c ; tDF = 2b ; tEF = a2 ;
tD =
a2 , (s , b)
=
b ,2 c
;
tE =
2b , (s , c)
=
a ,2 c
;
tF =
2c , (s , a)
=
b ,2 a
: We need to check whether, for some combination of +, , signs, we have c(b , a) a(b , c) b(a , c) = 0 :
But this is immediate! It follows from the converse to the GPT that there exists a circle that touches each of D, E , F and . Since the circle passing through D, E , F is the ninepoint circle of the triangle, it follows that and the ninepoint circle are tangent to one another. One would surmise that the GPT should provide a neat proof of the following theorem due to Victor Thebault: Let 4ABC have circumcircle ,, let D be a point on BC , and let
1 and 2 be the two circles lying within , that are tangent to , and also to AD and BC . Then the centres of 1 and 2 are collinear with the incentre of 4ABC . I have, however, not been able to nd such a proof, and I leave the problem to the interested reader. We note in passing that Thebault's theorem provides yet another proof of Theorem 1. References: [1] R.A. Johnson, Advanced Euclidean Geometry, Dover, 1960. Acknowledgements: I thank the referee for making several valuable comments that helped tidy up the presentation of the paper.
54
THE SKOLIAD CORNER No. 12 R.E. Woodrow
This issue we feature Part I of the 1995{96 Alberta High School Mathematics Competition, written November 21, 1995. My thanks go to Professor T. Lewis, the University of Alberta for forwarding me a copy. Students have 90 minutes to complete their contest. It is mostly written by students in Grade XII, but has often been won by students in earlier grades.
ALBERTA HIGH SCHOOL MATHEMATICS COMPETITION Part I
November 21, 1995 (Time: 90 minutes)
1. A circle and a parabola are drawn on a piece of paper. The number of regions they divide the paper into is at most A. 3 B. 4 C. 5 D. 6 E. 7. 2. The number of dierent primes p > 2 such that p divides 712 , 2 37 , 51 is A. 0 B. 1 C. 2 D. 3 E. 4. 3. Suppose that your height this year is 10% more than it was last year, and last year your height was 20% more than it was the year before. By what percentage has your height increased during the last two years? A. 30 B. 31 C. 32 D. 33 E. none of these. 4. Multiply the consecutive even positive integers together until the product 2 4 6 8 becomes divisible by 1995. The largest even integer you use is A. between 1 and 21 B. between 21 and 31 C. between 31 and 41 D. bigger than 41 E. nonexistent, since the product never becomes divisible by 1995. 5. A rectangle contains three circles as in the diagram, all tangent to the rectangle and to each other. If the height of the rectangle is 4, then the width of the rectangle 4 p2 is p p 4 A. 3 + 2 2 B. 4 + 3 C.p5 + 2 3 2 D. 6 E. 5 + 10.
55
6. Mary Lou works a full day and gets her usual pay. Then she works some overtime hours, each at 150% of her usual hourly salary. Her total pay that day is equivalent to 12 hours at her usual hourly salary. The number of hours that she usually works each day is A. 6 B. 7:5 C. 8 D. 9 E. not uniquely determined by the given information. 7. A fair coin is tossed 10; 000 times. The probability p of obtaining at least three heads in a row satis es A. 0 p < 41 B. 14 p < 12 C. 12 p < 34 D. 34 p < 1 E. p = 1: 8. In the plane, the angles of a regular polygon with n sides add up to less than n2 degrees. The smallest possible value of n satis es: A. n < 40 B. 40 n < 80 C. 80 n < 120 D. 120 n < 160 E. n 160. 9. A cubic polynomial P is such that P (1) = 1, P (2) = 2, P (3) = 3 and P (4) = 5. The value of P (6) is A. 7 B. 10 C. 13 D. 16 E. 19. 10. The positive numbers x and y satisfy xy = 1. The minimum value of x14 + 4y14 is A. 12 B. 58 C. 1 D. 45 E. no minimum. 11. Of the points (0; 0), (2; 0), (3; 1), (1; 2), (3; 3), (4; 3) and (2; 4), at most how many can lie on a circle? A. 3 B. 4 C. 5 D. 6 E. 7. 12. The number of dierent positive integer triples (x; y;z) satisfying the equations x2 + y , z = 100 and x + y2 , z = 124 is: A. 0
B. 1
C. 2
D. 3
E. none of these.
13. Which of the following conditions does not guarantee that the convex quadrilateral ABCD is a parallelogram? A. AB = CD and AD = BC B. \A = \C and \B = \D C. AB = CD and \A = \C D. AB = CD and AB is parallel to CD E. none of these.
56
14. How many of the expressions
x3 + y4; x4 + y3; x3 + y3; and x4 , y4; are positive for all possible numbers x and y for which x > y ? A. 0
B. 1
C. 2
D. 3
E. 4.
15. In triangle ABC , the altitude from A to BC meets BC at D, and the altitude from B to CA meets AD at H . If AD = 4, BD = 3 and CDp= 2, then the length of HD is p p A. 25 B. 23 C. 5 D. 52 E. 3 2 5 . 16. Which of the folowing is the best approximation to (23 , 1)(33 , 1)(43 , 1) (1003 , 1) ? (23 + 1)(33 + 1)(43 + 1) (1003 + 1)
A. 35
B. 33 50
C. 333 500
D. 35;;333 000
;333 E. 33 50;000 .
Last issue we gave the SHARP U.K. Intermediate Mathematical Challenge, written February 2, 1995. Here are answers. 1. E 2. C 3. D 4. E 5. A 6. B 7. C 8. B 9. D 10. A 11. E 12. D 13. C 14. D 15. B 16. A 17. A 18. C 19. C 20. D 21. D 22. E 23. B 24. B 25. A That completes this month's Skoliad Corner. I need materials of a suitable level to build up a bank of contests. Please send me suitable materials as well as comments, criticisms, and suggestions. I would like to have some feedback too about how your students do with these materials.
57
THE OLYMPIAD CORNER No. 172 R.E. Woodrow
All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. Because we are now publishing eight numbers of the Corner rather than ten, I am giving two Olympiad Sets for your pleasure. Besides, here in Canada it is winter, and quite cold, so having a stock of problems to contemplate in a warm spot is a good idea. Both of the sets we give this number were collected by Georg Gunther, Sir Wilfred Grenfell College, Corner Brook, when he was Canadian Team Leader to the IMO at Istanbul. Many thanks to him for gathering a wide sample of contests. We begin with the Telecom 1993 Australian Mathematical Olympiad.
TELECOM 1993 AUSTRALIAN MATHEMATICAL OLYMPIAD Paper 1 Tuesday, 9th February, 1993 (Time: 4 hours)
1. In triangle ABC , the angle ACB is greater than 90. Point D is the
foot of the perpendicular from C to AB ; M is the midpoint of AB ; E is the point on AC extended such that EM = BM ; F is the point of intersection of BC and DE ; moreover BE = BF . Prove that \CBE = 2\ABC . 2. For each function f which is de ned for all real numbers and satis es f (x; y) = x f (y) + f (x) y (1) and f (x + y) = f (x1993) + f (y1993) (2) p determine the value f ( 5753). 3. Determine all triples (a1; a2; a3 ), a1 a2 a3, of positive integers in which each number divides the sum of the other two numbers. 4. For each positive integer n, let
f (n) = [2pn] , [pn , 1 + pn + 1]: Determine all values n for which f (n) = 1. Note: If x is a real number, then [x] is the largest integer not exceeding x.
58
Paper 2
Wednesday, 10th February, 1993 (Time: 4 hours)
5. Determine all integers x and y that satisfy (x + 2)4 , x4 = y3:
6. In the acuteangled triangle ABC , let D, E, F be the feet of altitudes through A, B , C , respectively, and H the orthocentre. Prove that
AH + BH + CH = 2: AD BE CF 7. Let n be a positive integer, a1; a2; : : : ; an positive real numbers and s = a1 + a2 + + an . Prove that n ai n s , ai X n and X n(n , 1): i=1 s , ai n , 1 i=1 ai
8. The vertices of triangle ABC in the x,y plane have integer coordinates, and its sides do not contain any other points having integer coordinates. The interior of ABC contains only one point, G, that has integer coordinates. Prove that G is the centroid of ABC . Next we give the Final Round of the Japan Mathematical Olympiad.
JAPAN MATHEMATICAL OLYMPIAD Final Round  11 February, 1993 (Time: 4.5 hours)
1. Suppose that two words A and B have the same length n > 1 and that the rst letters of them are dierent while the others are the same. Prove that A or B is not periodic. 2. Let d(n) be the largest odd number which divides a given number n. Suppose that D(n) and T (n) are de ned by D(n) = d(1) + d(2) + + d(n); T (n) = 1 + 2 + + n:
Prove that there exist in nitely many positive numbers n such that 3D(n) = 2T (n).
59
3. In a contest, x students took part, and y problems were posed. Each student solved y=2 problems. For every problem, the number of students who solved it was the same. For each pair of students, just three problems were solved by both of them. Determine all possible pairs (x; y ). Moreover, for each (x; y ), give an example of the matrix (aij ) de ned by aij = 1 if the ith student solved the j th problem and aij = 0 if not. 4. Five radii of a sphere are given so that no three of them are in a common plane. Among the 32 possible choices of an end point from each segment, nd out the number of choices for which the 5 points are in a hemisphere. 5. Prove that there exists a positive constant C (independent of n, aj ) which satis es the inequality 0max x2
n Y
j =1
jx , aj j C n 0max x1
n Y
j =1
jx , aj j
for any positive integer n and any real numbers a1 ; : : : ; an . Last issue we gave a set of six Klamkin Quickies. Here are his \quick" solutions. Many thanks go to Murray Klamkin, the University of Alberta, for sending them to me.
SIX KLAMKIN QUICKIES
1. Which is larger
p
p
( 3 2 , 1)1=3
p
p
3 or 1=9 , 3 2=9 + 3 4=9? Solution. That they are equalpis an identity of Ramanujan. p 3 Letting x = 1=3 and y = 3 2=3, it suces to show that p (x + y)( 3 2 , 1)1=3 = x3 + y3 = 1;
or equivalently that
p
p
( 3 2 + 1)3( 3 2 , 1) = 3;
which follows by expanding out the left hand side. For other related radical identities of Ramanujan, see Susan Landau, How to tangle with a nested radical, Math. Intelligencer, 16 (1994), pp. 49{54. 2. Prove that
a b c b c a 1 1 1 3 min b + c + a ; a + b + c (a + b + c) a + b + c ;
60 where a, b, c are sides of a triangle. Solution. Each of the inequalities
a b c 1 1 1 3 b + c + a (a + b + c) a + b + c ; 3 ab + cb + ac (a + b + c) a1 + 1b + 1c ;
follow from their equivalent forms (which follow by expansion): (b + a , c)(c , a)2 + (c + b , a)(a , b)2 + (a + c , b)(b , c)2 0; (b + c , a)(c , a)2 + (c + a , b)(a , b)2 + (a + b , c)(b , c)2 0: 3. Let ! = ei=13. Express 1,1! as a polynomial in ! with integral coecients. Solution. We have 2 = (1 , ! 13) = 1 + ! + !2 + + ! 12;
(1 , ! ) (1 , ! ) 13 0 = (1(1++!! )) = 1 , ! + !2 , + ! 12:
Adding or subtracting, we get
1
2 4 12 (1 , ! ) = 1 + ! + ! + + ! = ! + !3 + + !11: More generally, if ! = ei=(2n+1), 1 = 1 + ! 2 + ! 4 + + ! 2n : (1 , !)
4. Determine all integral solutions of the simultaneous Diophantine equations x2 + y 2 + z 2 = 2w2 and x4 + y 4 + z 4 = 2w4. Solution. Eliminating w we get 2y2z2 + 2z2x2 + 2x2y 2 , x4 , y 4 , z 4 = 0 or
(x + y + z)(y + z , x)(z + x , y )(x + y , z ) = 0; so that in general we can take z = x + y . Note that if (x; y;z; w) is a solution, so is (x; y; z; w) and permutations of the x, y , z . Substituting back, we get x2 + xy + y2 = w2:
61 Since (x; y;w) = (1; ,1; 1) is one solution, the general solution is obtained by the method of Desboves, that is, we set x = r + p, y = ,r + q and 2 +pq +q 2 ) ( p w = r. This gives r = (q,p) . On rationalizing the solutions (since the equation is homogeneous), we get x = p2 + pq + q2 + p(q , p) = q2 + 2pq; ,y = p2 + pq + q2 , q(q , p) = p2 + 2pq; w = p2 + pq + q2; z = q2 , p2:
5. Prove that if the line joining the incentre to the centroid of a triangle is parallel to one of the sides of the triangle, then the sides are in arithmetic progression and, conversely, if the sides of a triangle are in arithmetic progression then the line joining the incentre to the centroid is parallel to one of the sides of the triangle. Solution. Let A, B, C denote vectors to the respective vertices A, B , C of the triangle from a point outside the plane of the triangle. Then the incentre I and the centroid G have the respective vector representations I and G, where (aA + bB + cC) ; G = (A + B + C) ; I= (a + b + c)
3
(where a, b, c are sides of the triangle). If G , I = k(A, B), then by expanding out
(b + c , 2a , k0)A + (a + c , 2b + k0 )B + (a + b , 2c)C = 0; where k0 = 3k(a + b + c). Since A, B , C are linearly independent, the coecient of C must vanish so that the sides are in arithmetic progression. Also then k0 = b + c , 2a = 2b , a , c. Conversely, if 2c = a + b, then G , I = 3(A6(,a+B)(b+bc,)a) , so that GI is parallel to the side AB . 6. Determine integral solutions of the Diophantine equation
x,y + y,z + z ,w + w,x = 0 x+y y+z z +w w+x
(joint problem with Emeric Deutsch, Polytechnic University of Brooklyn). Solution. It follows by inspection that x = z and y = w are two solutions. To nd the remaining solution(s), we multiply the given equation by the least common denominator to give
P (x;y; z; w) = 0;
62 where P is the 4th degree polynomial in x, y , z , w which is skew symmetric in x and z and also in y and w. Hence,
P (x;y; z; w) = (x , z)(y , w)Q(x; y;z; w);
where Q is a quadratic polynomial. On calculating the coecient of x2 in P , we get 2z (y , w). Similarly the coecient of y 2 is ,2w(x , z ), so that
P (x; y;z; w) = 2(x , z)(y , w)(xz , yw): Hence, the third and remaining solution is given by xz = yw.
Next we turn to the readers' solutions of problems from earlier numbers of the Corner. Let me begin by thanking Beatriz Margolis, Paris France; Bob Prielipp, University of WisconsinOshkosh, USA; Toshio Seimiya, Kawasaki, Japan; D.J. Smeenk, Zaltbommel, the Netherlands and Chris Wildhagen, Rotterdam, the Netherlands, for sending in nice solutions to some of the problems of the 1994 Canadian and 1994 U.S.A. Mathematical Olympiads. As we publish \ocial" solutions to the former, and refer readers to an MAA publication for the latter, I normally do not publish these solutions. Next, correspondence from Murray Klamkin, the University of Alberta,
led under the September number of the Corner, contains comments about problems from several numbers. 3. [1993: 5, 1994: 69] 1991 British Mathematical Olympiad. ABCD is a quadrilateral inscribed in a circle of radius r. The diagonals AC , BD meet at E. Prove that if AC is perpendicular to BD then EA2 + EB2 + EC 2 + ED2 = 4r2: () Is it true that, if () holds, then AC is perpendicular to BD? Give a reason for your answer. Klamkin's Comment. A simpler solution than that published, plus a generalization, is given in Crux, 1989, p. 243, #1. 3. [1994: 184] 1st Mathematical Olympiad of the Republic of China (Taiwan). If x1 ; x2 ; : : : ; xn are n nonnegative numbers, n 3 and x1 + x2 + + xn = 1 prove that x21x2 + x22x3 + + x2nx1 4=27. Klamkin's Comment. This problem appeared as problem 1292 in The Math. Magazine, April 1988. Now we turn to readers' solutions of problems proposed to the Jury, but not used, at the 34th International Mathematical Olympiad at Istanbul. A booklet of \ocial solutions" was issued by the organizers. Because I do not have formal permission to reproduce these solutions, I will only discuss readers' solutions that are dierent.
63
2. [1994: 216] Proposed by Canada.
Let triangle ABC be such that its circumradius R = 1. Let r be the inradius of ABC and let p be the inradius of the orthic triangle A0 B 0 C 0 of triangle ABC . Prove that p 1 , 1=3(1 + r)2 . Solution by D.J. Smeenk, Zaltbommel, The Netherlands. 1 instead of In the rst instance I misread the problem: I read 3(1 + r)2 1 (1 + r)2 ! A 3
C0
B0
C A0 Now rABC = R(cos + cos + cos , 1). From R = 1, we obtain r = cos + cos + cos , 1. Similarly, B
p = 12 [cos( , 2) + cos( , 2 ) + cos( , 2 ) , 1] = 2 cos cos cos : We are to show that p + 13 (1 + r)2 1. Equivalently,
Now
(1)
2 cos cos cos + 13 (cos + cos + cos )2 1:
(2)
,1 < cos cos cos 18 ; ) 1 < cos + cos + cos 32 :
(3)
[Bothema, Geom. Inequalities, 2.24, resp. 2.16] Now 2 18 + 13 ( 23 )2 = 1, so it is clear that (1) holds. 3. [194: 216] Proposed by Spain. Consider the triangle ABC , its circumcircle k of centre O and radius R, and its incircle of centre I and radius r. Another circle Kc is tangent to the sides CA, CB at D, E , respectively, and it is internally tangent to k. Show that I is the midpoint of DE .
64 Solution by D.J. Smeenk, Zaltbommel, The Netherlands. Assume (see gure) that > . Let F be the centre of Kc and its radius. Then F lies on the production of AI . Now FD = , so
AF = sin : 2
Also AO = R, OF = R , , and \FAO = ' = 12 ( , ). A =2 '
R G D
=2 =2
I
0
O R,p
E
F C
B
Apply the Law of Cosines to AFO: OF 2 = AF 2 + AO2 , 2AF AO cos '; or
, 2 2 R cos (R , )2 = sin2 + R2 , 2 sin 2 ; 6= 0:
From this we obtain
or Thus
2
2
2R cos ,2 ; ,2R + = sin2 , cos +2 2
cos2 2 = 4R sin 2 sin 2 : sin =2 sin2 =2 sin sin 2 2 2 = r = FD: = 4R sincos 22 cos2 2
65 Now let DE intersect AF at I 0 . Then DI 0 = DF cos 2 = cosr 2 . Let G be the foot of the perpendicular from I 0 to AB . Then I 0 G = I 0 D cos 2 = cosr 2 cos 2 = r. So I 0 coincides with I . 7. [1994: 217] Proposed by Israel. The vertices D, E , F of an equilateral triangle lie on the sides BC , CA, AB respectively of a triangle ABC . If a, b, c are the respective lengths of these sides and S is the area of ABC , prove that p p DE 2 2 S fa2 + b2 + c2 + 4 3 Sg,1=2: Comment by Murray S. Klamkin, The University of Alberta. This problem is equivalent to problem #624 of Crux [1982: 109{110]. 11. [1994: 241] Proposed by Spain. Given the triangle ABC , let D, E be points on the side BC such that \BAD = \CAE . If M and N are respectively, the points of tangency with BC of the incircles of the triangles ABD and ACE, show that
1 + 1 = 1 + 1 : BM MD NC NE
Solution by D.J. Smeenk, Zaltbommel, The Netherlands. 1 + 1 = 1 + 1 , or equivalently We are to show MB MD NC NE
BD NC NE = CE MB MD: We denote \BAD = \CAE = '.
(1)
A q
' '
B
q
I1
q
q
+'
M D q
q
+'
I2
EN q
q
q
C
Applying the law of sines to ABD we obtain
' ; BD = sin(c sin
+ ')
: AD = sin(c sin
+ ')
(2)
b sin ' ; CE = sin( + ')
: AE = sin(b sin + ')
(20 )
The sine law for ACE gives
66 From ABD: and from ACE :
2MB = AB + BD , AD 2MD = ,AB + BD + AD; 2NC = AC + CE , AE 2NE = ,AC + CE + AE:
From this we see that to show (1) we must show
BD(AC + CE , AE)(,AC + CE + AE) = CE(AB + BD , AD)(,AB + BD + AD);
or equivalently
BD(CE2 , AC 2 , AE2 + 2AC AE) = CE(BD2 , AB 2 , AD2 + 2AB AD): Now use the law of cosines in ACE , and in ABD to obtain CE2 2 , AC2 2 , AE22 = ,2ACAE cos ' BD , AB , AD = ,2ABAD cos ':
(3) (4)
Combining (3) and (4) we see that we must show
BD AC AE(1 , cos ') = CE AB AD(1 , cos '): As 1 , cos ' = 6 0, we nd with (2) and (2') that we must verify that c sin ' b b sin = b sin ' c c sin ; sin( + ) sin( + ' sin( + ') sin( + ') and as b sin = c sin , this holds.
13. [1994: 241] Proposed by India.
A natural number n is said to have the property P if, whenver n divides an , 1 for some integer a, n2 also necessarily divides an , 1. (a) Show that every prime number has property P . (b) Show there are in nitely many composite numbers n that possess property P . Solution by E.T.H. Wang, Sir Wilfrid Laurier University, Waterloo, On
tario.
(a) Suppose n = p is a prime such that ap 1 mod p. Since ap a mod p by Fermat's Little Theorem, we have a 1 mod p. Thus a = kp+1 p ,1 = (kp+1)p ,1 = for some ,integer k . Hence, by the Binomial Theorem a , , (kp)p + p1 (kp)p,1 + + p,p 1 kp. Since p,p 1 = p and p 2, it follows that p2 j ap , 1. (b) We show that all composite numbers of the form n = 2p, where p is an odd prime have property P . Suppose 2p j a2p , 1. Then p j (a2 )p , 1
67 which, in view of (a), implies p2 j a2p , 1. On the other hand, 2 j a2p , 1 implies 2 j (ap , 1)(ap + 1). Since ap , 1 and ap + 1 have the same parity, they must both be even, and hence 4 j (ap , 1)(ap +1). Since gcd(4; p2 ) = 1, 4p2 j a2p , 1 follows. Remarks. (1) The fact that all prime numbers satisfy property P is well known. In fact, using exactly the same argument as in the proof of (a) above, one can show easily that if ap bp mod p, then ap bp mod p2 . (See e.g. Ex. 13 on page 190 of Elementary Number Theory and its Applications by Kenneth H. Rosen, 3rd Edition). (2) n = 4 also has property P . For suppose that 4 j a4 , 1. Then 4 j (a2 , 1)(a2 + 1), which implies that a is odd. Since any odd square is congruent to 1 modulo 8, we have 8 j a2 , 1, which, together with 2 j a2 +1, yields 16 j (a2 , 1)(a2 + 1). (3) In view of (2) and our proof above, n = 8 is the rst natural number which need not possess property P . Indeed, it does not since 38 , 1 = 6560 is divisible by 8 but not by 64. (4) It might be of interest to characterize all natural numbers with property P .
We nish this number of the Corner with a solution to one of the IMO problems from the 35th IMO in Hong Kong. 2. [1994: 244] ABC is an isosceles triangle with AB = AC . Suppose that (i) M is the midoint of BC and D is the point on the line AM such that OB is perpendicular to AB; (ii) Q is an arbitrary point on the segment BC dierent from B and C ; (iii) E lies on the line AB and F on the line AC such that E , Q and F are distinct and collinear. Prove that OQ is perpendicular to EF if and only if QE = QF . Solution by D. J. Smeenk, Zaltbommel, the Netherlands. (=)). See Figure 1. Assume OQ ? EF . We want to show QE = QF . Note that quadrilateral OBEQ is inscribed on the circle with diameter OE. Thus \OEQ = \OBQ = \OQB = 2 . Also OFCQ is cyclic and thus \OFQ = \OCQ = \OAC = 2 . Together these give \OEQ = \OFQ and OQ ? EF , so QE = QF .
68
A r
=2 =2
E
r
=2
B
r
M
=2
r
Q
r
O
=2
C
r
r
=2
F
r
Figure 1. ((=). See Figure 2. Assume that QE = QF . We want to show that OQ ? EF . Let D lie on AC with QD parallel to AB . As QE = QF we see that D is the midpoint of AF . As AB = AC we have DQ = DC . (1) Also,
QE = 2DQ = 2CD
(2)
BE = AB , AE:
(3)
From (2) and (3)
BE = AC , 2CD = (AC , CD) , CD = AD , CD = DF , CD: Thus BE = CF . Draw OB , OE , OC , and OF .
(4)
69
A r
E
r
D
r
B
r
M
r
O
Q
r
C
r
r
Then
F
r
Figure 2.
OBE = COF [BE = CF; OB = OC; \OBE = \OCF = n=2]:
(5)
Now this implies that OE = OF , QE = QF giving OQ ? EF . Remark. This direction could be shortened by using the law of sines in triangles BQF and CQF , but the given solution is, I think, more elementary, and therefore more elegant. That completes the material we have available for this number. The Olympiad Season is fast approaching. Please collect your contests and send them to me. Also send me your nice solutions to problems posed in the Corner.
THE ACADEMY CORNER No. 2
This will appear in a future issue.
70
IMO95 PUZZLES
In the September 1995 issue of CRUX, two puzzles were given about the logo for the 1995 IMO. The second puzzle, attributed there to Mike Dawes of the University of Western Ontario, asked: Suppose that you take a physical model of this logo, and manipulate it to turn the in nity sign into a circle. What does the circle turn into? It turns out that the same question occurred to several other people at the same time, including some members of the Netherlands team at IMO 95. Less than half an hour prior to the start of the competition, the 1995 IMO logo inspired one of the contestants (RvL) to pose the following problem: Consider the logo as a knot composed of a blue string forming the in nitysign and a red string forming the zero. Can the knot be rearranged so that the blue string forms the zero and the red string forms the in nitysign (without using scissors)? The contestant solved the problem shortly after the competition and, upon returning home, wrote a program together with one of the other members of his team (DG). Ronald van Luijk (rmluijk@cs.ruu.nl) and Dion Gijswijt. The program was sent by the Leader of the Netherlands team (Johannes Notenboom) to the EditorinChief (in his role as Chief Operating Ocer of the 1995 IMO). We were so impressed by the program that we have had it made available on the IMO95 WWW site. You may download this program (zipped) by going to: http://camel.math.ca/IMO/IMO95/. Ronald has pointed out the the IMO 95 logo is, in fact, the Whitehead Link. See, for example, L.H. Kauman's book: On Knots, Annals of Mathematics Studies, 115, Princeton University Press, 1987, p. 14.
71
BOOK REVIEWS Edited by ANDY LIU
Assessing Calculus Reform Eorts : A Report to the Community, edited by James Leitzel and Alan C. Tucker, 1995. Paperback, 100+ pages, US$18.00, ISBN 0883850931. Preparing for a New Calculus, edited by Anita Solow, 1994. Paperback, 250+ pages, US$24.00, ISBN 0883850923. Both published by The Mathematical Association of America, Washington, DC 20036. Reviewed by Jack Macki, University of Alberta. Anyone who attended the January 1995 joint meetings in San Francisco could not help but be struck by the contrast between the evident malaise on the research side of the meeting and the enthusiasm, energy and high quality of the discourse on the educational side. These two books, very dierent in nature and content, together represent enormously persuasive documentation of the ongoing revolution in teaching and curriculum design at the University/College level. They describe the past and present of reform in undergraduate calculus, and evaluate the situation in a thorough and realistic manner. Assessing Calculus Reform Eorts is one of the best presentations ever of the history, present activities and future plans of the movement for reform in the design and teaching of the undergraduate mathematics curriculum. The history begins with the founding of the Committee on the Undergraduate Program (CUP) in 1953. The authors describe the debate between Anthony Ralston and Ron Douglas which led to the famous Tulane workshop in 1986 (with 25 participants, four of whom were research mathematicians). They follow the entire development of the reform movement, carefully describing the key players, and the roles of NRC and NSF. They present all kinds of surprising (at least to this reviewer) information along the way  did you know that in 1985 the IEEE, deeply dissatis ed with existing texts, published a calculus book? In the nine years since that seminal Tulane conference, reform has clearly moved into a central position in the mathematical life of North America. The eort of key individuals to convert the research community to believers (and activists) makes fascinating reading, beginning with a widely disseminated article by Peter Lax (UME Trends, May, 1990) which was highly critical of the attitudes of the majority of research mathematicians. Very determined department chairs at Stony Brook and Michigan convinced their departments to adopt reform texts. Said Don Lewis, chair at Michigan, \An NSF grant providing two months of summer salary is icing on the cake. The cake is calculus." The editors are very tactful in their presentation, but the lack of interest in education of the majority of research mathematicians in the United States
72 (and Canada?) comes through quite clearly. They present the results of four surveys carried out to date, and discuss the issue of \cosmetic" versus \real" change. The bottom line: By Spring 1994, threequarters of responding departments (1048 respondents) had some reform under way, onequarter of the respondents were conducting major reforms. The Graduate Record Exam has been changed to re ect reform. The NSF continues to fund reform initiatives at the level of about $2.7 million per year. Sales of reform texts have increased to 108,700 annually (552 institutions) in Fall, 1994 (not counting the 375 high schools which use them). The main part of this book has only 44 pages of text, and then concludes with 50 pages of important and highly readable appendices: data on enrolments, copies of surveys, textbytext description of reform materials presently available, and a very detailed list of NSFfunded projects. If you have colleagues who poohpooh the reform movement (and who doesn't?), get them started on the path to rightness with this book. If they can read it and remain unwilling to look at change, order them cons, they probably died a long time ago! Preparing for a New Calculus is the proceedings of a conference held at the University of Illinois in April, 1993. Eighty individuals attended, 40 of them associated with projects at colleges or universities, 25 of them associated with projects at the high school level, seven from community college projects, and eight from organizations like the MAA, NSF, NCTM, etc., and publishers. Part I contains seven background papers, which were provided to participants before the conference. The rst paper cites some impressive gures: well over 100 schools use the Harvard Consortium Project materials; 27% of all collegelevel calculus students in the State of Washington are in a reform course. All seven papers oer several important observations based on experience, for example: There is no correct way to do a reform course, there are many ways to treat calculus eectively. Reform courses have fewer topics with much richer content. The \rule of three" (graphical, numerical, analytical) should be replaced by the \rule of four" (writing) or \ ve" (oral). It is mindless to ask a student who has access to a calculator to approxip mate 17 using a dierential, but one can instead study Euler's method for solving the logistic dierential equation before the students know any integration theory.
73
Reform makes appropriate use of technology; technology does not equal reform. While most mathematics professors believe that learning is transmitted (I call this the \I am God" theory of learning), most evidence points to the correctness of the constructive theory of learning, which emphasizes that most learning is constructed by the learner in response to challenges to re ne or revise what he/she already knows in order to cope with new situations. Sally Berenson, North Carolina State University: \Telling is not teaching, listening is not learning".
Never underestimate the importance and diculty of educating faculty
for real change. Teaching reform courses is very hard work. Creating a cooperative learning environment is not easy, but pays o for all students, especially minorities. It is very useful to carry openended longterm applied problems through one or, even better, several courses. At the U.S. Military Academy at West Point, they begin with a few core applied problems and proceed to attack them with noncalculus methods, leading to a semester's study of dierence equations and the associated linear algebra. Calculus starts in term two! K. Stroyan of the University of Iowa starts his course by asking \Why is it we can eradicate polio and smallpox, but not measles and rubella?" You get the picture, there is just a tremendous amount of experience and insight available in these seven articles  and wonderful quotes: The chairman of physics at Duke, Larry Evans, when asked to comment on changes in the teaching of calculus: \There is nowhere to go but up." Other important themes: downgrading of exams, strong emphasis on (and evaluation of) writing, emphasis on cooperative learning. And don't expect all students to love the changes  students who are successful in standard courses often react negatively to being saddled with lab partners and being evaluated in unfamiliar ways. In fact, several authors emphasize that student questionnaires cannot be the sole means of teaching evaluation  one needs to talk to students who have gone down the road a way in order to get a fair picture. The second part of Preparing for a New Calculus reports on the workshops, one each on the topics of content, teaching strategies, and institutional context. Each report has a long and useful list of suggestions and observations. Among them: Characterize and reward that which constitutes eective teaching.
74
Allocate time as a resource to support involved faculty. Every person involved with evaluating sta should read Ernest Boyer's
Carnegie Foundation Report \Scholarship Reconsidered". The third section consists of ve contributed papers of very high quality on topics ranging from calculator courses, to the gateway exam at Michigan. The nal piece is a thoughtful article by Peter Renz of Academic Press on Publishers, Innovation and Technology, which should be required reading for all university mathematics professors. Many of the articles in this book have extensive lists of very timely and useful references. All in all, Preparing for a New Calculus is an excellent detailed introduction to calculus reform, to be read after Assessing Calculus Reform Eorts. Anyone who can read them both and not be interested in and excited about reform doesn't need a con, they've been dead too long!
Introducing the new Associate EditorinChief For those of you who do not know Colin, here is a short pro le: Born: Bedford, England 1 Educated Bedford Modern School University of Birmingham, England University of London, England Employment Queen's College, Nassau, Bahamas Sir John Talbot's Grammar School, Whitchurch, England Memorial University of Newfoundland, Canada Mathematical Mathematical Education Interests History and Philosophy of Science Nineteenth century Astrophysics
1
Bedford is halfway between Oxford and Cambridge (just an average place).
75
PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be typewritten or neatly handwritten, and should be mailed to the EditorinChief, to arrive no later that 1 October 1996. They may also be sent by email to cruxeditor@cms.math.ca. (It would be appreciated if email proposals and solutions were written in LATEX, preferably in LATEX2e). Graphics les should be in epic format, or plain postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication.
2114. Proposed by Toshio Seimiya, Kawasaki, Japan.
ABCD is a square with incircle ,. A tangent ` to , meets the sides AB and AD and the diagonal AC at P , Q and R respectively. Prove that AP + AR + AQ = 1: PB RC QD
2115. Proposed by Toby Gee, student, The John of Gaunt School, Trowbridge, England. Find all polynomials f such that f (p) is a prime for every prime p. 2116. Proposed by Yang Kechang, Yueyang University, Hunan, China. A triangle has sides a; b; c and area F . Prove that p 3 4 5 25 5(2F )6
abc
When does equality hold?
27
:
76
2117. Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a triangle with AB > AC , and the bisector of \A meets BC at D. Let P be an interior point of the side AC . Prove that \BPD < \DPC .
2118. Proposed by Paul Yiu, Florida Atlantic University, Boca Raton, Florida, USA. The primitive Pythagorean triangle with sides 2547 and 40004 and hypotenuse 40085 has area 50945094, which is an 8digit number of the form abcdabcd. Find another primitive Pythagorean triangle whose area is of this form. 2119. Proposed by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore. (a) Show that for any positive integer m 3, there is a permutation of m 1's, m 2's and m 3's such that (i) no block of consecutive terms of the permutation (other than the entire permutation) contains equal numbers of 1's, 2's and 3's; and (ii) there is no block of m consecutive terms of the permutation which are all equal. (b) For m = 3, how many such permutations are there?
2120. Proposed by Marcin E. Kuczma, Warszawa, Poland.
Let A1 A3 A5 and A2 A4 A6 be nondegenerate triangles in the plane. For i = 1; : : : ; 6 let `i be the perpendicular from Ai to line Ai,1Ai+1 (where of course A0 = A6 and A7 = A1 ). If `1 ; `3; `5 concur, prove that `2 ; `4 ; `6 also concur.
2121. Proposed by Krzysztof Chelminski, Technische Hochschule Darmstadt, Germany; and Waldemar Pompe, student, University of Warsaw, Poland. Let k 2 be an integer. The sequence (xn ) is de ned by x0 = x1 = 1 and k xn+1 = xxn + 1 for n 1: n,1 (a) Prove that for each positive integer k 2 the sequence (xn ) is a sequence of integers. (b) If k = 2, show that xn+1 = 3xn , xn,1 for n 1. (c) Note that for k = 2, part (a) follows immediately from (b). Is there an analogous recurrence relation to the one in (b), not necessarily linear, which would give an immediate proof of (a) for k 3?
77
2122. Proposed by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario. Little Sam is a unique child and his math marks show it. On four tests this year his scores out of 100 were all twodigit numbers made up of eight dierent nonzero digits. What's more, the average of these scores is the same as the average if each score is reversed (so 94 becomes 49, for example), and this average is an integer none of whose digits is equal to any of the digits in the scores. What is Sam's average? 2123. Proposed by Sydney Bulman{Fleming and Edward T. H. Wang, Wilfrid Laurier University, Waterloo, Ontario. It is known (e.g., exercise 23, page 78 of Kenneth H. Rosen's Elementary Number Theory and its Applications, Third Edition) that every natural number greater than 6 is the sum of two relatively prime integers, each greater than 1. Find all natural numbers which can be expressed as the sum of three pairwise relatively prime integers, each greater than 1. 2124. Proposed by Catherine Shevlin, Wallsend, England.
Suppose that ABCD is a quadrilateral with \CDB = \CBD = 50 and \CAB = \ABD = \BCD. Prove that AD ? BC .
D
C A
B
Mathematical Literacy 1. Who said: \To be able to read the great book of the universe, one must
rst understand its language, which is that of mathematics". 2. In referring to \the unreasonable eectiveness of mathematics in the natural sciences", who wrote: \The miracle of the appropriateness of the language of mathematics for the formulation of the laws of physics is a wonderful gift which we neither understand nor deserve".
78
SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems. 1827. [1993: 78; 1994:57] Proposed by Sefket Arslanagic, Berlin, Germany, and D.M. Milosevic, Pranjani, Yugoslavia. In commenting on the solutions submitted, the editor asked for a short proof of X bc (4R + r)2
s,a =s+
s
:
Two very nice and very dierent solutions have been received. I. Solution by TOSHIO SEIMIYA, Kawasaki, Japan. Since tan(A=2) = r1 =s, and using a result in R.A. Johnson, Advanced Euclidean Geometry, p. 189, we obtain
X
tan(A=2) =
X
r1=s = (4R + r)=s: X Since A + B + C = , we then have tan(B=2)tan(C=2) = 1, which leads to X 2 X tan(A=2) = ,1 + sec2 (A=2): 4R + r 2 X 2 Together, we now have sec (A=2) = 1 + . s bc = s bc 2 Since s,a s(s , a) = s sec (C=2), we have
X 2 X bc (4R + r)2 : = s sec ( A= 2) = s + s,a s
and.
II. Solution by Waldemar Pompe, student, University of Warsaw, PolSince
X
a = 2s, P bc = s2 + r2 + 4Rr, and abc = 4sRr, we get: X X (s , b)(s , c) = 3s2 , 4s2 + bc = r2 + 4Rr:
Using this, we obtain
X 1 r2 + 4Rr r2 + 4Rr = r + 4R : = = s , a (s , a)(s , b)(s , c) sr2 sr
79 Therefore
X bc X 1 X 1 1 s , a = abc a(s , a) = 4Rr a+ s , a 2 2 + 4Rr r + 4R + sr = 4Rr s +4rsRr 2 2 2 = s + r +s4Rr + 4Rr +s 16R = s + (4R s+ r) :
2000. [1994: 286] Proposed by Marcin E. Kuczma, Warszawa, Poland.
A 1000{element set is randomly chosen from f1; 2; : : : ; 2000g. Let p be the probability that the sum of the chosen numbers is divisible by 5. Is p greater than, smaller than, or equal to 1=5? Comment by Stan Wagon, Macalester College, St. Paul, Minnesota, USA. The answer is that ,400 1 4 200 = 1 + 410,482: p = 5 + 5 ,2000 5 1000 See Stan Wagon and Herbert S. Wilf, When are subset sums equidistributed modulo m?, Electronic Journal of Combinatorics 1 (1994). In particular, we obtained a necessary and sucient condition that the t{ subsets of [n] be equidistributed mod m. That condition is:
t > n mod d for all d dividing m (except d = 1), where refers to the least nonnegative residue.
In the case at hand, m = 5, so only d = 5 need be considered, and then t = n = 0, so equidistribution fails. While this does not directly answer problem 2000 (since it gives no information about the speci c remainder 0 mod 5), the paper does discuss many intriguingopen questions related to the mod m distribution of subset sums. Thus it strikes me that, because some of your readers were successful at generalizing the problem as stated, they would be interested in this reference. Indeed, perhaps some characterization of the quadruples (i;t; n; m), such that the set of t{subsets of [n] whose mod m sum is i has less than average size, is possible. The problem arose from lottery considerations. When are the tickets in a lottery equidistributed with respect to the mod m value of their sums?
80
2011. [1995: 52] Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a triangle with incentre I . BI and CI meet AC and AB at D and E respectively. P is the foot of the perpendicular from I to DE, and IP meets BC at Q. Suppose that IQ = 2IP . Find angle A.
Solution by the proposer. Let X and Y be the feet of perpendiculars from I to BC and AB , then IX = IY = r; where r is the inradius of 4ABC . We put \ABI = \IBC = , \ACI = \ICB = , \EIB = , \IDE = x, and \IED = y . Then we have + = and x + y = . As \IQC = \IBQ + \BIQ = \IBQ + \DIP = + 2 , x, we get r = IX = IQ sin \IQC = IQ sin( + 2 , x), so that r = IQ cos( , x) (1) As \AEI = \EBI + \EIB = + , we get r = IY = IE sin \AEI = IE sin( + ): (2) Because PI = IE sin y; and IQ = 2PI , we have from (1) r = 2PI cos( , x) = 2IE sin y cos( , x): (3) From (2) and (3) we have (cancelling IE ),
sin( + ) = 2 sin y cos( , x) = sin( + y , x) + sin(x + y , ) = sin( + y , x) + sin( , ): Therefore sin( + y , x) = sin( + ) , sin( , ) = 2 cos sin ; thus we obtain
sin cos(y , x) + cos sin(y , x) = 2 cos sin :
(4) Similarly we have (exchanging for , and x for y , y for x, simultaneously) sin cos(x , y ) + cos sin(x , y ) = 2 cos sin , or sin cos(y , x) , cos sin(y , x) = 2 cos sin : (5) Multiplying (4) by sin and (5) by sin , we get
(sin cos + cos sin ) sin(y , x) = 0 or sin( + ) sin(y , x) = 0, that is sin sin(y , x) = 0: Since sin > 0, we have sin(y , x) = 0, therefore y = x, and cos(y , x) = 1: Hence 1we get from (4) sin = 2 cos sin : 1Because sin > 0 we get cos = 2 : Thus we have = 60 . As = 90 , 2 \A; we obtain \A = 60 :
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UKand KEEWAI LAU, Hong Kong. There was one incorrect solution.
81
2016. [1995: 53] Proposed by N. Kildonan, Winnipeg, Manitoba.
Recall that 0:19 stands for the repeating decimal 0:19191919 : : : , for example, and that the period of a repeating decimal is the number of digits in the repeating part. What is the period of (a) 0:19 + 0:199 ; (b) 0:19 0:199 ? Solution by Christopher J. Bradley, Clifton College, Bristol, UK (a) We have
199 19 10101 + 199 1001 = 391118 0:19 + 0:199 = 19 99 + 999 = 999999 999999 = 0:391118; so the period is 6. (b) Since
199 ; 0:19 0:199 = 19 99 999
we look for a number with all nines which is a multiple of 999 99. In order to be a multiple of 999 it must have 3k nines for some integer k. When this number is divided by 999, the quotient has k ones interspersed with pairs of zeros: 1001001 : : : 1001. In order that this quotient be divisible by 99 as well, k must be divisible by 9 and must be even (using the well known rules for divisibility by 9 and 11). Hence the required number will have 3 9 2 = 54 nines, so the period is 54. That the number has the full period of 54 results from the fact that 19 and 199 are both coprime to 99 and 999 ensuring no fortuitous cancellations. Also solved by HAYO AHLBURG, Benidorm, Spain; SEFKET Berlin, Germany; NIELS BEJLEGAARD, Stavanger, Norway; ARSLANAGIC, FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and MARIA ASCENSION LOPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; MIGUEL OCHOA, Logro~no, Spain; ADRIAN CHAN, student, Upper ANGEL CABEZON Canada College, Toronto, Ontario; TIM CROSS, Wolverley High School, Kidderminster, UK; KEITH EKBLAW, Walla Walla, Washington, USA; JEFFREY K. FLOYD, Newnan, Georgia, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus Y, Ferris State University, Big Rapids, Michigan, USA; tria; VACLAV KONECN DAVID E. MANES, State University of New York, Oneonta, NY, USA; J. A. MCCALLUM, Medicine Hat, Alberta; P. PENNING, Delft, the Netherlands; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; HEINZ{JURGEN SEIFFERT, Berlin, Germany; DAVID R. STONE, Georgia Southern University, Statesboro; PANOS E. TSAOUSSOGLOU, Athens, Greece; EDWARD T. H. WANG, Wilfrid Laurier University, Waterloo, Ontario; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer.
82 Part (a) only was solved by CHARLES ASHBACHER, Cedar Rapids, Iowa, USA; TOBY GEE, student, The John of Gaunt School, Trowbridge, England; and JOHN S. VLACHAKIS, Athens, Greece. Part (b) only was solved by KEEWAI LAU, Hong Kong. Konecny and Perz had solutions which were comparable to Bradley's in their simplicity and minimal dependence on calculation. Cross notes that the product in (b) equals
3781 = 3780 + 1 ; 98901 98901 98901 where the rst fraction is 140=3663 = 0:038220, and the second fraction is
the \rather interesting" recurring decimal
0:000010 111121 222232 333343 444454 555565 666676 777787 888899 ! Bellot and Lopez mention the following general result of J. E. Oliver which is quoted (for an arbitrary number of fractions) on page 166 of Dickson's History of the Theory of Numbers, Volume 1: if x0 =x and z 0 =z are periodic fractions, with periods a and b respectively, then (x0 =x)(z 0 =z ) has a period of
xz [a; b]; [x; z] where [a; b] is the least common multiple of a and b. This formula, which can be more simply written as (x; z )[a; b] where (x; z ) is the greatest common divisor of x and z , gives precisely 54 in our case. However, perhaps \pe
riod" here was intended to mean any period rather than just the smallest period, because the formula fails in many cases. For example, for the fractions 11=3 and 1=11, which have periods 1 and 2 respectively, the formula gives (3; 11)[1; 2] = 1 2 = 2 as the period for the product (11=3)(1=11) = 1=3, which has period only 1 of course. Unfortunately, the reference given for Oliver's result is page 295 of Math. Monthly, Vol. 1, 1859; this cannot be the familiar American Math. Monthly, which only started publishing in 1894. Can any reader supply us with more information?
2017. [1995: 53] Proposed by D. J. Smeenk, Zaltbommel, the Netherlands. We are given a xed circle and two xed points A and B not lying on . A variable circle through A and B intersects in C and D. Show that the ratio AC AD BC BD
is constant. [This is not a new problem. A reference will be given when the solution is published.]
83 Solution by Toshio Seimiya, Kawasaki, Japan. Let be a xed circle passing through A; B and intersecting at X and Y . Then XY , CD, and AB are all parallel or concurrent at the radical centre of the three circles. Case 1. XY kAB: If XY kAB then CDkAB , so AC = BD and AD = BC: Hence
AC AD = 1 (which is a constant). BC BD Case 2. XY 6 k AB . Let P be the intersection of XY with AB , then P is a xed point (the radical centre) and CD passes through P . Since either \CAD = \CBD, or \CAD + \CBD = 180, we get [ACD] = AC AD ; (1) [BCD] BC BD where [UV W ] denotes the area of triangle UV W . Let A0 , B 0 be the feet of perpendiculars from A, B to CD, then 0 AA kBB0; and AA0 = AP : (2) BB0 BP [ACD] = AA0 , we have from (1) and (2) Because [BCD] BB0 AC AD = AP = constant. BC BD BP The proposer tells us that the problem comes from a 1939 book by Dr. P. Molenbroek. (Note that \CAD and \DBC might be supplementary rather than equal, so sin \CAD = sin \DBC still holds, but the triangles CAD and DBC are not necessarily similar.) Also solved by: CLAUDIO ARCONCHER, Jundia, Brazil; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; DAN PEDOE, Minneapolis, Minnesota, USA; P. PENNING, Delft, the Netherlands; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; there were four incorrect solutions.
84
2018. [1995: 53] Proposed by Marcin E. Kuczma, Warszawa, Poland.
How manyPpermutations (x1 ; : : : ; xn ) of f1; : : : ; 2g are there such that the cyclic sum ni=1 jx1 , xi+1 j (with xn+1 = x1 ) is (a) a minimum, (b) a maximum? Solutionby Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USA. (a) Let j and k be such that xj = 1 and kk = n. For all m, de ne xn+m = xm . Then we have kX ,1 jX ,1 n X jxi , xi+1j = jxi , xi+1j + jxi , xi+1j i=1 i=j i=k jxk , xj j + jxj , xk j by the triangle inequality
= 2n , 2:
Equality holds if and only if both sequences fxk = n; xk+1; : : : ; xj,1; xj = 1g fxk ; xk,1 ; : : : ; xj+1; xj = 1g are monotonic decreasing. Each element of fn , 1; n , 2; : : : ; 3; 2g may be in either the rst or second sequence, but not both. Choosing which elements are in these sequences determines the positions of all of fn , 1; n , 2; : : : ; 2; 1g uniquely. Thus there are n2n,2 permutations with minimal sum 2n , 2. (b) Consider the eect of a single element xi on the sum. If xi > xi+1 and xi > xi,1 , then the two terms in the sum involving xi are (xi , xi+1 ) and (xi , xi,1 ). Thus the term xi contributes 2xi to the sum. Similarly, if xi < xi+1 and xi < xi,1 , then the term xi contributes ,2xi to the sum. Further, if xi is less than one of xi,1 , xi+1 , and greater than the other, then xi has no contribution to the sum. Suppose that j elements have no contribution to the sum. There are n pairs of elements xi , xi+1 . In each pair, one number is greater and one is less than the other. Thus there are n,2 j numbers xi which contribute 2xi to the sum, and n,2 j numbers xi which contribute ,2xi to the sum. If we can arrange the numbers xi which contribute 2xi to the sum to be the largest n,j numbers, and the numbers xi which contribute ,2xi to the sum to be 2 the smallest n,2 j numbers, then a maximum value is clearly attained with j as small as possible. This is indeed possible. If n is even (n = 2k) and j = 0, we must have that fxi ; xi+2 ; : : : ; xi,4 ; xi,2 g is a permutation of fk +1; k +2; : : : ; 2kg for i = 0 or i = 1, and the other k numbers must be a permutation of f1; 2; : : : ; kg. Otherwise, for some xi, we would have xi < xi+1 < xi+2 which is a contradiction. This gives a same maximal sum of 2(k + 1 + k + 2 + : : : + 2k) , 2(1 + 2 + : : : + k) = 2(k2) = n2=2, for any permutation
85 of fk + 1; k + 2 : : : ; 2kg and f1; 2; : : : ; kg. So there are 2(k!)2 possible permutations with maximal sum if n = 2k. If n is odd (n = 2k + 1), then j must be odd, so j is at least 1. Placing the middle element k +1 in one of the 2k +1 possible positions gives permutations of f1; 2; : : : ; kg and fk +2; k +3; : : : ; 2k +1g in alternating positions. This gives the same maximal sum, (n2 , 1)=2, for every such permutation. Hence there are 2(2k + 1)(k!)2 possible permutations with maximal sum of n = 2k + 1. Also solved by NEILS BEJLEGAARD, Stavanger, Norway; TOBY GEE, student, The John of Gaunt School, Trowbridge, England, part (a) only; P. PENNING, Delft, the Netherlands; and the proposer.
2019. [1995: 53] Proposed by P. Penning, Delft, the Netherlands.
In a plane are given a circle C with diameter `, and a point P within C but not on `. Construct the equilateral triangles that have one vertex at P , one on C , and one on `. Solutionby Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USA. Assume that we have a solution triangle, and consider the eect of rotating the circle C by 60 about P . We can do this in two directions, clockwise or counterclockwise. The vertex of the equilateral triangle on C is rotated to the vertex on the original diameter. Since each rotated circle intersects the diameter only once (or twice, if ` is to be interpreted as the line containing the given diameter), this vertex is now known and we can easily nd the vertex on C of the equilateral triangle. We see that there are two such equilateral triangles (four if the extension of the diameter is allowed). To construct the pair of rotated circles, nd their centres as the third vertices of the equilateral triangles that have PO as base, where O is the centre C ; their radius is half the length of the given diameter. Also solved by CLAUDIO ARCONCHER, Jundia, Brazil; NIELS BEJLEGAARD, Stavanger, Norway; CHRISTOPHER J. BRADLEY, CliftonCollege, Bristol, UK; PETER HURTHIG, Columbia College, Burnaby, BC; D. J. SMEENK, Zaltbommel, the Netherlands; and the proposer. One incorrect solution was received. Most of the solvers used an argument similar to the featured solution. Bradley remarked that he has seen the rotation technique used to solve the analogous problem in which a line parallel to ` is given instead of the circle C . Bejlegaard went even further, pointing out that the given triangle could have any prescribed shape (instead of equilateral) and that C and ` could be any two curves for which the points of intersection of ` with the rotated image of C could be constructed.
86
2020. [1995: 53] Proposed by Christopher J. Bradley, Clifton College, Bristol, UK. Let a, b, c, d be distinct real numbers such that a + b + c + d = 4 and ac = bd: b c d a
Find the maximum value of
a + b + c + d: c d a b
Solutionby Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USA. b c b 1 a Let x = and y = . Then by ac = bd we have = = ,
b c b c y d a x d = c = 1, a = b a b y c xy and d = c d = x . Thus we are to nd the maximum y 1 x value of xy + + + subject to the conditions that a, b, c, d are distinct x xy y and
1
() x + y + x1 + y1 = 4: 1
z 1 + x = ef . x xy y
Let e = x + and f = y + . Then xy + +
x
y
By the Arithmetic MeanGeometric Mean Inequality, we have t + 1t 2 if t > 0 and t + 1t ,2 if t < 0. By (), we have that x and y cannot both be negative. If both are positive, then () implies x = y = 1 or a = b = c, a contradiction. Hence exactly one of x and y is negative. Assume, without loss of generality, that x > 0, y < 0. Then we get f ,2; e = 4p, f 6 and so ef ,12. Equality p holds, for example, when a = 3 + 2 2; b = 1; c = ,1 and d = ,(3 + 2 2). Berlin, Germany; NIELS Also solved by SEFKET ARSLANAGIC, BEJLEGAARD, Stavanger, Norway; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; PETER HURTHIG, Columbia College, Burnaby, BC; VACLAV Y, Ferris State University, Big Rapids, Michigan, USA; KEEWAI KONECN LAU, Hong Kong; DAVID E. MANES, SUNY at Oneonta, Oneonta, New York, USA; BEATRIZ MARGOLIS, Paris, France; P. PENNING, Delft, the Nether lands; HEINZJURGEN SEIFFERT, Berlin, Germany; JOHN VLACHAKIS, Athens, Greece; and the proposer. Ten incorrect or incomplete solutions were also received. (Is this a record?)
87
a b c d
Many of them showed that + + + 4 and then claimed erroneously c d a b that 4 is actually the maximum value. From Bosley's solutions above, it is not dicult to show that p2)thek; ,upper (3 bound , 12 is attained if and only if ( a; b; c; d ) equals ( k; (3 2 p p p 2 2)k; ,k) or (k; ,k; ,(3 2 2)k; (3 2 2)k) for some k 6= 0. This was shown by Flanigan and Mane. 661 Arslanagic conjectured that the minimum value of the given sum is ,27 !
900
2021. [1995: 89] Proposed by Toshio Seimiya, Kawasaki, Japan.
P is a variable interior point of a triangle ABC , and AP , BP , CP meet BC , CA, AB at D, E, F respectively. Find the locus of P so that
[PAF ] + [PBD] + [PCE ] = 12 [ABC ];
where [XY Z ] denotes the area of triangle XY Z . I. Solution by P. Penning, Delft, the Netherlands. First, note that [PAE ] + [PFB ] + [PDC ] = 21 [ABC ].
C
E
P
D
A
B
F Let AF=AB = 12 , x, BD=BC = 21 , y , CE=CA = 12 , z . From Ceva's Theorem, we get: , 1 , x , 1 , y, 1 , z = , 1 + x , 1 + y, 1 + z 2 2 2 2 2 2 or
x + y + z = 4xyz:
(1)
From the gure, we can see that , 1 , x [ABC ] = [PAF ]+[PAE]+[PEC ] = [PAE]+ 1 [ABC ],[PBD] : 2 2 Thus x[ABC ] = [PBD] , [PAE]: (2)
88 Similarly, we have
y[ABC ] = [PCE] , [PFB]; z[ABC ] = [PAF ] , [PDC ]:
(3) (4)
Adding (2), (3), and (4) gives
(x + y + z)[ABC ] = 12 [ABC ] , 12 [ABC ]; so that x + y + z = 0. This, together with (1), gives xyz = 0, so that x = 0 or y = 0 or z = 0.
Thus the locus consists of the three medians of the triangle (excluding the end points). II. Solution by the Austin Academy Problem Solvers, Austin, Texas, USA. Assign masses a, b, c to the vertices A, B , C , respectively, so that P is the centre of mass and a + b + c = 2. [PAF ] = FP AF = c b . Then [ABC ] CF AB (a + b + c) (a + b) There are similar expression for PBD and PCE . So we need to nd the locus of P such that
bc + ac + ab = (a + b + c) ( = 1): (a + b) (b + c) (a + c) 2
Cross multiplying and simplifying leads to
(a , b)(b , c)(c , a)(a + b + c) = 0: But a + b + c = 2, so we must have at least one of a = b, b = c and c = a.
Now, this is exactly when P lies on a median of triangle ABC (excluding the end points). Berlin, Germany; CARL BOSLEY, Also solved by SEFKET ARSLANAGIC, student, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUEL AMENGUAL COVAS, Cala Figuera, Mallora, Spain; PETER HURTHIG, Columbia College, Burnaby, BC; KEEWAI LAU, Hong Kong; ASHISH KR. SINGH, Student, Kanpur, India; D. J. SMEENK, Zaltbommel, the Netherlands; HOE TECK WEE, student, Hwa Chong Junior College, Singapore; and the proposer.
89
2022. [1995: 89] Proposed by K. R. S. Sastry, Dodballapur, India. Find the smallest integer of the form
A?B; B
where A and B are threedigit integers and A?B denotes the sixdigitinteger formed by placing A and B side by side. Solution by M. Parmenter, Memorial University of Newfoundland, St. John's, Newfoundland. I claim that the answer to the problem is 121. A ? B = 121, so this number Note that when A = 114 and B = 950, then B is actually obtained. We must therefore show that in any other case when D = A ? B is an integer, then D 121.
B
A ? B = (1000A + B) = 1000A + 1, so we must now show B 1000A B B that whenever E = B is an integer, then E 120. A Let B = GX where G = gcd(B; 1000). We will show that whenever is X 1000 A an integer, then F = G X 120. Note that since B < 1000, we 1000 . Also, recall that we are only interested in cases when A is have X < X an integer. G Observe that
If G = 1, 2, 4, 5 or 8, then
1000 > 120, and we are done.
G
A
If G = 10, then X < 100, so 2 (since A 100). In this case, F 200, and we are done. X
A 3. Thus F 150, and we are done. X A If G = 25, then X < 40 and 3. Thus F 120, and we are done. X A If G = 40, then X < 25 and 5. Thus F 125, and we are done. X A If G = 50, then X < 20 and 6. Thus F 120, and we are done. X A If G = 100, then X < 10 and 12. (Note that x 9). Thus X F 120, and we are done. If G = 20, then X < 50 and
90
A
If G = 125, then X < 8 and 15. Thus F 120, and we are X done.
A
If G = 200, then X < 5 and 25. (Note that x 4). Thus F 125, and we are done. X
A
If G = 250, then X < 4 and 34. Thus F 136, and we are X done.
1000A
If G = 500, then B = 500 and B = 2A 200, and we are now all done. Note that I am assuming that A is an \honest" threedigit integer, that is 100 A. Otherwise, there is a trivial solution: A = 001, B = 500, and A ? B = 3. B Also solved by CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, OCHOA, Logro~no, Spain; JEFFREY R. FLOYD, UK; MIGUEL ANGEL CABEZON Newnan, Georgia, USA; TOBY GEE, student, the John of Gaunt School, Trowbridge, England; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEEWAI LAU, Hong Kong; J. A. MCCALLUM, Medicine Hat, Alberta; STEWART METCHETTE, Culver City, California, USA; P. PENNING, Delft, the Netherlands; the SCIENCE ACADEMY PROBLEM SOLVERS; Austin, Texas, USA; DAVID STONE and BILL MEISEL, Georgia Southern University, Statesboro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; HOE TECK WEE, student, Hwa Chong Junior College, Singapore; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer. Some solvers made use of computers to search for all possible solutions. One incorrect solution was submitted. Janous suggested an extension to:
A
Let 2 N. Determine the integerminimum, n of , where A B and B are ndigit numbers.
91
2023. [1995: 89] Proposed by Waldemar Pompe, student, University of Warsaw, Poland. Let a, b, c, d, e be positive numbers with abcde = 1. (a) Prove that a + abc + b + bdc + c + cde 1 + ab + abcd 1 + bc + bcde 1 + cd + cdea + 1 +dde+ +deadeab + 1 +eea+ +eabeabc 10 3:
(b) Find a generalization! Solution to (a) by Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USA. Let x1 = a, x2 = ab, x3 = abc, x4 = abcd and x5 = abcde = 1. Multiply the second, the third, the fourth and the fth fractions on the left by a , ab , abc and abcd respectively. Then the expression on the left becomes: a ab abc abcd x1 + x3 + x2 + x4 + x3 + x5 x5 + x2 + x4 x1 + x3 + x5 x2 + x4 + x1 + x x+4 x+ x+1 x + x x+5 x+ x+2 x : 3 5 2 4 1 3 Add 1 to each of the ve fractions. Then the desired inequality is equivalent to: 5 ! X xi x + x1 + x + x + x1 + x + x + x1 + x 5 2 4 1 3 5 2 4 1 i=1
+ x + x1 + x + x + x1 + x 25 3; 3 5 2 4 1 3
which follows by applying the Arithmetic Mean { Harmonic Mean inequality to the second factor on the left. Equality holds if and only if the ve denominators are all equal, which happens if and only if x1 = x2 = x3 = x4 = x5 , which is true if and only if a = b = c = d = e = 1. Solution to (b) by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore. Suppose that a1 ; a2 ; : : : ; an are positive real numbers such that a1 a2 : : : an = 1, where n > 1. For each i = 1; 2; : : : ; n, de ne ai+n = ai, i+Y j ,1 and for i; j = 1; 2; : : : ; n, let Ai;j = ar . Further, for each i = r =i 1; 2; : : : ; n, de ne Ai;j +n = Ai;j . Let I be any nonempty subset of S = f0; 1; 2; : : : ; n , 1g. Then the
92 generalized inequality is:
0 X A n B j 2S,I i;j X B@ X Ai;k i=1 k2I
The proof is as follows: let u =
X j 2I nX ,1 k=0
Ai;j Ai;k
X j 2I nX ,1
=
k=0
1 CC n (n , jI j) : A jI j nX ,1 r=0
A1;r. Then
X
A1;i,1Ai;j
=
A1;i,1Ai;k
j 2I nX ,1 k=0
X = u1 A1;j+i,1;
A1;j+i,1 A1;k+i,1
j 2I
so that
0 n B B X B B i=1 B @
X j 2I nX ,1 k=0
Ai;j Ai;k
1 CC n 1 X n XX 1 CC = X A1;j+i,1 = u A1;j+i,1 CA i=1 u j 2I j 2I i=1 X = u1 u = jI j: j 2I
By the Arithmetic Mean { Harmonic Mean inequality, we have:
0 XA n B j 2S i;j X B@ X Ai;k i=1 k2I
and thus
0 X A n B j 2S,I i;j X B@ X Ai;k i=1 k2I
1 C C A
n X i=1
2 nX j 2I X k2S
Ai;j
2 = njI j ;
Ai;k
1 CC n2 , n = n (n , jI j) : A jI j jI j
Note that part (a) is the special case when n = 5 and I = f0; 2; 4g.
93 Besides Bosley and Wee, both parts were also solved by VEDULA N. MURTY, Andhra University, Visakhapatnam, India; and the proposer. Part (a) only was solved by SABIN CAUTIS, Earl Haig Secondary School, North York, Ontario.
2024. [1995: 90] Proposed by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. It is a known result that if P is any point on the circumcircle of a given triangle ABC with orthocentre H , then (PA)2 + (PB )2 + (PC )2 , (PH )2 is a constant. Generalize this result to an ndimensional simplex. Solution by Christopher J. Bradley, Clifton College, Bristol, UK. Let O be the centre of a circumhypersphere, let R be the radius, and let A1 , A2 , : : : ; An be the vertices. De ne H by the vector expression: OH =
n X
i=1
OAi :
Also, let P be any point on the surface of the circumhypersphere such that jOP j2 = R2. Then we claim that Let OP = ~x. Then n X
n X
(PAi)2 , (PH )2 = constant.
i=1
(PAi)2 , (PH )2 =
i=1
n X i=1
(~x , OAi) : (~x , OAi)
, ~x ,
n X
!
n X
!
OAi : x~ , OAi i=1 n X = n j~xj2 , 2~x: OAi + OA2i i=i i=1 n n X n X X , j~xj2 + 2~x: OAi , OAi:OAj i=1 i=1 j =1 n X n X 2 = (n , 1)R , OAi:OAj = constant.
i=1 n X
i=1 j =1
CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA, WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria, and the proposer had essentially the same solution.
94
2027. [1995: 90] Proposed by D. J. Smeenk, Zaltbommel, the Netherlands. Quadrilateral ABCD is inscribed in a circle ,, and has an incircle as well. EF is a diameter of , with EF ? BD. BD intersects EF in M and AC in S. Show that AS : SC = EM : MF . Solution by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore. Note: It is necessary that A and E lie on the same side of the line BD, otherwise the result is false. Since ABCD has an incircle, we have AB + CD = BC + AD, or, equivalently, BC , CD = AB , AD. Thus C F
B S
A E
M D
AS = AreaABD = AB AD sin(\BAD) AC Area BCD BC CD sin(\BCD) AB AD = BC CD (since \BAD + \BCD = 180 ) ,(BC , CD)2 , BD2 2 AB AD = 2BC CD ((AB , AD)2 , BD2 ) ,BC 2 + CD2 , 2BC CD , BD2 AB AD = 2BC CD (AB 2 + AD2 , 2AB AD , BD2) BC 2 + CD2 , BD2 , 2BC CD CD = AB2 + AD22BC , BD2 , 2AB AD 2AB AD cos( \BCD) , 1 = cos(\BAD) , 1 ; using the cosine rule, applied to triangles ABD and BCD.
95 Since EF is a diameter, we have BE + DF = DE + BF , and so, by an argument similar to the above, we get EM = cos(\BFD) , 1 : MF cos(\BED) , 1 Since ABCD is a cyclic quadrilateral, we have \BED = \BAD and \BFD = \BCD, and hence that AS : SC = EM : MF . Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEEWAI LAU, Hong Kong; P. PENNING, Delft, the Netherlands; SCIENCE ACADEMY PROBLEM SOLVERS, Austin, Texas, USA; TOSHIO SEIMIYA, Kawasaki, Japan; ASHISH KR. SINGH, student, Kanpur, India; and the proposer. These solvers assumed implicitly that A and E lay on the same side of the line BD.
2028. [1995: 90] Proposed by Marcin E. Kuczma, Warszawa, Poland.
If n m k 0, are integers , such that n + m , k + 1 is a power of 2, prove that the sum nk + mk is even. Solution by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore. For any nonnegative integer t, let the binary representation of t be (ar ar,1 a0)2, where ai 2 f0; 1g, i = 0; 1; : : : ; r. Let f (t) be the largest integer such that 2 divides t. Let g (t) be the sum of the digits in the binary representation of t. Then
f (t!) = =
1t X
r X
(ar ar,1 aj )2 j = j =1 j =1 2 ,1 r jX X
j =1 s=0
2s =
r X
j =1
aj (2j , 1) = t , g(t):
From this we get
n k = g(n , k) + g(k) , g(n): ,, Similarly, f mk = g (m , k) + g (k) , g(m). Let n + m , k + 1 = 2 , that is, n + m , k = 2 , 1. Then, by considering the binary representations of n, m , k, 2 , 1, m, and n , k, it is clear that g(n) + g(m , k) = = g(n , k) + g(m): Thus g (n , k) + g (k) , g (n) = g (m , k) + g (k) , g (m), that is, f ,,n = f ,,m. Hence 2j,n , 2j,m, so ,n + ,m is even. f
k
k
k
k
k
k
96 Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; HEINZJURGEN SEIFFERT, Berlin, Germany; SOFIYA VASINA, student, University of Arizona, Tucson, USA; and the proposer. Kuczma notes that the problem can be solved without any computation as follows: Look at Pascal's triangle, rows 0 through 2 , modulo 2, and visualize it geometrically as an equilateral triangle with black and white spots (zeros and ones). It has three (geometric) symmetry axes, and each one of these symmetries preserves the spot design. This fact can be considered as known. And, if not, it suces to draw the \row 0 through 7 triangle" and notice that it consists of three copies of the (twice) smaller triangle formed by rows 0 through 3, the remaining quarter in the centre of the gure being
lled with zeros. The base row is all ones. These observations provide a scheme for an induction proof of the claim for rows 0 through 2 , 1. The symmetry with respect to one of the oblique axes is precisely the contents of the problem, expressed in algebraic terms.
Comment by the EditorinChief Choosing a solution to print is not an easy task. Each collection of submitted solutions is assigned to a member of the Editorial Board, and these editors work independently of one another. The choice of which solution to highlight is left entirely to that editor. Sometimes it happens, as in this issue, that several solutions are chosen from the same solver. I would like to assure all subscribers that every submission is very important to CRUX, and that we encourage everyone to submit solutions, as well as proposals for problems, articles for publication, and contributions to the other corners. Every subscriber is very important to us and we really value all contributions. And while I am on the subject of contributions, please continue to send in proposals for problems. We publish 100 per year, and we do not have too many in reserve at this time. Without your contributions, there would be no CRUX.
97
Dissecting Triangles into Isosceles Triangles Daniel Robbins Sudhakar Sivapalan
student, Ecole Secondaire Beaumont, Beaumont student, Harry Ainlay Composite High School, Edmonton
Matthew Wong
student, University of Alberta, Edmonton Problem 2 of Part II of the 19931994 Alberta High School Mathematics Competition (see CRUX [20:65]) goes as follows: An isosceles triangle is called an amoeba if it can be divided into two isosceles triangles by a straight cut. How many dierent (that is, not similar) amoebas are there? All three authors wrote that contest. Afterwards, they felt that the problem would have been more meaningful had they been asked to cut nonisosceles triangles into isosceles ones. We say that a triangle is ndissectible if it can be dissected into n isosceles triangles where n is a positive integer. Since we are primarily interested in the minimum value of n, we also say that a triangle is ncritical if it is ndissectible but not mdissectible for any m < n. The isosceles triangles themselves are the only ones that are 1dissectible, and of course 1critical. Note that, in the second de nition, we should not replace \not mdissectible for any m < n" by \not (n , 1)dissectible". It may appear that if a triangle is ndissectible, then it must also be mdissectible for all m > n. However, there are two exceptions. The solution to the contest problem, which motivated this study, shows that almost all 1dissectible triangles are not 2dissectible. We will point out later that some 2dissectible ones are not 3dissectible. On the other hand, it is easy to see that all 1dissectible triangles are 3dissectible. Figure 1 illustrates the three cases where the vertical angle is acute, right, and obtuse respectively. A A A Z A ZZ A
(a)
,@ @ , ,@ , , , @
(b) Figure 1.
HH H S S HH S H
(c)
98 We now nd all 2dissectible triangles. Clearly, such a triangle can only be cut into two triangles by drawing a line from a vertex to the opposite side, as illustrated in Figure 2.
B
A
SS S S S S 2 SS 2
D
Figure 2.
C
Note that at least one of \ADB and \ADC is nonacute. We may assume that \ADB 90 . In order for BAD to be an isosceles triangle, we must have \BAD = \ABD. Denote their common value by . By the Exterior Angle Theorem, \ADC = 2. There are three ways in which CAD may become an isosceles triangle. Case 1. Here \ACD = \ADC = 2, as illustrated in Figure 2. Then \CAD = 180 , 4 > 0. This class consists of all triangles in which two of the angles are in the ratio 1:2, where the smaller angle satis es 0 < < 45 . Of these, only the (36 ; 72 ; 72 ) triangle is 1dissectible, but it turns out that every triangle here is 3dissectible. Case 2. Here, \CAD = \ADC = 2. Then \CAB = 3 and \ACD = 180 , 4 > 0. This class consists of all triangles in which two of the angles are in the ratio 1:3, where the smaller angle satis es 0 < < 45 . Of 540 ; 540 ) triangles are 1these, only the (36 ; 36 ; 108 ) and the ( 180 7 ; 7 7 dissectible. It turns out that those triangles for which 30 < < 45 , with a few exceptions, are not 3dissectible. Case 3. Here, \ACD = \CAD. Then their common value is 90 , so that \CAB = 90 . This class consists of all right triangles. Of these, only the (45; 45; 90 ) triangle is 1dissectible, and it turns out that every triangle here is 3dissectible. We should point out that while our three classes of 2dissectible triangles are exhaustive, they are not mutually exclusive. For instance, the (30; 60; 90 ) triangle appears in all three classes, with the same dissection. The (20; 40; 120) triangle is the only one with two dierent dissections. We now consider 3dissectible triangles. Suppose one of the cuts does not pass through any vertices. Then it divides the triangle into a triangle and a quadrilateral. The latter must then be cut into two triangles, and this cut must pass through a vertex. Hence at least one cut passes through a vertex.
99 Suppose no cut goes from a vertex to the opposite side. The only possible con guration is the one illustrated in Figure 1(a). Since the three angles at this point sum to 360, at least two of them must be obtuse. It follows that the three arms have equal length and this point is the circumcentre of the original triangle. Since it is an interior point, the triangle is acute. Thus all acute triangles are 3dissectible. In all other cases, one of the cuts go from a vertex to the opposite side, dividing the triangle into an isosceles one and a 2dissectible one. There are quite a number of cases, but the argument is essentially an elaboration of that used to determine all 2dissectible triangles. We leave the details to the reader, and will just summarize our ndings in the following statement. Theorem. A triangle is 3dissectible if and only if it satis es at least one of the following conditions: 1. 2. 3. 4. 5.
It is an isosceles triangle. It is an acute triangle. It is a right triangle. It has a 45 angle. It has one of the following forms: (a) (, 90 , 2, 90 + ), 0 < < 45; (b) (, 90 , 32 , 90 + 2 ), 0 < < 60 ; (c) (, 360 , 7, 6 , 180 ), 30 < < 45 ; (d) (2, 90 , 32 , 90 , 2 ), 0 < < 60; (e) (3, 90 , 2, 90 , ), 0 < < 45 ; (f) (180 , 2, 180 , , 3 , 180 ), 60 < < 90 ; (g) (180 , 4, 180 , 3, 7 , 180 ), 30 < < 45 . 6. Two of its angles are in the ratio p:q , with the smaller angle strictly between 0 and r , for the following values of p, q and r:
p 1 1 1 1 1 1 2 3 3 q 2 3 4 5 6 7 3 4 5 r 60 30 22.5 30 22.5 22.5 60 67.5 67.5
100 The fact that all right triangles are 2dissectible is important because every triangle can be divided into two right triangles by cutting along the altitude to its longest side. Each can then be cut into two isosceles triangles by cutting along the median from the right angle to the hypotenuse, as illustrated in Figure 3. It follows that all triangles are 4dissectible, and that there are no ncritical triangles for n 5. S S S S HH S HH S HH S HH S
Figure 3.
We can prove by mathematical induction on n that all triangles are ndissectible for all n 4. We have established this for n = 4. Assume that all triangles are ndissectible for some n 4. Consider any triangle. Divide it by a line through a vertex into an isosceles triangle and another one. By the induction hypothesis, the second can be dissected into n triangles. Hence the original triangle is (n + 1)dissectible.
Announcement For the information of readers, we are saddened to note the deaths of two mathematicians who are wellknown to readers of CRUX. Leroy F. Meyers died last November. He was a long time contributor to CRUX. See the February 1996 issue of the Notices of the American Mathematical Society. D.S. Mitrinovic died last March. He is wellknown for his work on Geometric Inequalities, in particular, being coauthor of Geometric Inequalities and Recent Advances in Geometric Inequalities. See the July 1995 issue of the Notices of the American Mathematical Society.
101
THE SKOLIAD CORNER No. 13 R.E. Woodrow
This issue we feature the Eleventh W.J. Blundon Contest, written February 23, 1994. This contest is sponsored by the Department of Mathematics and Statistics of Memorial University of Newfoundland and is one of the \Provincial" contests that receives support from the Canadian Mathematical Society
THE ELEVENTH W.J. BLUNDON CONTEST February 23, 1994
1. (a) The lesser of two consecutive integers equals 5 more than three times the larger integer. Find the two integers. (b) If 4 x 6 and 2 y 3, nd the minimum values of (x , y )(x + y ). 2. A geometric sequence is a sequence of numbers in which each term, after the rst, can be obtained from the previous term, by multiplying by the same xed constant, called the common ratio. If the second term of a geometric sequence is 12 and the fth term is 81=2, nd the rst term and the common ratio. 3. A square is inscribed in an equilateral triangle. Find the ratio of the area of the square to the area of the triangle. 4. ABCD is a square. Three parallel lines l1 , l2 and l3 pass through A, B and C respectively. The distance between l1 and l2 is 5 and the distance between l2 and l3 is 7. Find the area of ABCD. 5. The sum of the lengths of the three sides of a right triangle is 18. The sum of the squares of the lengths of the three sides is 128. Find the area of the triangle. 6. A palindrome is a word or number that reads the same backwards and forwards. For example, 1991 is a palindromic number. How many palindromic numbers are there between 1 and 99; 999 inclusive? 7. A graph of x2 ,2xy +y2 ,x+y = 12 and y2 ,y ,6 = 0 will produce four lines whose points of intersection are the vertices of a parallelogram. Find the area of the parallelogram. 8. Determine the possible values of c so that the two lines x , y = 2 and cx + y = 3 intersect in the rst quadrant. 9. Consider the function f (x) = 2xcx+3 , x 6= ,3=2. Find all values of c, if any, for which f (f (x)) = x.
102
10. Two numbers are such that the sum of their cubes is 5 and the sum of their squares is 3. Find the sum of the two numbers. Last issue we gave the problems of Part I of the Alberta High School Mathematics Competition, which was written Tuesday November 21, 1995. This month we give the solutions. How well did you do?
ALBERTA HIGH SCHOOL MATHEMATICS COMPETITION Part I: Solutions
November 21, 1995 (Time: 90 minutes)
1. A circle and a parabola are drawn on a piece of paper. The number of regions they divide the paper into is at most A. 3 B. 4 C. 5 D. 6 E. 7. Solution. (D) The parabola divides the plane into two regions. The circle intersects the parabola in at most four points, so that it is divided by the parabola into at most four arcs. Each arc carves an existing region into two. 2. The number of dierent primes p > 2 such that p divides 712 , 2 37 , 51 is A. 0 B. 1 C. 2 D. 3 E. 4. Solution. (D) We have 712 , 362 , 51 = (71 + 37)(71 , 37) , 51 =
3 17(36 2 , 1).
3. Suppose that your height this year is 10% more than it was last year, and last year your height was 20% more than it was the year before. By what percentage has your height increased during the last two years? A. 30 B. 31 C. 32 D. 33 E. none of these. Solution. (C) Suppose the height was 100 two years ago. Then it was
120 a year ago and 132 now.
4. Multiply the consecutive even positive integers together until the product 2 4 6 8 becomes divisible by 1995. The largest even integer you use is A. between 1 and 21 B. between 21 and 31 C. between 31 and 41 D. bigger than 41 E. nonexistent, since the product never becomes divisible by 1995.
103 Solution. (C) All factors of 1995 are distinct and odd, with the largest one being 19. Hence the last even number used is 38. 5. A rectangle contains three circles as in the diagram, all tangent to the rectangle and to each other. If the height of the rectangle is 4, then the width of the rectangle is 4 p p p 4 2 2 2 A. 3 + 2 2 C.p5 + 3 B. 4 + 3 D. 6 E. 5 + 10. Solution. (A) Let O be the centre of the large circle, P that of one of the small circles, and Q the point of tangency of the small circles. Then \PQO p = 180, PQ = 1 and OP = 1 + 2. By Pythagoras' Theorem, OQ = 2 2. 6. Mary Lou works a full day and gets her usual pay. Then she works some overtime hours, each at 150% of her usual hourly salary. Her total pay that day is equivalent to 12 hours at her usual hourly salary. The number of hours that she usually works each day is A. 6 B. 7:5 C. 8 D. 9 E. not uniquely determined by the given information. Solution. (E) Suppose Mary Lou usually works x hours per day but y on that day. All we know is x + 32 (y , x) = 12 or x + 3y = 24.
7. A fair coin is tossed 10; 000 times. The probability p of obtaining at least three heads in a row satis es A. 0 p < 14 B. 14 p < 12 C. 12 p < 34 D. 34 p < 1 E. p = 1: Solution. (D) Partition the tosses into consecutive groups of three, discarding the last one. If we never get 3 heads in a row, none of the 3333 groups can consist of 3 heads. The probability of this is ( 78 )3333, which is clearly less than 14 . In fact, 12
7 8
2 1310722 = 1 : = 117649 < 812 236 8
8. In the plane, the angles of a regular polygon with n sides add up to less than n2 degrees. The smallest possible value of n satis es: A. n < 40 B. 40 n < 80 C. 80 n < 120 D. 120 n < 160 E. n 160.
104 Solution. (E) The sum of the angles is exactly (n , 2)180. From 2 180 < nn, 2 = n + 2 + n ,4 2 < n + 3;
we have n > 177. 9. A cubic polynomial P is such that P (1) = 1, P (2) = 2, P (3) = 3 and P (4) = 5. The value of P (6) is A. 7 B. 10 C. 13 D. 16 E. 19. Solution. (C) By the Binomial Theorem, P (5) = 4P (4) , 6P (3) + 4P (2) , P (1) = 9. It follows that P (6) = 5P (5) , 10P (4) + 10P (3) , 5P (2) + P (1) = 16. 10. The positive numbers x and y satisfy xy = 1. The minimum value of x14 + 4y14 is A. 12
B.
5 8
C. 1
E. no minimum.
D. 45
Solution. (C) We have
1 + 1 = 1 , 1 2 + 1 1; x4 4y4 x2 2y2 x2y2 with equality if and only if x2 = 2y 2. 11. Of the points (0; 0), (2; 0), (3; 1), (1; 2), (3; 3), (4; 3) and (2; 4), at most how many can lie on a circle? A. 3 B. 4 C. 5
D. 6
E. 7.
Solution. (C) Since (0; 0), (1; 2) and (2; 4) are collinear, a circle passes through at most two of them. Since (2; 0), (3; 1), (3; 3) and (4; 3) are not concyclic, a circle passes through at most three of them. The circle with centre (1; 2) and passing through (0; 0) also passes through (2; 0), (3; 1), (3; 3) and (2; 4). 12. The number of dierent positive integer triples (x; y;z) satisfying the equations x2 + y , z = 100 and x + y2 , z = 124 is: A. 0 B. 1 C. 2 D. 3 E. none of these. Solution. (B) Subtraction yields 24 = x + y 2 , x2 , y = (y , x)(y + x , 1). Note that one factor is odd and the other even, and that the rst is smaller than the second. Hence either y , x = 1 and y + x , 1 = 24, or y , x = 3 and y + x , 1 = 8. They lead to (x; y;z) = (12; 13; 57) and (3; 6; ,85) and (3; 6; ,85) respectively. However, we must have z > 0.
105
13. Which of the following conditions does not guarantee that the convex quadrilateral ABCD is a parallelogram? A. AB = CD and AD = BC B. \A = \C and \B = \D C. AB = CD and \A = \C D. AB = CD and AB is parallel to CD E. none of these. Solution. (C) Let \ABD = \BAD = \DCB = 40 and \CBD = 80 . Then ABCD is not a parallelogram. Let E on BC be such that \CDE = 40 . Then triangles BAD and CDE are congruent, so that AB = CD. It is easy to see that the other three conditions do guarantee parallelograms. 14. How many of the expressions x3 + y4; x4 + y3; x3 + y3; and x4 , y4; are positive for all possible numbers x and y for which x > y ? A. 0 B. 1 C. 2 D. 3
E. 4.
Solution. (A) We have x3 + y 4 < 0 for x = , 14 and y = , 31 . Each of the other three expressions is negative if x = 0 and y < 0. 15. In triangle ABC , the altitude from A to BC meets BC at D, and the altitude from B to CA meets AD at H . If AD = 4, BD = 3 and CD = 2, then the length of HD is p p p A. 25 D. 52 B. 23 C. 5 E. 3 2 5 . Solution. (B) Note that \CAD = 180 , \BCA = \CBE . Hence triangles CAD and HBD are similar, so that
HD = CD : BD AD
16. Which of the following is the best approximation to (23 , 1)(33 , 1)(43 , 1) (1003 , 1) ? (23 + 1)(33 + 1)(43 + 1) (1003 + 1) A. 35
B. 33 50
C. 333 500
D. 35;;333 000
;333 . E. 33 50;000
106 Solution. (C) The given expression factors into
(2 , 1)(22 + 2 + 1)(3 , 1)(32 + 3 + 1) (100 , 1)(1002 + 100 + 1) : (2 + 1)(22 , 2 + 1)(3 + 1)(32 , 3 + 1) (100 + 1)(1002 , 100 + 1) Since ((n + 2) , 1) = n + 1 and (n + 1)2 , (n + 1) + 1 = n2 + n + 1, cancellations yield
(2 , 1)(3 , 1)(1002 + 100 + 1) = 10101 : (22 , 2 + 1)(99 + 1)(100 + 1) 15150
That completes the Skoliad Corner for this issue. Send me contest materials, as well as your comments, suggestions, and desires for future directions for the Skoliad Corner.
Citation As was announced in the February 1996 issue of CRUX, Professor Ron Dunkley was appointed to the Order of Canada. This honour was bestowed on Ron by the GovernorGeneral of Canada, the His Excellency The Right Honourable Romeo LeBlanc, in midFebruary, and we are pleased to publish a copy of the ocial citation: Professor Ronald Dunkley, OC
A professor at the University of Waterloo and founding member of the Canadian Mathematics Competition, he has dedicated his career to encouraging excellence in students. He has trained Canadian teams for the International Mathematics Olympiad, authored six secondary school texts and chaired two foundations that administer signi cant scholarship programs. An inspiring teacher, he has stimulated interest and achievement among students at all levels, and provided leadership and development programs for teachers across the country.
107
THE OLYMPIAD CORNER No. 173
R.E. Woodrow All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. The rst Olympiad problems that we give in this issue are the problems of the Selection Tests for the Romanian Team to the 34th International Mathematical Olympiad. My thanks go to Georg Gunther, Sir Wilfred Grenfell College for collecting them and sending them to us while he was Canadian team leader at the IMO at Istanbul, Turkey.
SELECTION TESTS FOR THE ROMANIAN TEAM, 34th IMO. Part I  Selection Test for Balkan Olympic Team 1. Prove that the sequence Im (zn), n 1, of the imaginary parts of
the complex numbers zn = (1 + i)(2+ i) (n + i) contains in nitely many positive and in nitely many negative numbers. 2. Let ABC be a triangle inscribed in the circle O(O;R) and circumscribed to the circle J (I; r). Denote d = RRr . Show that there exists +r a triangle DEF such that for any interior point M in ABC there exists a point X on the sides of DEF such that MX d. 3. Show that the set f1; 2; : : : ; 2ng can be partitioned in two classes such that none of them contains an arithmetic progression with 2n terms. 4. Prove that the equation xn + yn = (x + y)m has a unique integer solution with m > 1, n > 1, x > y > 0.
Part II  First Contest for IMO Team 1st June, 1993
1. Find the greatest real number a such that
+ pz 2y+ x2 + p 2z 2 > a y +z x +y
p
x
2
2
is true for all positive real numbers x, y , z . 2. Show that if x, y, z are positive integers such that x2 + y2 + z2 = 1993, then x + y + z is not a perfect square.
108
3. Each of the diagonals AD, BE and CF of a convex hexagon ABCDEF determine a partition of the hexagon into quadrilaterals having the same area and the same perimeter. Does the hexagon necessarily have a centre of symmetry? 4. Show that for any function f : P (f1; 2; : : : ; ng) ! f1; 2; : : : ; ng there exist two subsets, A and B , of the set f1; 2; : : : ; ng, such that A 6= B and f (A) = f (B ) = maxfi j i 2 A \ B g.
Part III  Second Contest for IMO Team 2nd June, 1993
1. Let f : (0; 1) ! R be a strict convex and strictly increasing function. Show that the sequence ff (n)gn1, does not contain an in nite arithmetic progression. 2. Given integer numbers m and n, with m > n > 1 and (m; n) = 1, nd the gcd of the polynomials f (X ) = X m+n , X m+1 , X + 1 and g(X ) = X m+n + X n+1 , X + 1. 3. Prove that for all integer numbers n, with n 6, there exists an npoint set M in the plane such that every point P in M has at least three other points in M at unit distance to P . 4. For all ordered 4tuples (n1; n2; n3; n4) of positive integer numbers with ni 1 and n1 + n2 + n3 + n4 = n, nd the 4tuples for which the n! number 2l, where n1!n2!n3!n4! l = n21 + n22 + n23 + n24 + n1n2 + n2n3 + n3n4;
has a maximum value.
Part IV  Third Contest for IMO Team 3rd June, 1993
1. The sequence of positive integers fxn gn1 is de ned as follows: x1 = 1, the next two terms are the even numbers 2 and 4, the next three terms are the three odd numbers 5, 7, 9, the next four terms are the even numbers 10, 12, 14, 16 and so on. Find a formula for xn . 2. The triangle ABC is given and let D, E, F be three points such \ = that D 2 (BC ), E 2 (CA), F 2 (AB ), BD = CE = AF and BAD \ = A\ CBE CF . Show that ABC is equilateral.
109
3. Let p be a prime number, p 5, andZp = f1; 2; : : : ; p , 1g. Prove
that for any partition with three subsets of Zp there exists a solution of the equation
x + y z mod p;
each term belonging to a distinct member of the partition. As a national Olympiad, we have the Final Round of the Czechoslovak Mathematical Olympiad, 1993.
CZECHOSLOVAK MATHEMATICAL OLYMPIAD 1993 Final Round 1. Find all natural numbers n for which 7n , 1 is a multiple of 6n , 1. 2. A 19 19 table contains integers so that any two of them lying on
neighbouring elds dier at most by 2. Find the greatest possible number of mutually dierent integers in such a table. (Two elds of the table are considered neighbouring if they have a common side.) 3. A triangle AKL is given in a plane such that j\ALK j > 90 + j\LAK j. Construct an equilateral trapezoid ABCD, AB ? CD, such that K lies on the side BC , L on the diagonal AC and the outer section S of AK and BL coincides with the centre of the circle circumscribed around the trapezoid ABCD. 4. A sequence fang1n=1 of natural numbers is de ned recursively by a1 = 2 and an+1 = the sum of 10th powers of the digits of an , for all n1 1. Decide whether some numbers can appear twice in the sequence fan gn=1 . 5. Find all functions f : Z! Zsuch that f (,1) = f (1) and
f (x) + f (y) = f (x + 2xy) + f (y , 2xy) for all integers x, y .
6. Show that there exists a tetrahedron which can be partitioned into eight congruent tetrahedra, each of which is similar to the original one. Ah, the ling demons are at it again. When I attacked a rather suspicious looking pile of what I thought were as yet un led solutions to 1995 problems from the Corner, I found a small treasuretrove of solutions to various problem sets from 1994, and some comments of Murray Klamkin about earlier material that he submitted at the same time. The remainder of this column will be devoted to catching up on this backlog in an attempt to bring things up to the November 1994 issue. First two comments about solutions from 1992 numbers of the corner.
110
2. [1990: 257; 1992: 40] 1990 Asian Paci c Mathematical Olympiad.
Let a1 ; a2 ; : : : ; an be positive real numbers, and let Sk be the sum of the products of a1 ; a2 ; : : : ; an taken k at a time. Show that 2 SkSn,k nk a1a2 : : : an; for k = 1; 2; : : : ; n , 1:
Comment by Murray S. Klamkin, University of Alberta. It should be noted that the given inequality, as well as other stronger ones, follows from the known and useful Maclaurin inequalities, and that "
Sk ,
#1=k
n k
is a nonincreasing sequence in k, and with equality i all the ai 's are equal. 2. [1992: 197] 1992 Canadian Mathematical Olympiad. For x; y;z 0, establish the inequality
x(x , z)2 + y(y , z)2 (x , z)(y , z)(x + y , z);
and determine when equality holds. Comment by Murray S. Klamkin, University of Alberta. It should be noted that the given inequality is the special case = 1 of Schur's inequality
x (x , y)(x , z) , y(x , z)(x , y) + z(z , x)(z , y) 0: For a proof, since the inequality is symmetric, we may assume that x y z, or that x z y. Assuming the former case, we have x(x , y)(x , z) , y(y , z)(x , y) + z(z , x)(z , y) x (x , y)(y , z) , y(x , z)(x , y) + z(z , x)(z , y) 0: We have assumed 0. For < 0, we have (yz ),(x , y )(x , z ) , (zx), (y , z )(x , y) + (xy ),(x , z )(y , z ) (yz ),(x , y )(x , z) , (zx),(y , z )(x , z ) + (xy ), (x , z )(y , z ) 0: The case x z y goes through in a similar way. Note that if is an even integer, x, y , z can be any real numbers.
111 Next a comment about a solution from the February 1994 number. 6. [1994: 43; 1992: 297] Vietnamese National Olympiad. Let x, y , z be positive real numbers with x y z . Prove that
x2 + y2z + z2x x2 + y2 + z2: y x y
Comment by Murray S. Klamkin, University of Alberta. We give a more direct solution than the previous ones and which applies in many cases where the constraint conditions are x y z . Let z = a, y = a + b, x = a + b + c where a > 0, and b;c 0. Substituting back in the inequality, multiplying by the least common denominator and combining the cubic terms, we get
(a + b + c)3(ab + b2 ) + a3 c(a + b + c) (a + b)3(ab + ac): On inspection, for every term in the expansion of the right hand side there is a corresponding term on the left hand side, which establishes the inequality. Amongst the solutions sent in were three solutions by Klamkin to problems 3, 4 and 7 of the 1992 AustrianPolish Mathematics Competition. We discussed reader's solutions to these in the December number [1995: 336{ 340]. My apologies for not mentioning his solutions there. He also sent in a complete set of solutions to the 43rd Mathematical Olympiad in Poland, for which we discussed solutions to most of the problems in the February 1996 Corner [1996: 24{27]. One problem solution was not covered there and we next give his solution to ll that gap. 5. [1994: 130] 43rd Mathematical Olympiad in Poland. The rectangular 2ngon is the base of a regular pyramid with vertex S. A sphere passing through S cuts the lateral edges SAi in the respective points Bi (i = 1; 2; : : : ; 2n). Show that n X
i=1
SB2i,1 =
n X
i=1
SB2i:
Solution by Murray S. Klamkin, University of Alberta. Let S be the origin (0; 0; 0) of a rectangular coordinate system and let the coordinates of the vertices Ak of the regular 2ngon be given by (r cos k ; r sin k ; a), k = 1; 2; : : : ; 2n where k = k=n. A general sphere through S is given by
(x , h)2 + (y , k)2 + (z , l)2 = h2 + k2 + l2 :
Since the parametric equation of the line SAk is given by
112
x = tr cos k ; y = tr sin k ; z = ta;
its intersection with the sphere is given by
(tr cos k , h)2 + (tr sin k , k)2 + (ta , l)2 = h2 + k2 + l2 : Solving for t, t = 0 corresponding to point S and k + al) : t = (hr cos k(r+2 +krasin 2) p Since SBk = t r2 + a2 , the desired result will follow if X
cos 2k,1 =
X
X
X
cos 2k and sin 2k,1 = sin 2k (where the sums are over k = 1; 2; : : : ; n). Since in the plane (cos 2k,1 ; sin 2k,1 ); k = 1; 2; : : : ; n; P are the vertices ofPa regular ngon, cos 2k,1 and P sin 2k,1 vanish P both and the same for cos 2k and sin 2k . Next we give a comment and alternative solution to a problem discussed in the May 1994 Corner. 7. [1994: 133; 1993: 6667] 14th AustrianPolish Mathematical Olympiad. For a given integer n 1 determine the maximum of the function
2 + x2n,1 f (x) = x + x(1 + xn)2 over x 2 (0; 1) and nd all x > 0 for which this maximum is attained.
Comment by Murray S. Klamkin, University of Alberta. Here is a more compact solution than the previously published one. We show that the maximum value of f (x) is attained for x = 1, by establishing the inequality
4(x + x2 + x2n,1 ) (2n , 1)(1 + 2xn + x2n ): This is a consequence of the majorization inequality [1], i.e., if F (x) is convex and the vector (x1 ; x2 ; : : : ; xn ) majorizes the vector (y1; y2; : : : ; yn ), then
F (y1) + F (y2) + + F (yn) F (x1) + F (x2) + + F (xn): Note that for a vector X of n components xi in nonincreasing order to majorize a vector Y of n components yi in nonincreasing order, we must have
113
p X i=1
xi
p X i=1
yi for p = 1; 2; : : : ; n , 1 and
X
xi =
X
yi
and we write X Y . Since xt is convex in t, we need only show that
X = (2n; 2n;: : : ; 2n; n; n; : : : ; n; 0; 0; : : : ; 0) Y = (2n , 1; : : : ; 2n , 1; 2n,2; : : : ; 2n,2;: : : ; 1; : : : ; 1) where for X there are 2n,1 components of 2n; 2(2n , 1) components of n; (2n , 1) components of 0, while for Y there are 4 components of each of (2n , 1); (2n , 2); : : : ; 1. It follows easily that X and Y have the same number of components and that the sums of their components are equal. As for the rest, actually all we really need show is that
4[(2n , 1) + (2n , 2) + + (n + 1)] = 6n2 , 12n < (2n , 1)(2n) + (2n , 3)(n) = 6n2 , 5n: Reference. 1. A.W. Marshall, I. Olkin, Inequalities: Theory of Majorization and Its Applications, Academic Press, NY, 1979.
Now we turn to some more solutions to problems proposed to the jury but not used at the IMO at Istanbul. Last number we gave solutions to some of these. My \found mail" includes another solution to 2 [1994: 216] by Murray Klamkin, University of Alberta, and solutions to 13 [1994: 241] by Bob Prielipp, University of WisconsinOshkosh and by Cyrus C. Hsia, student, Woburn Collegiate Institute, Scarborough, Ontario. 1. [1994: 216] Proposed by Brazil. 2 Show that there exists a nite set A R such that for every X 2 A there are points Y1 ; Y2 ; : : : ; Y1993 in A such that the distance between X and Yi is equal to 1, for every i. Solution by Cyrus C. Hsia. We will prove the following proposition Pn : There exists a nite set A R2 such that for every point X in A there are n points Y1; Y2; : : : ; Yn in A such that the distance between X and Yi is equal to 1, for every i, and n is an integer greater than 1. (By mathematical induction on n). For the case n = 2 just take any two points a unit distance apart. For n = 3 just take any three points a unit distance apart from each other (i.e. any equilateral triangle of side 1 has vertices with this property). The proposition is satis ed in these two cases.
114 Suppose that the proposition Pn is true for n = k points. Let there be L ( nite) points that satisfy the proposition. Since there are a nite number of points in Ak there are a nite number of unit vectors formed between any two points of distance 1 unit apart. Now choose any unit vector, ~v, dierent from any of the previous ones. Consider the set of 2L points formed by the original L points translated by ~v and the original L points. (No point can be translated onto another point by our choice of ~v). Now by the induction hypothesis, every point in the original L points was a unit distance from k other points. But the translation produced two such sets with each point from one set a unit distance from its translated point in the other. Thus every point of the 2L points are at least a unit distance from k + 1 other points, which means the proposition is true for n = k + 1. Induction complete.
6. [1994: 217] Proposed by Ireland.
Let n, k be positive integers with k n and let S be a set containing distinct real numbers. Let T be the set of all real numbers of the form x1 + x2 + xk where x1; x2 ; : : : ; xk are distinct elements of S. Prove that T contains at least k(n , k) + 1 distinct elements. Solution by Cyrus C. Hsia. The problem should say: Let n, k be positive integers with k n and let S be a set containing n distinct real numbers. Let T be the set of all real numbers of the form x1 + x2 + + xk where x1 ; x2 ; : : : ; xk are distinct elements of S . Prove that T contains at least k(n , k)+1 distinct elements. WOLOG let x1 < x2 ; < xn since all xi are distinct. Then consider the k(n , k) + 1 increasing numbers
x1 + x2 + x3 + + xk,1 + xk < x1 + x2 + x3 + +xk,1 + xk+1 < < x1 + x2 + x3 + + xk,1 + xn < x1 + x2 + x3 + + xk,2 + xk + xn < x1 + x2 + x3 + +xk+1 + xn < < x1 + x2 + x3 + + xn,1 + xn < x1 + x2 + + xk,1 + xn,1 + xn < x1 + x2 + + xk,1 +xn,1 + xn < < x1 + x2 + + xn,2 + xn,1 + xn . . .
(
n , k + 1) (
n , k)
(
n , k) . . .
< x2 + xn,k+2 + + xn,1 + xn < x3 + xn,k+2 + +xn,1 + xn < < xn,k+1 + xn,k+2 + + xn,1 + xn (n , k ) There are at least (n , k + 1) + (k , 1)(n , k) = k(n , k) + 1 distinct
numbers. 9. [1994: 217] Proposed by Poland. Let Sn be the number of sequences (a1 ; a2 ; : : : an ), where ai 2 f0; 1g, in which no six consecutive blocks are equal. Prove that Sn ! 1 when n ! 1.
115 Solution by Cyrus C. Hsia. \No six consecutive blocks are equal" interpretation: There is no sequence with the consecutive numbers 0; 0; 0; 0; 0; 0 or 1; 1; 1; 1; 1; 1 anywhere. Consider the blocks 0; 1 and 1; 0. If the sequences were made only with these then we cannot have six consecutive blocks equal. Let Tn be the number of such sequences for n even (and ending with 0 or 1 for n odd). For example T2 = 2 f0; 1 or 1; 0g T3 = 4 f0; 1; 0 or 0; 1; 1 or 1; 0; 0 or 1; 0; 1g Thus Tn = 2d n2 e . Now Tn < Sn since any 1 in a Tn sequence can be changed to a 0 to form a new sequence in Sn which was not counted in Tn . E.g. 1; 0; 0; 1; 0; 1 in Tn changes to 1; 0; 0; 0; 0; 1 which is in Sn . Note: this change cannot produce six consecutive blocks equal. Thus as n ! 1, Tn = 2d n2 e ! 1. And since Sn > Tn , Sn ! 1. 18. [1994: 242] Proposed by the U.S.A. Prove that
a b c d 2 b + 2c + 3d + c + 2d + 3a + d + 2a + 3b + a + 2b + 3c 3 for all positive real numbers a, b, c, d. Solution by Cyrus C. Hsia. Using the CauchySchwartzBuniakowski inequality, we have
a b c d + + + b + 2c + 3d c + 2d + 3a d + 2a + 3b a + 2b + 3c (a(b + 2c + 3d) + b(c + 2d + 3a) + ) (a + b + c + d)2 ) S(4)(ab + a + ad + bc + bd + cd) (a + b + c + d)2;
where
a b c d S = b + 2c + 3d + c + 2d + 3a + d + 2a + 3b + a + 2b + 3c : Now a2 + b2 2ab from the AMGM inequality. Likewise for the other ve
pairs we have the same inequality. Adding all six pairs gives
Therefore as required.
3(a2 + b2 + c2 + d2) 2(ab + ac + ) ) 3(a + b + c + d)2 8(ab + ac + ): (a + b + c + d)2 ( 38 (ab + ac + ) 2 ; S 4( ab + ac + ) 4(ab + ac + ) 3
116 Comment by Murray Klamkin, University of Alberta. A generalization of this problem appeared recently in School Science & Mathematics as problem #4499. The generalization is the following: Let n be a natural number greater than one. Show that, for all positive numbers a1 ; a2 ; : : : ; an ,
ai 2 ai+1 + 2ai+2 + + (n , 1)ai+n,1 n , 1 ; with equality if and only if a1 = a2 = an . Here all the subscripts greater than n are reduced modulo n. For n = 2 and n = 3, the stated inequality X
reduces to and
a1 + a2 2 a2 a1
a1 a2 a3 a2 + 2a3 + a3 + 2a1 + a1 + 2a2 1; respectively. The special case n = 4 was proposed to the jury for the 34th International Mathematical Olympiad.
That's all the space I have this issue. Send me your nice solutions, and your regional and national Olympiads!
Mathematical Literacy 1. Which Victorian physicist characterised nonnumerical knowledge as \meagre and unsatisfactory". 2. Who said, in 1692, \There are very few things which we know; which are not capable of being reduc'd to a Mathematical Reasoning". 3. Which mathematician is reputed to have examined (and passed) Napoleon at the Ecole Militaire in 1785. 4. Which mathematician was said to have: \frequented low company, with whom he used to guzzle porter and gin".
117
THE ACADEMY CORNER No. 2 Bruce Shawyer
All communications about this column should be sent to Professor Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7 In the rst corner, I mentioned the APICS (Atlantic Provinces Council on the Sciences) annual undergraduate contest. This contest is held at the end of October each year during the Fall APICS Mathematics Meeting. As well as the competition, there is a regular mathematics meeting with presentations, mostly from mathematicians in the Atlantic Provinces. An eort is made to ensure that most of the talks are accessible to undergraduate mathematics students. In fact, a key talk is the annual W.J. Blundon Lecture. This is named in honour of Jack Blundon, who was Head of Department at Memorial University of Newfoundland for the 27 years leading up to 1976. Jack was a great support of CRUX, and long time subscribers will remember his many contributions to this journal.
APICS Mathematics Contest 1995 Time allowed: three hours
1. Given the functions g : R ! R and h : R ! R, with g (g (x)) = x for every x 2 R, and a real number such that jj 6= 1, prove that there exists exactly one function f : R ! R such that f (x) + f (g(x)) = h(x) for every x 2 R. 2. A solid fence encloses a square eld with sides of length L. A cow is in a meadow surrounding the eld for a large distance on all sides, and is tied to a rope of length R attached to a corner of the fence. What area of the meadow is available for the cow to use? 3. Find all solutions to
(x2 + y )(x + y 2) = (x , y )3
where x and y are integers dierent from zero.
,2n
4. For what positive integers n, is the nth Catalan number, n , odd?
n+1
118 5. N pairs of diametrically opposite points are chosen on a circle of radius 1. Every line segment joining two of the 2N points, whether in the same pair or not, is called a diagonal. Show that the sum of the squares of the lengths of the diagonals depends only on N , and nd that value. 6. A nite pattern of checkers is placed on an in nite checkerboard, at most one checker to a square; this is Generation 0. Generation N is generated from Generation (N , 1), for N = 1, 2, 3; : : : , by the following process: if a cell has an odd number of immediate horizontal of vertical neighbours in Generation (N , 1), it contains a checker in GenerationN ; otherwise it is vacant. Show that there exists an X such that GenerationX consists of at least 1995 copies of the original pattern, each separated from the rest of the pattern by an empty region at least 1995 cells wide. 7. A and B play a game. First A chooses a sequence of three tosses of a coin, and tells it to B ; then B chooses a dierence sequence of three tosses and tells it to B . Then they throw a fair coin repeatedly until one sequence or the other shows up as three consecutive tosses. For instance, A might choose (head, tail, head); then B might choose (tail, head, tail). If the sequence of tosses is (head, tail, tail, head, tail), then B would win. If both players play rationally (make their best possible choice), what is the probability that A wins?
119
BOOK REVIEWS Edited by ANDY LIU
e The Story of a Number by Eli Maor. Published by Princeton University Press, Princeton NJ, 1994, ISBN 0691033900, xiv+223 pages, US$24.95. Reviewed by Richard Guy, University of Calgary. As the author says in his Preface: The story of has been extensively told, no doubt because its history goes back to ancient times, but also because much of it can be grasped without a knowledge of advanced mathematics. Perhaps no book did better than Petr Beckmann's A History of , a model of popular yet clear and precise exposition. The number e fared less well. It continues to fare less well. Many of us are searching for ways to bring mathematics to a much broader audience. This book is a good opportunity missed. There are too many formulas, and they appear too early; even in the Preface there are many. Moreover, they are rarely displayed (only one out of more than a dozen in the Preface), which makes the text hard to read  particularly for the layman. Several chapters are no more than excerpts from standard calculus texts. Chap. 4, Limits. Chap. 10, dy=dx = y , Chap. 12, Catenary. Chaps. 13 & 14, Complex Numbers; even including a treatment of the CauchyRiemann and Laplace equations. This is a history book, but the history is often invented, either consciously or unconsciously (Pythagoras experimenting with strings, bells and glasses of water; Archimedes using parabolic mirrors to set the Roman eet ablaze; Descartes watching a y on the ceiling; `quotations' of Johann and Daniel Bernoulli which are gments of E. T. Bell's fertile imagination), or just plain wrong. Pascal's dates (16231662) are given, and an illustration of \his" triangle from a work published nearly a century before Pascal was born, and another from a Japanese work more than a century after he died. The author, elsewhere, mentions that it appeared in 1544 in Michael Stifel's Arithmetica integra; all of which leaves the reader bewildered. The triangle was known to the Japanese some centuries before, to the Chinese some centuries before that, and to Omar Khayyam before that. After saying that it is unlikely that a member of the Bach family met one of the Bernoullis, the author gives a conversation between J. S. Bach and Johann Bernoulli. This piece of ction distracts the reader from the real topic, which is the mathematics of musical scales. No connection with e is mentioned.
120 One of the various examples that is given of the occurrence of the exponential function, or of its inverse, the natural logarithm, is the WeberFechner law, which purports to measure the human response to physical stimuli. This is probably also regarded as a piece of ction by most modern scientists. It has far less plausibility than the gas laws, say. There are many missed opportunities: more examples of damping, the distribution of temperature and pressure in the earth's atmosphere, alternating currents, radioactive decay and carbon dating; and why not mention the distribution of the prime numbers  they are as relevant and probably of more interest to the personinthestreet than many of the topics covered. Let us look at some of the examples that are given. Newton's law of cooling says that the temperature (dierence) is proportional to its derivative. There is no mention of its eclipse by the work of Dulong & Petit and the StefanBoltzmann law [1], presumably because they have little connexion with e. The parachutist whose air resistance is proportional to his velocity; in practice the resistance is at least quadratic and probably more complicated [2]. Yet another example is the occurrence of the logarithmic spiral in art and in nature, without any hint of the controversial aspects of the subject. There is the oftquoted and illustrated capitulumpof a sun ower; though the current wisdom [3] is that Fermat's spiral, r = a , which does not involve e, provides a more realistic model. Incidentally, this spiral is essentially identical to the orbit of a nonrelativistic charged particle in a cyclotron; there are area, (resp. energy), considerations which make this the `right' spiral for these two applications. Population growth gets only a onesentence mention, whereas there are important mathematical and social lessons to be learnt from this topic. The author is not afraid of quoting whole pages of calculus, so why not the following? There are many good examples of exponential growth, or decay, in everyday life. But none of them is perfect. Look at a perturbation of the equation for the exponential function, a particular case of Bernoulli's equation:
dp = kp , p2 dt which, after multiplying by ekt =p2 , we may write as !
d ekt = ekt dt p (Yes! Integrating factors and even characteristic equations occur on pp. 104{ 105.) Integrate and divide by ekt :
121
1 = Ce,kt +
p
k
As the time tends to in nity, the population tends to k=. If is small, this is large, but at least it is nite, and there is some hope for our planet. What is the p2 term? It is roughly proportional to the number of pairs of people, and the term represents the eect of competition. But what if is negative? What if we substitute cooperation for competition? As we approach the nite time (1=k) ln(Ck=(,)) the population tends to in nity! No wonder that capitalism is more successful than communism. There is a good deal of irrelevant padding, presumably in an attempt to make the book `popular': the cycloid, the lemniscate, Euler's formula for polyhedra, the NewtonLeibniz controversy, and even Fermat's Last Theorem, though Andrew Wiles has a wrong given name in the Index. There are misprints: `Appolonius' (p. 66; he does not appear in the Index), `audibile' (p. 112);  no mathematics book is free of misprints. But there are worse than misprints:  `each additional term brings us closer to the limit (this is not so with a series whose terms have alternating signs)' (p. 36); `can be thought of as the sum of in nitely many small triangles' (p. 54) and Note 1 on p. 93 might lead many readers to infer that a continuous function has a derivative. Let us hope that the book will spark the interest of nonmathematicians, but the fear is that it will con rm their suspicions that we mathematicians just juggle symbols which the rest of the world has little hope of comprehending. REFERENCES 1. Charles H. Draper, Heat and the Principles of Thermodynamics, Blackie & Son, London and Glasgow, 1893, pp. 221{222. 2. Horace Lamb, Dynamics, Cambridge Univ. Press, 1914, p. 290. 3. Helmut Vogel, A better way to construct the sun ower head, Math. Biosciences, 44 (1979), pp. 179{189.
122
PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be typewritten or neatly handwritten, and should be mailed to the EditorinChief, to arrive no later that 1 November 1996. They may also be sent by email to cruxeditor@cms.math.ca. (It would be appreciated if email proposals and solutions were written in LATEX, preferably in LATEX2e). Graphics les should be in epic format, or plain postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication.
2125 Proposed by Bill Sands, University of Calgary, Calgary, Alberta.
At Lake West Collegiate, the lockers are in a long rectangular array, with three rows of N lockers each. The lockers in the top row are numbered 1 to N , the middle row N + 1 to 2N , and the bottom row 2N + 1 to 3N , all from left to right. Ann, Beth, and Carol are three friends whose lockers are located as follows: @ , , @
:::
@ , , @
:::
@ , , @
By the way, the three girls are not only friends, but also nextdoor neighbours, with Ann's, Beth's, and Carol's houses next to each other (in that order) on the same street. So the girls are intrigued when they notice that Beth's house number divides into all three of their locker numbers. What is Beth's house number?
123
2126 Proposed by Bill Sands, University of Calgary, Calgary, Alberta.
At Lake West Collegiate, the lockers are in a long rectangular array, with three rows of N lockers each, where N is some positive integer between 400 and 450. The lockers in the top row were originally numbered 1 to N , the middle row N + 1 to 2N , and the bottom row 2N + 1 to 3N , all from left to right. However, one evening the school administration changed around the locker numbers so that the rst column on the left is now numbered 1 to 3, the next column 4 to 6, and so forth, all from top to bottom. Three friends, whose lockers are located one in each row, come in the next morning to discover that each of them now has the locker number that used to belong to one of the others! What are (were) their locker numbers, assuming that all are threedigit numbers?
2127 Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is an acute triangle with circumcentre O, and D is a point on the minor arc AC of the circumcircle (D 6= A; C ). Let P be a point on the side AB such that \ADP = \OBC , and let Q be a point on the side BC such that \CDQ = \OBA. Prove that \DPQ = \DOC and \DQP = \DOA.
2128 Proposed by Toshio Seimiya, Kawasaki, Japan.
ABCD is a square. Let P and Q be interior points on the sides BC and CD respectively, and let E and F be the intersections of PQ with AB and AD respectively. Prove that \PAQ + \ECF < 54 :
USA.
2129? Proposed by Stanley Rabinowitz, Westford, Massachusetts, p
For n > 1 and i = ,1, prove or disprove that
1 4i
4n X
k=1
gcd (k;n)=1
ik tan k 4n
is an integer.
2130 Proposed by D. J. Smeenk, Zaltbommel, the Netherlands.
A and B are xed points, and ` is a xed line passing through A. C is a variable point on `, staying on one side of A. The incircle of ABC touches BC at D and AC at E. Show that line DE passes through a xed point.
124
2131 Proposed by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore. Find all positive integers n > 1 such that there exists a cyclic permutation of (1; 1; 2; 2; : : : ; n; n) satisfying: (i) no two adjacent terms of the permutation (including the last and rst term) are equal; and (ii) no block of n consecutive terms consists of n distinct integers. 2132 Proposed by Sefket Arslanagic, Berlin, Germany.
Let n be an even number and z a complex number. Prove that the polynomial P (z ) = (z + 1)n , z n , n is not divisible by z 2 + z + n. 2133 Proposed by K. R. S. Sastry, Dodballapur, India. Similar nonsquare rectangles are placed outwardly on the sides of a parallelogram . Prove that the centres of these rectangles also form a nonsquare rectangle if and only if is a nonsquare rhombus. 2134? Proposed by Waldemar Pompe, student, University of Warsaw, Poland. Let fxn g be an increasing sequence of positive integers such that the sequence fxn+1 , xn g is bounded. Prove or disprove that, for each integer m 3, there exist positive integers k1 < k2 < : : : < km , such that xk1 ; xk2 ; : : : ; xkm are in arithmetic progression. 2135 Proposed by Joaqun Gomez Rey, IES Luis Bu~nuel, Alcorcon, Madrid, Spain. Let n be a positive integer. Find the value of the sum bn= X2c k=1
(,1)k(2n , 2k)! : (k + 1)!(n , k)!(n , 2k)!
2136 Proposed by G. P. Henderson, Campbellcroft, Ontario.
Let a; b; c be the lengths of the sides of a triangle. Given the values of p = P a and q = P ab, prove that r = abc can be estimated with an error of at most r=26.
2137 Proposed by Aram A. Yagubyants, Rostov na Donu, Russia.
Three circles of (equal) radius t pass through a point T , and are each inside triangle ABC and tangent to two of its sides. Prove that: (i) t =
2R , R+2
(ii) T lies on the line segment joining the centres of the circumcircle and the incircle of ABC .
125
SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems. 2015. [1995: 53 and 129 (Corrected)] Proposed by ShiChang Chi and Ji Chen, Ningbo University, China. Prove that
sin(A) + sin(B) + sin(C )
p
1 + 1 + 1 27 3 ; A B C 2
where A, B , C are the angles (in radians) of a triangle. Editor's comment on the featured solution by Douglass L. Grant, University College of Cape Breton, Sydney, Nova Scotia, Canada. [1996: 47] There is a slight and very subtle aw in the published solution. To correct this, all that is required is to replace the open domain S by a closed domain. The error is a very natural one, and has been made in the past by many others. We refer readers to Mathematics Magazine, 58 (1985), pp. 146{150, for several examples illustrating this subtle point.
2025. [1995: 90] Proposed by Federico Ardila, student, Massachusetts Institute of Technology, Cambridge, Massachusetts, USA. (a) An equilateral triangle ABC is drawn on a sheet of paper. Prove that you can repeatedly fold the paper along the lines containing the sides of the triangle, so that the entire sheet of paper has been folded into a wad with the original triangle as its boundary. More precisely, let fa be the function from the plane of the sheet of paper to itself de ned by x is on the same side of BC as A is, fa(x) = xthe if,re ection of x about line BC , otherwise, (fa describes the result of folding the paper along the line BC ), and analogously de ne fb and fc . Prove that there is a nite sequence fi1 ; fi2 ; : : : ; fin , with each fij = fa , fb or fc , such that fin (:: : (fi2 (fi1 (x))) : : : ) lies in or on the triangle for every point x on the paper. (b) Is the result true for arbitrary triangles ABC ?
126 Solution to (a) by Catherine Shevlin, Wallsend, England. We shall show that it is possible by reversing the problem: we start with the triangle ABC and unfold three copies of it along the lines containing the edges of the original triangle, to create a larger shape. This process of unfolding shapes is repeated as illustrated in the diagram. We see how the plane is covered by the sequence of shapes, fSk g. The areas of the shapes 3k2 , 3k + 2 ! 1 increase: the area of Sk is 1 + 3 + 6 + : : : + 3(k , 1) = 2 as k ! 1.
::: S1
S2
S3
S4
S5
:::
We now show that the minimum distance from a point on the boundary of the shape to the centre increases without bound. Let O be the centre of the original triangle. Then, proceeding away from triangle ABC in each of the six principal directions is a sequence of triangles, as illustrated below. For convenience, we rename triangle ABC as A1 A2 A3 .
A1
A3
O
A5
q
::: :::
A6 : : : The distance OAk is the minimum distance of the kth repetition from O. This is easy to calculate in terms of the side of the original triangle. It is clear that OAk ! 1 as k ! 1. It is easy to see that the area covered includes a circle of radius OAk . A2
A4
Hence, the area covered by the repetitions tends to in nity, and so it will cover any nite area. Thus the sheet of paper will be covered by the foldings. Part (a) was also solved by the proposer. One incorrect attempt was received. Noone sent in anything on part (b), so this problem remains open.
127
2026. [1995: 90] Proposed by Hiroshi Kotera, Nara City, Japan. One white square is surrounded by four black squares: Two white squares are surrounded by six black squares: Three white squares are surrounded by seven or eight black squares:
What is the largest possible number of white squares surrounded by n black squares? [According to the proposer, this problem was on the entrance examination of the junior high school where he teaches!] Solution by Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USA.
gion.
The largest possible number of white squares is 8 2k2 , 2k + 1 if n = 4k; > > < 2k2 , k if n = 4k + 1; 2 2 k if n = 4k + 2; > > : 2k2 + k if n = 4k + 3: Consider an arrangement of black squares surrounding some white re
Three black squares that are horizontally or vertically adjacent can be changed as follows to increase the number of white squares surrounded by one as shown below.

Suppose there are two pairs of black squares that are horizontally adjacent. Then we can shift all the black squares between these pairs down as shown and still keep [at least] the same number of white squares surrounded, as shown below.

128 [Similarly we can assume there are not two pairs of vertically adjacent black squares.] Thus the maximum number of white squares surrounded can be obtained when there are either one horizontal and one vertical pair of adjacent black squares or when there are no horizontally or vertically adjacent black squares. [For example, it is impossible to have one horizontal pair and no vertical pair of adjacent black squares: just consider the usual chessboard colouring of the squares.  Ed.] Label a set of diagonally adjacent squares as shown below. 1
1 2
1
1 2
1 2
1 2
1
2
1
2
1
2
1
2
1
2
1 2 1 2
1 2
2
2 1
1 2
1 2
2 1
1 2
1 2
2 1
1 2
1 2
2 1
1 2
1 2
2 1
1 2
2
2 1
1
1
2 1
2
2
Then if all the black squares are diagonally adjacent, we see that the squares alternate from type 1 to type 2 and that there must be an even number of them. Thus, if there is an odd number of black squares, we have two black squares that are vertically adjacent and two black squares that are horizontally adjacent. Connect the centres of adjacent squares. Then for a maximal number of white squares surrounded, this polygon must be convex, since if it is not convex, we can increase the number of white squares surrounded, as shown below.

Thus the black squares must form a rectangular formation, as in the second part of the gure above, if the number of black squares is even. If there are 2a + 2b black squares, a + 1 along two opposite sides and b + 1 along the other two sides, the number of white squares enclosed can be found to be ab +(a , 1)(b , 1) = 2ab , a , b +12, so the maximum value is 2k2 , 2k +1 if there are 4k black squares and 2k if there are 4k + 2 black squares, for each k > 0.
129 [Editor's note. With 2a+2b = 4k the number of white squares is 2ab,2k+1, which is maximized when a = b = k; when 2a + 2b = 4k + 2 the number of white squares is 2ab , 2k, which is maximized when fa; bg = fk; k + 1g.] If the number of black squares is odd, there are two squares that are horizontally adjacent and two squares that are vertically adjacent. We can remove one of these squares and shift the others to form a closed gure as shown below.
This second gure must be a rectangle, so we can add on at most a , 1 squares, where a + 1 is the number of squares in one of the largest sides of the rectangle. Therefore if there are 4k + 1 black squares, we can surround 2k2 , k white squares; with 4k + 3 black squares, we can surround 2k2 + k white squares. [Editor's note. Here are some details. Letting the resulting black rectangle have a + 1 squares on one side and b + 1 on the other, where a b, there are 2a + 2b black squares surrounding 2ab , a , b + 1 white squares as before. Thus in the original gure there are 2a + 2b + 1 black squares surrounding 2ab , b white squares. Now if the number of black squares is 4k +1, then a + b = 2k, and 2ab , b is maximized when a = b = k; thus the number of white squares surrounded is 2k2 , k. Similarly, if the number of black squares is 4k + 3, then a + b = 2k + 1, and 2ab , b is maximized when a = k + 1, b = k; and this time the number of white squares surrounded is 2(k + 1)k , k = 2k2 + k.] Also solved by HAYO AHLBURG, Benidorm, Spain; RICHARD I. HESS, Rancho Palos Verdes, California, USA; R. DANIEL HURWITZ, Skidmore College, Saratoga Springs, New York; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KOJI SHODA, Nemuro City, Japan; UNIVERSITY OF ARIZONA PROBLEM SOLVING LAB, Tucson; and the proposer. One incorrect and one incomplete solution were also received. Bosley and Janous actually made similar minor errors at the end of their solutions. Bosley's has been corrected in the above writeup.
2029?. [1995: 91] Proposed by Junhua Huang, the Middle School
Attached To Hunan Normal University, Changsha, China. ABC is a triangle with area F and internal angle bisectors wa, wb, wc. Prove or disprove that
p wbwc + wcwa + wawb 3 3F:
130 Solution by KeeWai Lau, Hong Kong. The inequality is true. Denote, as usual, the semiperimeter, inradius and circumradius by s, r and R respectively. In the following, sums and products are cyclic over A, B, C , and/or over a, b, c, as is appropriate. We need the following identities:
wa F X sin2(A) X sin(A) X
sin(A) sin(B ) = Y sin(A) = X
X
= = = =
cos(A) =
cos(A) cos(B) = Y
Y
cos(A) =
cos A ,2 B
=
2ab b + c cos(A=2); etc., 1 ab sin(C ); etc., 2 s2 , 4Rr , r2 ; 2 s ; 2R R2 s + 4Rr + r2 ; 4R2 sr ; 2R2 R + r; R r2 + s2 , 4R2 ; 4R2 2 s , 4R2 , 4Rr , r2 ; 4R2 2 s + 2Rr + r2 : 8R2
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
We also need the inequality:
16Rr , 5r2 s2 4R2 + 4Rr + 3r2:
(11)
Identities (1) and (2) are wellknown. Identities (3) through (9) can be found on pages 55{56 Identity (10) can be obtained from (5) and (8) by of [1]. P P A , B 1 rewriting cos 2 as 4 (1 + sin(A) sin(B ) + cos(A) cos(B )). Inequality (11) is due to J.C. Gerretsen, and can be found on page 45 of [1].
131 From (1) and (2), we have
wawb = 8 X c2 cos(A=2) cos(B=2) F (c + a)(c + b) sin(C ) X 2 sin(C ) = cos((B , C )=2)cos((A , C )=2) : Thus, it is sucient to prove that p3 Y X 3 sin(A) cos((B , C )=2) 2 cos((B , C )=2): P
(12)
By squaring both sides, we see that (12) is equivalent to
4
X
sin2 (A) cos2((B , C )=2) X +8 sin(A) sin(B ) cos((B , C )=2)cos((A , C )=2) Y 27 cos2((B , C )=2):
To prove (13), it suces to show that
4
Since
4
(13)
X
sin2(A) cos2((B , C )=2) Y X +8 cos((B , C )=2) sin(A) sin(B ) Y ,27 cos2((B , C )=2) 0:
(14)
X
sin2(A) cos2((B , C=2)) X X = 2 sin2(A) + cos(B ) cos(C ) Y X Y X , cos(A) cos(A) + sin(A) sin(A) ;
we use (3) { (10), to obtain that (14) is equivalent to
11s4 + (22r2 , 20rR , 64R2)s2 , 148r2 R2 , 20r3 R + 11r4 0:
(15) Let x = s and denote the left side of (15) by f (x). Then f (x) is a convex function of x. In view of (11), in order to prove (15), it is sucient to show both f (16Rr , 5r2) 0; (16) 2 2 f (4R + 4Rr + 3r ) 0: (17) Now 2
and
f (16Rr , 5r2) = ,4r(R , 2r)(256R2 , 155rR + 22r2);
f (4R2 + 4Rr + 3r2) = ,4(R , 2r)(20R3 + 36rR2 + 45r2R + 22r3): Since R 2r, both (16) and (17) hold, and the desired inequality is proved.
132
Reference [1.] D.S. Mitrinovic, J.E. Pecaric and V. Volenec, Recent Advances in Geometric Inequalities, Kluwer Academic Publishers, 1989. No other solutions were received. etokrzyski, Poland. 2030. Proposed by Jan Ciach, Ostrowiec Swi
For which complex numbers s does the polynomial z 3 , sz 2 + sz , 1 possess exactly three distinct zeros having modulus 1? I. Combination of solutions by F.J. Flanigan, San Jose State University, and the late John B. Wilker, University of Toronto. Denote by S the set of such complex numbers s. We oer two parametrizations of S (neither injective), the second of these yielding a description of S as the interior of a certain curvilinear triangle (hypocycloid or deltoid) inscribed in the disc jz j 3. Since the three zeros have modulus 1 and product equal to unity, they may be written as ei ; ei ; e,i(+ ) with 0 ; < 2 . It follows that
s = ei + ei + e,i(+ ):
(1)
Moreover, the three zeros will be distinct if and only if 6= and neither 2 + nor 2 + is an integer multiple of 2. (We note in passing that the fact the coecient of z in the cubic is s does not impose a further restriction.) From (1) we see that jsj 3 with equality if and only if = = ,( + ) mod 2. We improve (1) by noting that ei + ei = ei(+ )=2 (ei(, )=2 + , i ( ,
)=2 ) = 2 cos ei where = ( + )=2 and = ( , )=2. Thus e we have
s = 2 cos ei + e,2i:
(2)
Equation (2) enables us to visualize the parameter set S as follows. Fix and let cos varyi through its full range: ,1 cos 1. Then the complex numbers 2 cos e lie on a line segment of length 4 centred at the origin in the direction of the vector ei . Thus the points s for this xed lie on a line segment of length 4 joining P = e,2i + 2ei to P 0 = e,2i , 2ei. This segment is centred at the point e,2i and makes an angle with the xaxis. The set S is the union of all these (overlapping) segments PP 0. As varies, the path of either endpoint of the segment is the curvilinear triangle with vertices at 3; 3ei=3, and 3e,i=3 , namely,
133
z(t) = 2eit + e,2it (where P = z () and P 0 = z ( + )). One veri es easily that this curve is a hypocycloid with three cusps: it is the locus of a point P xed to the circumference of a circle (whose diameter is PP 0 ) that is rolling clockwise around the inside of the circle jz j = 3 as runs from 0 to 2 . This curve is called a deltoid because its shape resembles the Greek letter ([2], pp 73{ 79). De ne P 00 = 2 cos(3)ei + e,2i . Note that P 00 lies on PP 0 , and since
P 00 = (e3i + e,3i)ei + e,2i = e4i + 2e,2i = z(,2); it lies on the deltoid as well. Moreover, P 00 is the only point of the deltoid in the interior of the segment PP 0 (as is easily veri ed, or see [2], p. 75), so that PP 0 is tangent to the deltoid at P 00 . It follows that all points inside the deltoid are in S (which is the union of these diameters). Note further that P (the point where = 0) corresponds to = , P 0 (where = ) to = + 2, and P 00 (where = 3) to = ,2 ; consequently, thei values of s on the deltoid correspond to multiple zeros of the given cubic (e = ei
or ei = e,i(+ ) ). Since the proposal excludes multiple zeros we conclude that S is precisely the interior of the region bounded by the deltoid. Editor's comment (by Chris Fisher). It is certainly clear (from (2)) that all points between P and P 0 on the segment PP 0 lie in S . Many of those whose solutions described S geometrically concluded that the interior of the segment joining the points P and P 0 automatically lies inside the deltoid, since P and P 0 lie on the boundary. However, further argument seems to be required: since S is not a convex region we have no guarantee against an interval of points on PP 0 that (like P 00 ) lie outside the interior of the deltoid. II. Solution by Kurt Fink and Jawad Sadek, Northwest Missouri State University. Let P (z ) = z 3 , sz 2 + sz , 1 and let P 0 (z ) = 3z 2 , 2sz + s be its derivative. A result of A. Cohn, specialized to P (z ) (see [3], p. 206, Exercise 3), states that the zeros of P (z ) lie on the unit circle and are simple if and only if the zeros of P 0 (z ) lie in jz j < 1 or, equivalently, the zeros of the polynomial sz 2 , 2sz + 3 lie in jz j > 1. An application of Theorem 6.8b on p. 493 of [1], shows that (we omit the elementary calculations) this occurs if and only if jsj < 3 and jsj4 + 18jsj2 , 8<(s3 ) , 27 < 0. If s = x + iy , the condition becomes
134
(x2 + y2)2 + 18(x2 + y 2) , 8(x3 , 3xy 2) , 27 < 0:
This is the interior of a deltoid whose closure is contained in the disk of radius 3 centredpat the origin; it touches p the boundary circle at the points (3; 0); (,3=2; 3 3=2), and (,3=2; ,3 3=2). The closed deltoid region contains p the unit circle andp boundaries of these curves meet at (,1; 0), (1=2; 3=2) and (1=2; , 3=2). References 1. Peter Henrici, Applied and Computational Complex Analysis, Vol. 1, Wiley, New York, 1974. 2. E.H. Lockwood, A Book of Curves, Cambridge Univ. Press, Cambridge, 1963. 3. Morris Marden, Geometry of Polynomials, Amer. Math. Society, Princeton, 1985. Also solved by ED BARBEAU, University of Toronto, Toronto, Ontario; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; KEITH EKBLAW, Walla Walla, Washington, USA; JEFFREY K. FLOYD, Newnan, Georgia, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; P. PENNING, Delft, the Nether lands; HEINZJURGEN SEIFFERT, Berlin, Germany; and the proposer. Janous showed that if the polynomial
p(z) = zn , szn,1 + + (,1)n n X
eik , where 1; : : : ; n 2 [0; 2) and k=1 1 + + n = 0 (mod 2 ). Furthermore, if ak is the coecient of z k then, necessarily, (,1)n ak is the coecient of z n,k . The proposer outlined has n zeros of modulus 1, then s =
an argument that if the polynomial
p(z) = zn , szn,1 + sz , 1 has n zeros of modulus 1, then s belongs to the interior of the region bounded by the hypocycloid
, 1 eit + 1 e,i(n,1)t: z(t) = nn , 2 n,2
135
2031. [1995: 129] Proposed by Toshio Seimiya, Kawasaki, Japan. Suppose that , , are acute angles such that
sin( , ) + sin( , ) + sin( , ) = 0: sin( + ) sin( + ) sin( + )
Prove that at least two of , , are equal. All solvers had the same idea, so we present a composite solution. By dividing the rst term, top and bottom, by cos cos , and the other two terms similarly, the given condition is equivalent to
tan , tan + tan , tan + tan , tan = 0: tan + tan tan + tan tan + tan
By multiplying by the common denominator, this reduces to
(tan , tan )(tan , tan )(tan , tan ) = 0: Hence, at least two of tan , tan , tan are equal, and since , , are
acute angles, at least two of them must be equal. Berlin, Germany; CARL BOSLEY, stuSolved by SEFKET ARSLANAGIC, dent, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J. OCHOA, BRADLEY, Clifton College, Bristol, UK; MIGUEL ANGEL CABEZON Logro~no, Spain; SABIN CAUTIS, student, Earl Haig Secondary School, North York, Ontario; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; THEODORE CHRONIS, student, Aristotle University of Thessalonika, Thessalonika, Greece; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEEWAI LAU, Hong Kong; VICTOR OXMAN, Haifa University, Haifa, Israel; HEINZJURGEN SEIFFERT, Berlin, Germany; D. J. SMEENK, Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; HOE TECK WEE, student, Hwa Chong Junior College, Singapore; and the proposer.
2032. [1995: 129] Proposed by Tim Cross, Wolverley High School, Kidderminster, UK. Prove that, for nonnegative real numbers x, y and z , p
p
p
p
x2 + 1 + y2 + 1 + z2 + 1 6(x + y + z) :
When does equality hold? This problem attracted 25 correct solutions. Six solvers (marked y in the list of solvers) gave the generalization highlighted below. So we present a composite solution based on the submissions of several solvers.
136 We prove the generalization: for nonnegative real numbers fxk g, v u u t
n q X k=1
x2k + 1 2n
n X k=1
xk :
From (xk xj , 1)2 0, we get
x2kx2j + x2k + x2j + 1 x2k + 2xkxj + x2j : Hence
(x2k + 1)(x2j + 1) (xk + xj )2:
Taking the square root, and since the xk are nonnegative, we have q
q
2 (x2k + 1) (x2j + 1) 2(xk + xj ): Hence
n X n q X
q
(x2k + 1) (x2j + 1) 2(n , 1)
k=1 j =1
n X k=1
xk:
We now add the nonnegative quantities (xk , 1)2, to get n X k=1
Thus
(xk , 1)2 + n X k=1
n X n q X k=1 j =1
x2k + n
q
(x2k + 1) (x2j + 1) 2(n , 1)
n X n q X k=1 j =1
n X
q
(x2k + 1) (x2j + 1) 2n
k=1
But the left side is the square of n q X k=1
(x2k + 1);
and so the result follows. Equality holds when xk = 1 for all k such that 1 k n.
n X k=1
xk:
xk:
137 Berlin, Germany; yNIELS Solved by ySEFKET ARSLANAGIC, BEJLEGAARD, Stavanger, Norway; CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton OCHOA, Logro~no, Spain; College, Bristol, UK; yMIGUEL ANGEL CABEZON SABIN CAUTIS, student, Earl Haig Secondary School, North York Ontario; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; yTHEODORE CHRONIS, student, Aristotle University of Thessalonika, Greece; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; TOBY GEE, student, the John of Gaunt School, Trowbridge, England; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; PETER HURTHIG, Columbia College, Burnaby, BC; yWALTHER JANOUS, Ursul Y, Ferris State Uniinengymnasium, Innsbruck, Austria; yVACLAV KONECN versity, Big Rapids, Michigan, USA; SAI C. KWOK, Boulder, Colorado, USA; KEEWAI LAU, Hong Kong; VICTOR OXMAN, Haifa, Israel; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; SCIENCE ACADEMY PROBLEM SOLVERS, Austin, Texas, USA; yHEINZJURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary, Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; EDWARD T. H. WANG, Wilfrid Laurier University, Waterloo, Ontario; HOE TECK WEE, student, Hwa Chong Junior College, Singapore; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer. Four incorrect solutions were received. They all used calculus. Those solvers are referred to the comment on the solution to problem 2015, earlier in this issue.
2033. [1995: 129] Proposed by K. R. S. Sastry, Dodballapur, India.
The sides AB , BC , CD, DA of a convex quadrilateral ABCD are extended in that order to the points P , Q, R, S such that BP = CQ = DR = AS. If PQRS is a square, prove that ABCD is also a square. Solution by Cyrus Hsia, student, University of Toronto, Toronto, Ontario. Suppose that ABCD is not a rectangle. Then it must contain an interior angle greater than 2 . Without loss of generality, let \A > 2 . Since PQRS is a square, PQ = QR = RS = SP . Rotate QP about P clockwise until Q coincides with S. Let B0 be the new position of B. Now \B 0 PA = \B 0 PS + \SPA = \BPQ + \SPA (from rotation) = \SPQ = :
2
(Editor's note: Since PQRS is a square, this is just a rotation through
radians and the fact that \B 0 PA = follows directly.) 2
3 2
138 Since \A > 2 ; \SAP < 2 which implies that PB 0 = AS > h, where h is an altitude from S to AP . Then > \SB0P = \QBP; 2 so \ABC > 2 . Similar arguments show that \BCD > 2 and \CDA > 2 . Therefore, the sum of the interior angles of ABCD > 2 . Impossible! Therefore, ABCD is a rectangle. Since \SAP = \PBQ = \QCR = \RDS = 2 , the triangles ASP; BPQ; CQR; DRS are congruent and AB = BC = CD = DA. Therefore, ABCD is a square. Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; TOSHIO SEIMIYA, Kawasaki, Japan; D. J. SMEENK, Zaltbommel, the Netherlands; and the proposer. There was one incorrect solution. The proposer notes that his starting point was the analogous theorem for triangles. Bradley notes that the problem was set in the Second Selection examination for the 36th IMO in Bucharest in April 1995 and is attributed to L. PANAITOPOL. He also notes that two solutions are given, one trigonometrical, which is correct, the other, a pure solution, which is in error. Finally, both Bradley and Smeenk note that a simple modi cation of the proof works in the case of convex polygons of any number of sides.
2034. [1995: 130, 157] Proposed by Murray S. Klamkin and M. V. Subbarao, University of Alberta. (a) Find all sequences p1 < p2 < < pn of distinct prime numbers such that is an integer. (b) Can
1 + p1
1
1 + p1 1 + p1 2 n
1 1 1 1 + a2 1 + a2 1 + a2 1
2
n
be an integer, where a1 ; a2 ; : : : are distinct integers greater than 1? Solution to (a), by HeinzJurgen Seiert, Berlin, Germany. If the considered product is an integer, then pn j(pi + 1) for some i 2 f1; 2; : : : ; n , 1g. Since pi + 1 pn, it then follows that pn = pi + 1, which implies pi = 2 and n = 2, p1 = 2, p2 = 3. This is , pn =, 3. Thus, indeed a solution since 1 + 21 1 + 13 = 2. Solution to (b). In the following, we present four dierent solutions submitted by eight solvers and the proposers. In all of them, p denotes the given product, and it is shown that 1 < p < 2 and thus, p cannot be an integer. Clearly one may assume, without loss of generality, that 1 < a1 < a2 < : : : < an.
139 Solution I, by Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USA; KeeWai Lau, Hong Kong; and Kathleen E. Lewis, SUNY, Oswego, New York. Since 1 + x < ex for x > 0, we have 1 < p < exp a12 + a12 + : : : + a12 1 2 n 1 1 < exp 22 + 32 + : : :
2 = exp 6 , 1 1:90586 < 2:
Solution II, by Toby Gee, student, the John of Gaunt School, Trowbridge, England. Since a1 2, we have nY +1
nY +1 2 +1 1 k + 1 < nY k2 1 < p 1 + k2 = 2 k2 k=2 ! k=2 k=2 k , 1 ! nY +1 nY +1 , k k = 2(n + 1) < 2: = n+2 k=2 k + 1 k=2 k , 1
Solution III, by Walther Janous, Ursulinengymnasium, Innsbruck, Austria; Vaclav Konecny, Ferris State University, Big Rapids, Michigan, USA; HeinzJurgen Seiert, Berlin, Germany; and the proposers. 1 1 Y Y 1 1 1 1<p< 1+ 2 = 1 + 2 = sinh 1:838 < 2. t=2
t
t
2 t=1
2
Solution IV, Richard I. Hess, Rancho Palos Verdes, California, USA. 1 Y 1 Let p1 = 1 + 2 . Then clearly 1 < p < p1 = Q R where k=2
h
1 Y 1 1 Q= 1 + h2 1:6714 and R = 1 + k2 . k=2 k =11 Z 1 Z 1 1 X dx = 1 . Now, ln R = ln 1 + k12 < ln 1 + x12 dx < 2 10 10 10 x k=11 1 (Ed: This is because f (x) = ln 1 + is strictly decreasing on (0; 1) 10 Y
x2
and ln(1 + t) < t for t > 0). Therefore, R < e0:1 1:1052, and so p1 < (1:68)(1:11) = 1:8648 < 2. Part (a) was also solved by SABIN CAUTIS, student, Earl Haig Secondary School, North York, Ontario; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; and ASHISH KR. SINGH, student, Kanpur, India.
140 rael.
2036. [1995: 130] Proposed by Victor Oxman, Haifa University, Is
You are given a circle cut out of paper, and a pair of scissors. Show how, by cutting only along folds, to cut from the circle a gure which has area between 27% and 28% of the area of the circle. Solution. The response to this problem was 10 dierent solutions in the range [27:2535 : : : ; 27:8834 : : : ]. Except for the proposer's solution, all involved straight line cuts. The proposer's solution involves cutting along four curved lines, outlined by parts of the circle that have been folded inwards. The minimum number of (single) folds was four { Hurthig wondered if a prize was available for the least! Rather than give the details of any one, we shall present ten diagrams that illustrate the ingenuity of our readership, and leave you, the reader, to do the exact calculation for each. @ @ @ @ @ @ @ @
Gee, Godwin, Hsai In the above diagram, each line bisects an angle formed previously, until an angle of 35=64 is generated. ,@ , @ , ,, @ , ,@ , ,, , @ ,@ , , , , ,, , @, @ , , , @, , , @, , @ , @, Bosley
Godwin
Grant, Hurthig
@ , @ , @ , @, ,@ , @ , @ , @ Hess
Hurthig
Jackson
141
@ , @ , @ , @, ,@ , @ , @ , @ Perz
, , , , , , @ , @ , @ , @ , @ , @,
,@ , @ , @ , @ , @ , @ @ , @ , @ , @ , @ , @,
Tsaoussoglou
The proposer
Solved by CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; TOBY GEE; student, the John of Gaunt School, Trowbridge, England; DOUGLASS GRANT, University College of Cape Breton; Sydney, Nova Scotia; RICHARD I. HESS, Rancho Palos Verdes, California, USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; PETER HURTHIG, Columbia College, Burnaby, BC; DOUGLAS E. JACKSON, Eastern New Mexico University, Portales, New Mexico, USA; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer.
2037. [1995: 130] Proposed by Paul Yiu, Florida Atlantic University, Boca Raton, Florida, USA. The lengths of the base and a slant side of an isosceles triangle are integers without common divisors. If the lengths of the angle bisectors of the triangle are all rational numbers, show that the length of the slant side is an odd perfect square. Solution by David Doster, Choate Rosemary Hall, Wallingford, Connecticut, USA. Let ABC be isosceles, with b = c, and let the angle bisectors be AD, BE and CF . It is given that a and c are relatively prime integers. Using Stewart's Theorem, for example, one can show that the length of the angle bisector to AB (and hence to AC ) is p
p ac ( a + b + c )( a + b , c ) a c(2c + a) : CF = = a+b a+c
[This formula can also be derived from the formula for the angle bisector given in CRUX on [1995: 321].  Ed.] Also [by the Pythagorean Theorem]
p 2 2 AD = 4c 2, a :
Since CF and AD are both rational, we must have, for some positive integers u and v,
142
4c2 , a2 = v 2: Since gcd(a; c) = 1, it follows that gcd(a; 2c + a) = 1. Hence both c and 2c + a are perfect squares. Now, if c is even, then a, being relatively prime to c, is odd. From the equation 4c2 , a2 = v 2 we see that v too is odd. Thus a2 + v2 2 mod 4. But 4c2 0 mod 4, a contradiction. Thus c is an odd
c(2c + a) = u2
and
perfect square. Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUEL OCHOA, Logro~no, Spain; ADRIAN CHAN, student, Upper ANGEL CABEZON Canada College, Toronto, Ontario; RICHARD K. GUY, University of Calgary, Calgary, Alberta; RICHARD I. HESS, Rancho Palos Verdes, California, USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEEWAI LAU, Hong Kong; HEINZJURGEN SEIFFERT, Berlin, Germany; and the proposer. Four slightly incorrect solutions were also received, most of these assuming, without proof, that the altitude from the apex of the triangle is an integer. Amengual notes that the perimeter of the triangle is also a perfect square. This is contained in the above proof, since 2c + a is the perimeter. The proposer gave several examples of triangles satisfying the given property. The smallest has base 14 and slant sides 25.
2038. [1995: 130] Proposed by Neven Juric, Zagreb, Croatia.
Show that, for any positive integers m and n, there is a positive integer
k so that
pm + pm , 1n = pk + pk , 1:
Solution by Francisco Bellot Rosado, I.B. Emilio Ferrari, Valladolid, Spain and Maria Ascension Lopez Chamorro, I.B. Leopoldo Cano, Valladolid, Spain. We prove a more general result: Let a b be nonnegative real numbers, and let n 2 N. Let Then giving
2 n p = (a + b) ,2 (a , b) :
p2 + ,a2 , b2n = 14 ,(a + b)n + (a , b)n2 ;
143 q
p2 + (a2 , b2)n = 21 ,(a + b)n + (a , b)n = (a + b)n , p:
This given
q
p
p2 + (a2 , b2)n + p2 = (a + b)n: (1) p p In (1), set a = s, b = s , k, where s; k 2 N; s k, so that p2 is a
positive integer, and we have
ps + ps , kn = pp2 + pp2 , k2:
(2)
ps + ps , 1n = pp2 + pp2 , 1:
(3)
Letting k = 1 in (2), gives the result of the original problem here:
If we let s = 2 in (3), we get
p2 + 1n = pp2 + pp2 , 1;
which was a problem proposed by Hong Kong (but not used) at the 1989 IMO at Braunschweig, Germany. This version also appears as Elementary Problem E 950 (proposed by W.R. Ransom, Tufts College) in the American Mathematical Monthly 58 (1951) p. 566, on which we have based our solution. Berlin, Germany; CARL BOSLEY, Also solved by SEFKET ARSLANAGIC, student, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER OCHOA, J. BRADLEY, Clifton College, Bristol, UK; MIGUEL ANGEL CABEZON Logro~no, Spain; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; TOBY GEE, the John of Gaunt School, Trowbridge, England; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOHN G. HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; PETER HURTHIG, Columbia College, Burnaby, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Y, Ferris State University, Big Rapids, Michigan, Austria; VACLAV KONECN USA; VICTOR OXMAN, Haifa, Israel; P. PENNING, Delft, the Netherlands; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; CRISTOBAL SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon, Spain; SCIENCE ACADEMY PROBLEM SOLVERS; Austin, Texas, USA; HEINZJURGEN SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawasaki, Japan; PANOS E. TSAOUSSOGLOU, Athens, Greece; HOE TECK WEE, student, Hwa Chong Junior College, Singapore; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer. One incorrect submission was received.
144 Chan states that the problem is in the 1980 Romanian Mathematical Olympiad. Perz observes that the special case m = 10 occurs in CRUX [1977: 190], and, along with Hsia, notes that the case m = 2 was in the 1994 Canadian Mathematical Olympiad (see CRUX [1994: 151], [1994: 186])
2039?. [1995: 130] Proposed by Dong Zhou, Fudan University, Shang
hai, China, and Ji Chen, Ningbo University, China. Prove or disprove that
sin A + sin B + sin C 9p3 ; B C A 2
where A, B , C are the angles (in radians) of a triangle. [Compare with CRUX 1216 [1988: 120] and this issue!] Solution by Douglass L. Grant, University College of Cape Breton, Sydney, Nova Scotia. In problem 2015 [1995: 53 and 129 (Corrected), 1996: 47], it was shown that
p3 1 1 1 27 K = (sin A + sin B + sin C ) A + B + C 2 ;
where A, B , C are the angle of a triangle measured in radians, and that the lower bound is attained in the equilateral triangle. Let
F = sinAA + sinBB + sinCC ; G = sinBA + sinCB + sinAC ; H = sinCA + sinAB + sinBC : Then K =pF + G + H . By problem 1216 [1987: 53, 1988: 120], we have p 2 < F 923 , so that we have G + H 182 3 . But G can be obtained from H by a permutation of symbols. Thus each of G p and H has the same minimum, which will be at least 923 . But, in the equilateral triangle, G (and H ) achieves this value, so the required inequality is true and sharp. No other submission was received.
145
On the Sum of n Dice J.B. Klerlein
Department of Mathematics, Western Carolina University, Cullowhee, North Carolina, USA.
Introduction An interesting problem can be found in Tucker [2, p. 421] which we rephrase as follows. Suppose n distinct fair dice are rolled and S is the sum of their faces. Show that the probability that 2 divides S is 12 . There are several ways to solve this problem. One, which we will use, lends itself to generalization. Die Two 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 Die 3 4 5 6 7 8 9 One 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 Table 1. Possible Sums of Two Dice The result is clear for one die, and easily veri ed for two dice. See Table 1. The thirtysix entries in Table 1 each have the same probability. Since there are eighteen even sums among the thirtysix entries, the result follows for n = 2. A subproblem is, how to represent the sum of three, or more generally n, dice while maintaining the dierent frequencies for dierent sums. That is, the sum 3 is only obtained in one way using three dice, while the sum 15 can be obtained in ten ways with three dice. Fortunately, these frequencies are unnecessary for this solution. In Table 2, we label the rows by the possible sums of (n , 1) dice and the six columns by the possible values of the nth die. We note that, unlike the entries in Table 1, the entries in Table 2 are not equiprobable. However, the table does allow us to count the number of even sums among the 6n possible outcomes from rolling n dice. For suppose there are fn+2 ways to obtain the sum (n + 2) when (n , 1) dice are rolled. Then by looking at the row labelled by (n + 2), we can account for 3fn+2 even sums among the 6n possible outcomes for n dice, since there are three even entries in this row. Since each row of the table consists of six consecutive integers, there
146 Die n Sum of
n , 1)
(
dice
n,1 n n+1 n+2
n
n+2 n+3 n+4 n+5
4
n+3 n+4 n+5 n+6
5
n+1 n+2 n+3
n+1 n+2 n+3 n+4
n+4 n+5 n+6 n+7
n+5 n+6 n+7 n+8
. . .
. . .
. . .
. . .
. . .
. . .
. . .
n,7 n,6
6 6
1
n,6 n,5
6 6
2
3
n,5 n,4
6 6
n,4 n,3
6 6
6 6
n,3 n,2
Table 2. Possible Sums of n Dice
n,2 n,1
6 6
6
6
n,1 6n
are exactly three even entries in each row. Thus, we can sum the frequencies (fn,1 + fn + fn+2 + : : : + f6n,6 ) and multiply this sum by 3 to obtain the number of even sums among the 6n possible outcomes. As the sum of the frequencies is 6n,1 , we have the probability that 2 divides S is
3 6n,1 = 1 : 6n 2
As mentioned above, this method generalizes easily. Let us rst state two problems. Suppose n dice are rolled, and let Sn be the sum of their faces. Let P (kjSn) be the probability that k divides Sn . 1. Find values of k such that P (kjSn) = k1 for all n. 2. Find values of k and n for which P (kjSn) = 1 . k
Problem 1. Find values of k such that P (kjSn) = 1k for all n. As mentioned, each row in Table 2 consists of six consecutive integers. Thus, in each row, two entries are divisible by 3, and one entry is divisible by 6. It follows, in exactly the same manner as above, that P (3jSn) = 31 and P (6jSn) = 16 . Clearly, P (1jSn) = 1. Combining these observations with our rst result, we have that, for all n; P (kjSn) = k1 when k = 1; 2; 3; and 6. By considering n = 1 (or n = 3, if we wish to avoid the trivial case), we see that the only values of k that hold for all n are precisely k = 1; 2; 3; and 6. Problem 2. Find values of k and n for which P (kjSn) = k1 . Let n and k satisfy P (kjSn) = 1k . Since 6nP (kjSn ) is an integer, it follows that k divides 6n , that is, k is of the form 2s 3t for some nonnegative integers s and t. For 0 s; t 1 we have k = 1; 2; 3; or 6. These values of k, as we have seen, are precisely the solutions to Problem 1. Let us consider k = 4.
147 We shall show that P (4jSn) = 14 if and only if n 2 (mod 4). We note that in Table 2 each row does not contain the same number of multiples of 4. For this reason, we are led to consider another approach, in particular, the congruence classes depicted in Table 3. 1 2 3 4 5 6 0 1 2 3 0 1 2 1 2 3 0 1 2 3 2 3 0 1 2 3 0 3 0 1 2 3 0 1 Table 3. Congruence Classes Modulo 4 for n Dice We construct Table 3 in a similar fashion to Table 2, except that the congruence classes 0; 1; 2; and 3 (modulo 4) label the rows. The columns are still labelled by the face values of the nth die. The entry in the ith row and the j th column is the congruence class of the sum of an element from the congruence class of the ith row and the face value of the j th column. Let P (n; i) be the probability that the sum of n dice is congruent to i (mod 4) for i = 0; 1; 2; 3: Note that P (4jSn) is precisely P (n; 0). The following recurrence relations are a consequence of Table 3.
6P (n; 0) = 2P (n , 1; 3) + 2P (n , 1; 2) +P (n , 1; 1) + P (n , 1; 0) = P (n , 1; 3) + P (n , 1; 2) + 1 6P (n; 1) = 2P (n , 1; 3) + 2P (n , 1; 0) +P (n , 1; 1) + P (n , 1; 2) = P (n , 1; 3) + P (n , 1; 0) + 1 6P (n; 2) = 2P (n , 1; 0) + 2P (n , 1; 1) +P (n , 1; 2) + P (n , 1; 3) = P (n , 1; 0) + P (n , 1; 1) + 1 6P (n; 3) = 2P (n , 1; 1) + 2P (n , 1; 2) +P (n , 1; 3) + P (n , 1; 0) = P (n , 1; 1) + P (n , 1; 2) + 1 From these recurrence relations, we observe that P (n; 0) + P (n; 2) = 12 and P (n; 1) + P (n; 3) = 21 : 323 + 4 P (n , 4; 2). With a little iteration, we obtain P (n; 0) = 1296 1296
148 , Replacing the last term by 12 , P (n , 4; 0) we have, 1 , 4 P (n , 4; 0) : P (n; 0) = 14 + 1296 1296 It follows that P (4jSn) = 14 if and only if P (4jSn,4) = 41 . Now P (4jS2) = 14 , but P (4jSm) 6= 14 if m = 1; 3; or 4. Thus, P (4jSn ) = 14 if and only if n 2 (mod 4).
Conclusion
What about other possible k's (k = 2s 3t ) for which it may be that P (kjSn) = k1 for some n? To be honest, we don't know. We have checked those k's which are less than 54 for n 55. For k = 8 and 9 respectively, the only values of n 55 for which P (kjSn) = k1 are n = 3 and 2 respectively. For n 55; P (kjSn) = k1 never holds for 12 < k < 54. The last value for which we could report is k = 12. But rather than spoil the fun of the interested reader, we say no more.
References 1. Ne, J.D. Dice tossing and Pascal's triangle. TwoYear College Mathematics Journal, 13 (1982), pp. 311314. 2. Tucker, A. Applied Combinatorics, 2nd Ed., John Wiley and Sons, 1984.
Here are three curves that divide a circle into four equal areas.
Any other nice examples?
149
THE SKOLIAD CORNER No. 14
R.E. Woodrow First this month, an observation by an astute reader, Derek Kisman, student, Queen Elizabeth High School, Calgary, who noticed that the gure we gave for question 25 of the Sharp U.K. Intermediate Mathematical Challenge, 1995, [1996: 14] is missing the crucial upright F in the net. It should be in the top right corner! As a contest this month we give the European \Kangaroo" Mathematical Challenge, written Thursday 23 March, 1995. It is organized by the U.K. Mathematics Foundation and L'Association Europeennes Kangarou des Mathematiques. The contest is for students at about school year nine or below. This was written by about 5000 students in the U.K. My thanks go to Tony Gardiner of the U.K. Mathematics Foundation for sending the materials to Crux.
EUROPEAN \KANGAROO" MATHEMATICAL CHALLENGE 23 March 1995
A. 0
1. 1 9 9 5 , (1 + 9 + 9 + 5) makes:
'$ &%
B. 381 C. 481 D. 429 2. Which shape does not appear in this gure? A. circle @@,, B. square C. rightangled triangle ,@@ D. isosceles triangle , E. equilateral triangle
E. 995
3. The whole numbers from 1 to 1995 are alternately added and subtracted thus: 1 , 2 + 3 , 4 + 5 , 6 + + 1993 , 1994 + 1995. What is the result? A. 997 B. 1995 C. 998 D. 0 E. ,997 4. What is the angle between the hour hand and the minute hand of a clock at 1.30? A. 180 B. 120 C. 130 D. 150 E. 135 5. C1 is a circle of radius 6cm, C2 is a circle of radius 8cm. Jose wants the two circles to touch tangentially. He knows that there are two possibilities for the distance between their centres. What are these two distances? A. 3 and 4cm B. 2 and 8cm C. 2 and 14cm D. 6 and 8cm E. 6 and 14cm
150
6. A train 1km long is restricted to travel through a tunnel of length 1km at 1km/h. How long does it take the train to pass through the tunnel? A. 1 hour B. 1h 30m C. 2 hours D. 3 hours E. 1=2 hour 7. The angle x in the gure is equal to: A. 20 B. 22 C. 24 D. 26 E. 28
x
8. Which number is smallest?
A. 1995
B. 19 95
22
C. 1995
22 2x
64
D. 1995
E. 1995
9. What is the ratio of the perimeter of the shaded region to the circumference of the circle? 2 B. 4+ C. 4+ D. 4+ E. 41 A. 34 4 2 10. After two successive 20% reductions, a coat costs $320. What was its original price? A. $204 B. $400 C. $448 D. $500 E. $533 11. Nine people are sitting in a room; their average age is 25. In another room eleven people are gathered their average age is 45. If the two groups were to combine, what would their average age be? A. 70 B. 36 C. 35 D. 32 E. 20 12. How many squares are there in this gure?
A. 13 B. 14 C. 19 D. 21 E. 23
151
13. The quadrilateral PQRS is a rectangle; M is any point on the diagonal SQ. What can one say for sure about the two shaded regions? A. the upper area is larger P B. the lower area is larger Q C. they always have equal areas M D. the two areas are equal only if M is the midpoint of SQ E. insucient information S R
14. A metallic disc of diameter 20cm weighs 2:4kg. From it one cuts out a disc of diameter 10cm. What does it weigh? A. 1:2kg B. 0:8kg C. 0:6kg D. 0:5kg E. 0:4kg 15. What is the area (in unit squares) of the quadrilateral MNPQ? P , N, M HHH ,, H,Q
A. 9 B. 10 C. 11 D. 12 E. 13
16. Six hundred and twenty ve students enter a 100m competition. The track has ve lanes. At the end of each race the winner survives, while the other four are eliminated. How many races are needed to determine the champion sprinter? A. 98 B. 106 C. 125 D. 126 E. 156 17. In the gure O is the centre of
the circle and OA = BC . Which of the following relations holds? A. 2x = 3y B. x = 2y C. x = y D. x + y = 90 E. x + 2y = 180 A. 6
A
y x
B O
18. What is the sum of all the digits in the number 1095 , 95?
C
B. 29 C. 108 D. 663 E. 842 19. In a group of pupils, 40% of the pupils have poor eyesight; 70% of these wear glasses, while the other 30% wear contact lenses. If there are 21 pairs of glasses, which of these assertions is true? A. 45 pupils have poor eyesight B. 30 pupils have good eyesight C. there are 100 pupils in the group D. 10 pupils wear contact lenses E. none of the other assertions is true
152
20. One has to arrange four pawns in this gure so that each column contains one pawn, and each row contains at most one pawn. How many dierent arrangements are possible? A. 64 B. 28 C. 16 D. 8 E. 4 21. Let X = X 4 and X j Y = X + Y . Then 2 j 2 equals: A. 3 24 B. 29 C. 212 D. 220 E. 23 22. Bruce buys three ostriches, seven koalas and one kangaroo. Darren buys four ostriches, ten koalas and one kangaroo. Sheila buys one ostrich, one koala and one kangaroo. Bruce pays 3150 Australian dollars and Darren pays 4200 Australian dollars. How many Australian dollars must Sheila pay? A. 1700 B. 1650 C. 1200 D. 1050 E. 950 23. The diagonal of an isosceles trapezium is 16cm long, and makes an angle of 45 with the base. What is the area 16cm of the trapezium? 45 A. 64cm2 B. 96cm2 C. 128cm2 2 D. more information required E. 256cm 24. Each positive whole number which (in base 10) can be written as a string of 1's and 2's only is called simple. For example 22121 and 2222 are simple; 1021 is not simple. How many simple numbers are there less than one million? A. 63 B. 62 C. 127 D. 128 E. 126 25. What is the maximum possible number of points of intersection one can get with eight circles? A. 16 B. 32 C. 38 D. 44 E. 56 Next we give solutions to the contest given last issue, The Eleventh W.J. Blundon Contest.
THE ELEVENTH W.J. BLUNDON CONTEST February 23, 1994
1. (a) The lesser of two consecutive integers equals ve more than three times the larger integer. Find the two integers. Solution. Let the integers be n, n + 1. Then n = 3(n + 1) + 5 so n = ,4.
153 (b) If 4 x 6 and 2 y 3, nd the minimum values of (x , y )(x + y ). Solution. If 4 x 6 and 2 y 3, then (x , y )(x + y ) = x2 , y 2 2 4 , 32 = 16 , 9 = 5.
2. A geometric sequence is a sequence of numbers in which each term after the rst can be obtained from the previous term by multiplying by the same xed constant, called the common ratio. If the second term of a geometric sequence is 12 and the fth term is 81=2, nd the rst term and the common ratio. Solution. Let the rst term be a, and the ratio r. Then the sequence is a; ar; arr = ar2 ; ar3 ; ar4 ; ar5 ; : : : , with the nth term being arn,1 . So ar = 12 and ar4 = 81=2. Dividing gives 3 ar4 81=2 81 27 r = ar = 12 = 24 = 8 ; so r = 32 . Since ar = a 23 = 12, a = 8.
3. A square is inscribed in an equilateral triangle. Find the ratio of the area of the square to the area of the triangle. Solution. Let the equilateral triangle
ABC and the square DEFG be as shown. Now \ADG = \ABC = \AGD = \ACB = 60 so ADG is an equilateral triangle with side length s, the length of the side of the square. If DE is superimposed over GF a third equilateral triangle is formed with height s.
A D s=2 s B E
G F C p3 2
Now the area of an equilateral triangle with side length s is 4 s , while the area of an equilateral triangle with height s is p13 s2 . So the area of the square is s2 and the area of ABC is p3 p3 1 ! 2 2 s2 2
4 s + s + p3 = 1 + 4 + p3 s :
The ratio of the area of the square to the area of the triangle is then
p
p
1 = p4 3 = 4 3 (,4p3 + 7) = 28p3 , 48: 1 + 4p7 3 4 3 + 7 ,48 + 49
154
4. ABCD is a square. Three parallel lines l1 , l2 and l3 pass through A, B and C respectively. The distance between l1 and l2 is 5 and the distance between l2 and l3 is 7. Find the area of ABCD. Solution. Let the line perpendicular l1 to l2 through B meet l1 at E and E l2 at F respectively. Let l2 meet l2 5 AD at G. Then \BAE = \CBF A s and \AEB = 90 = \BFC so B l3 BAE is similar to BCF . If the 7 side length is s this gives s F s s G ps2 , 52 = 7 C D so s2 = 52 + 72 = 25 + 49 = 74.
5. The sum of the lengths of the three sides of a right triangle is 18. The sum of the squares of the lengths of the three sides is 128. Find the area of the triangle. Solution. a2 + b2 = c2 c b a + b + c = 18 a and a2 + b2 + c2 = 128: 2 2 So 2c = 128, c = 64 and c = 8. Thus a + b = 10, a2 + b2 = 64, so 2ab = (a + b)2 , (a2 + b2 ) = 100 , 64 = 36 36 The area 21 ab = 2ab 4 = 4 = 9.
6. A palindrome is a word or number that reads the same backwards and forwards. For example, 1991 is a palindromic number. How many palindromic numbers are there between 1 and 99; 999 inclusive? Solution. Let us nd the number nk of the properly k digit palindromic numbers. To be a k digit number the rst digit must be one of 1; 2; 3; : : : ; 9. The answer now depends whether k is even or odd. Now n1 = 9. If k > 1 is odd, k = 2l + 1, l 1 then nk = 9 10l. If k is even, k = 2l, l 1 then nk = 9 10l,1. For the answer we want n1 + n2 + n3 + n4 + n5 = 9 + 9 + 90 + 90 + 900 = 1098: 7. A graph of x2 ,2xy +y2 ,x+y = 12 and y2 ,y ,6 = 0 will produce
four lines whose points of intersection are the vertices of a parallelogram. Find the area of the parallelogram. Solution. Note that x2 , 2xy + y 2 , x + y = 12 is equivalent to (x , y )2 , (x , y ) , 12 = 0 or
155
((x , y ) , 4)(x , y + 3) = 0 Thus its graph is the two parallel lines x , y , 4 = 0 and x , y + 3 = 0. Also y 2 , y , 6 = 0 is equivalent to (y , 3)(y + 2) = 0, giving two horizontal lines y , 3 = 0 and y + 2 = 0. The sides of the parallelogram lie on these lines. It has base 7 and height 5. (The distance between the
horizontal lines.) 8. Determine the possible values of c so that the two lines x , y = 2 and cx + y = 3 intersect in the rst quadrant. Solution. For the curves to intersect in the rst quadrant we need x 0, y 0. Adding the equations gives (c + 1)x = 5, so we require c + 1 > 0, 5 into x , y = 2 gives or c > ,1. Substitution of x = c+1
y = c +5 1 , 2 = 3c,+21c
and we need 3 , 2c 0, so c 32 . The intersection is in the rst quadrant for ,1 < c 32 . 9. Consider the function f (x) = 2xcx+3 , x 6= ,3=2. Find all values of c, if any, for which f (f (x)) = x. Solution.
c cx f (f (x)) = cx2x+3
for x 6=
,3
2 2 2x+3 + 3 and f (x) 6= ,3=2, i.e. cx=(2x + 3) 6= ,3=2, x 6= ,9=(6 + 2c). For these excluded values f (f (x)) = x gives
c 2xcx+3
so
=x 2 2xcx+3 + 3
c2x = x(2cx + 3(2x + 3)) c2x = 2cx2 + 6x2 + 9x so equating coecients of x, x2 , etc. (since the polynomials are equal for in nitely many values of x) gives c2 = 9 and 2c = 6, so c = 3.
10. Two numbers are such that the sum of their cubes is 5 and the sum of their squares is 3. Find the sum of the two numbers. Solution. Let the numbers be x and y . We are given x3 + y 3 = 5 and 2 x + y2 = 3. Let x + y = A. From (x + y )3 = x3 + 3x2 y + 3xy 2 + y 3 = x3 + y 3 + 3xy (x + y ) we have A3 = 6 + 3xyA.
156 From (x + y )2 = x2 + y 2 + 2xy we have A2 = 3 + 2xy or xy = 1 (A2 , 3). 2 So we get A3 = 5 + 32 (A2 , 3)A or
1 A3 + 9 A + 5 = 0 and A3 + 9A + 10 = 0 2 2 this gives A = ,1.
That completes the Skoliad Corner for this issue. Send me your contest materials, comments, suggestions, and solutions.
Historical Titbit Taken from a 1950's University Scholarship Paper. A box contains fortyeight chocolates all exactly similar in appearance. There are four of each of twelve dierent sorts. How many must you take out to be certain of having at least one of each of
ve dierent sorts? Suppose that, having taken this number out, you eat four of them, and nd that they are all of one sort. How many must you now put back to be certain that the box contains at least two of each of eight dierent sorts?
157
THE OLYMPIAD CORNER No. 174
R.E. Woodrow All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. As a rst Olympiad set this number we give the third Team Competition, the Baltic Way 1994. The contest was written in Vilnius, Lithuania. Teams from Denmark, St. Petersburg, Poland, Latvia, Iceland, Lithuania, Estonia and Sweden participated. My thanks go to Georg Gunther for collecting this problem set when he was Canadian Team Leader to the IMO at Istanbul.
MATHEMATICAL TEAM CONTEST \BALTIC WAY  92" Vilnius, 1992  November 5{8 1. Let p, q be two consecutive odd prime numbers. Prove that p + q is a product of at least 3 positive integers > 1 (not necessarily dierent). 2. Denote by d(n) the number of all positive divisors of a positive integer n (including 1 and n). Prove that there are in nitely many n such that d(nn) is an integer. 3. Find an in nite nonconstant arithmetic progression of positive integers such that each term is neither a sum of two squares, nor a sum of two cubes (of positive integers). 4. Is it possible to draw a hexagon with vertices in the knots of an integer lattice so that the squares of the lengths of the sides are six consecutive positive integers? 5.4 Given that4 a2 4+ b2 4+ (a + b)24 = c2 + d2 + (c + d)2, prove that 4 a + b + (a + b) = c + d + (c + d) . ,1 , k = 2; 3; : : : ; 100, 6. Prove that the product of the 99 numbers kk33+1 2 is greater than 3 . 7. Let a = 1992p1992. Which number is greater: a
a
a
p
a9 > > > > = > > > > ;
1992 or 1992?
158
8. Find all integers satisfying the equation
2x (4 , x) = 2x + 4:
9. A polynomial f (x) = x3 + ax2 + bx + c is such that b < 0 and ab = 9c. Prove that the polynomial has three dierent real roots. 10. Find all fourth degree polynomials p(x) such that the following
four conditions are satis ed: (i) p(x) = p(,x), for all x, (ii) p(x) 0, for all x, (iii) p(0) = 1, (iv) p(x) has exactly two local minimum points x1 and x2 such that jx1 , x2j = 2. 11. Let Q+ denote the set of positive+ rational numbers. Show that there exists one and only one function f : Q ! Q+ satisfying the following conditions: (i) If 0 < q < 12 then f (q ) = 1 + f 1,q2q . (ii) If 1 < q 2 then f (q ) = 1 + f (q , 1). (iii) f (q ) f ( q1 ) = 1 for all q 2 Q+.
12. Let N denote the set of positive integers. Let ' : bijective function and assume that there exists a nite limit '(n) = L: lim n!1 n
N
! N be a
What are the possible values of L? 13. Prove that for any positive x1; x2 ; : : : ; xn, y1; y2; : : : ; yn the inequality n 1 X 4n2 Pn i=1 (xi + yi)2 i=1 xi yi holds. 14. There is a nite number of towns in a country. They are connected by one direction roads. It is known that, for any two towns, one of them can be reached from the other one. Prove that there is a town such that all the remaining towns can be reached from it. 15. Noah has 8 species of animals to t into 4 cages of the ark. He plans to put species in each cage. It turns out that, for each species, there are at most 3 other species with which it cannot share the accommodation. Prove that there is a way to assign the animals to their cages so that each species shares with compatible species. 16. All faces of a convex polyhedron are parallelograms. Can the polyhedron have exactly 1992 faces?
159
17. Quadrangle ABCD is inscribed in a circle with radius 1 in such a way that one diagonal, AC , is a diameter of the circle, while the other diagonal, BD, is as long as AB . The diagonals intersect in P . It is known that the length of PC is 25 . How long is the side CD? 18. Show that in a nonobtuse triangle the perimeter of the triangle is always greater than two times the diameter of the circumcircle. 19. Let C be a circle in the plane. Let C1 and C2 be nonintersecting circles touching C internally at points A and B respectively. Let t be a common tangent of C1 and C2 , touching them at points D and E respectively, such that both C1 and C2 are on the same side of t. Let F be the point of intersection of AD and BE . Show that F lies on C . 20. Let a b c be the sides of a right triangle, and let 2p be its perimeter. Show that p(p , c) = (p , a)(p , b) = S (the area of the triangle). As a second Olympiad set to give you recreation over the summer months, we give the 8th Iberoamerican Mathematical Olympiad written September 14{15, 1993 in Mexico.
8th IBEROAMERICAN MATHEMATICAL OLYMPIAD September 14{15, 1993 (Mexico) First Day  4.5 hours
1. (Argentina) Let x1 < x2 < < xi < xi+1 < be all the palindromic natural numbers, and for each i, let by yi = xi+1 , xi . How many distinct prime numbers belong to the set fy1 ; y2; y3 ; : : : g? 2. (Mexico) Show that for any convex polygon of unit area, there exists a parallelogram of area 2 which contains the polygon. 3. (Mexico) Let N = f1; 2; 3; : : : g. Find all the functions f : N ! N
such that (i) If x < y , then f (x) < f (y ) (ii) f (yf (x)) = x2 f (xy ), for all x, y belonging to N. Second Day  4.5 hours
4. (Spain) Let ABC be an equilateral triangle, and , its incircle. If D and E are points of the sides AB and AC , respectively, such that DE is tangent to ,, show that AD + AE = 1: DB EC
160
5. (Mexico) Let P and Q be distinct points of the plane. We denote m(PQ) the perpendicular bisector of the segment PQ. Let S be a nite
subset of the plane, with more than one element, which satis es the following properties: (i) If P and Q are distinct points of S , then m(PQ) intersects S . (ii) If P1 Q1, P2 ; Q2 and P3 Q3 are three distinct segments with extreme points belonging to S , then no point of S belongs simultaneously to the three lines m(P1Q1), m(P2Q2), m(P3Q3). Determine the number of possible points of S . 6. (Argentina) Two nonnegative integer numbers, a and b, are \cuates" (friends in Mexican) if the decimal expression of a + b is formed only by 0's and 1's. Let A and B be two in nite sets of nonnegative integers, such that B is the set of all the numbers which are \cuates" of all the elements of A, and A is the set of all the numbers which are \cuates" of all the elements of B . Show that in one of the sets A or B there are in nitely many pairs of numbers x, y such that x , y = 1. We now turn to the readers' comments and solutions to problems given in the December 1994 number of the Corner and the Nordic Mathematical Contest, 1992 [1994: 277]. 1. [1994: 277] Determine all real numbers x, y, z greater than 1, satisfying the equation p x + y + z + x ,3 1 + y ,3 1 + z ,3 1 = 2 px + 2 + y + 2 + pz + 2 :
Solutions by Sefket Arslanagic, Berlin, Germany; by Cyrus C. Hsia, student, Woburn Collegiate Institute, Toronto; by Chandan Reddy, Rochester, Michigan; Michael Selby, University of Windsor, Windsor, Ontario; and by D.J. Smeenk, Zaltbommel, the Netherlands. We give Hsia's solution. For a > 1, a 2 R we have the Arithmetic Mean{Geometric Mean inequality
2 2pa + 2; a , 1 + aa + ,1 2 with p equality if and only if a , 1 = aa+2 ,1 , or a , 3a , 1 = 0, giving a = (3 + 13)=2, since a > 1. For each of a = x, y , z we have this same result and adding them gives 2 + y + 2 + z + 2 2(px + 2 + py + 2 + pz + 2) x + y + z , 3 + xx + ,1 y,1 z,1 whence
x + y + z + x ,3 1 + y ,3 1 + z ,3 1 2(px + 2 + y + 2 + pz + 2) p
161
p
with equality if and only if x = y = z = (3+ p 13)=2. Since there is equality, the unique solution is x = y = z = (3 + 13)=2. 2. [1994: 277] Let n be an integer greater than 1 and let a1; a2; : : : ; an be n dierent integers. Prove that the polynomial f (x) = (x , a1 )(x , a2 ) : : : (x , an ) , 1 is not divisible by any polynomial of positive degree less than n and with integer coecients and leading coecient 1. Solutions by Cyrus C. Hsia, Woburn Collegiate Institute, Toronto, Ontario; by Murray S. Klamkin, University of Alberta, Edmonton, Alberta; by Chandan Reddy; Rochester, Michigan; and by Michael Selby, University of Windsor, Windsor, Ontario. We give the solution sent in by Selby and Klamkin's comment. Suppose p(x) divides f (x), where 1 deg p < n, and p is monic with integral coecients. Therefore f (x) = p(x)q (x) where q (x) is also monic, with integral coecients and 1 deg q < n. Now p(ai )q (ai) = ,1, i = 1; 2; : : : ; n. Since p(ai), q (ai) are integers, p(ai) = 1 and q(ai) = ,1 or p(ai) = ,1 and q(ai) = 1 for each i = 1; 2; : : : ; n. In either case p(ai ) + q (ai) = 0 for i = 1; 2; : : : ; n. Consider p(x) + q (x). This is a polynomial of positive degree less than n, since both p(x) and q(x) are monic with degree less than n. However p(ai) + q(ai) = 0 for n distinct values, but has positive degree less than n, an impossibility. Therefore no such p(x) exists. [Editor's Note.] Klamkin (whose solution was similar) points out that the problem is wellknown. 3. [1994: 277] Prove that among all triangles with given incircle, the equilateral one has the least perimeter. Solutionsby Sefket Arslanagic, Berlin, Germany; by Murray S. Klamkin, University of Alberta, Edmonton, Alberta; by Bob Prielipp, University of Wisconsin{Oshkosh, Wisconsin, USA; and by Chandon Reddy, Rochester, Michigan. We rst give Reddy's solution. Let the three sides be a, b, c. Then the inradius r times the semiperimeter is the area A = rs, r = 1 so, using Heron's formula p
s(s , a)(s , b)(s , c) = s
or so
(s , a)(s , b)(s , c) = ps
p
s = (s , a)(s , b)(s , c):
(1)
s = (s , a) + (s , b) + (s , c):
(2)
Also
162 We minimize the perimeter and therefore s by the AM{GM inequality and (1) and (2)
(s , a)(s , b)(s , c) (s , a) + (s ,3 b) + (s , c)
p
The RHS is minimized when we have equality, that is s , a = s , b = s , c, and a = b = c. Next we give Klamkin's comments and alternative approaches. Solution by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. This is a wellknown result. It follows immediately from the isoperimetric theorem for polygons, that is, of all ngons with given perimeter, the regular one has the maximum area and, dually, of all polygons of given area, the regular one has the least perimeter. For triangles we have the inequality p2 p20 = 12p3 F F0 where p and F denote the perimeter and area of a general triangle and p0 ; F0 correspond to the same for an equilateral triangle. Since F = rp=2 (r = inradius), we have p
p 6 3r
and with equality if and only if the triangle is equilateral. For another proof in terms of the angles A, B , C of the triangle, it is equivalent to establishing the known triangle inequality p cot A cot B cot C 2r 2 + 2 + 2 6 3 r
It also follows that of all ngons circumscribed to a given circle, the regular one has the least perimeter (and area as well). Comment: These results generalize to simplexes. In particular for the tetrahedron, wePhave the following known inequalities, wherePE = the sum of the 6 edges Ei, F = the sum of the areas of the 4 faces Fi , V = the volume and r = the inradius, Y E 6 Ei1=6 k1V 1=3; (1)
F 4 Fi1=4 k2V 2=3; Y
(2)
and there is equality if the tetrahedron is regular sop that the constants p k1, k2 are determined by taking Ei = 1. Then, Fi = 3=4 andkVk rF = 2=12. On multiplication of (1) and (2), we get EF k1 k2 V = 1 32 or that E k1k32r . This proves a 3dimensional extension of the given result.
163
4. [1994: 277] Peter has a great number of squares, some of them are black, some are white. Using these squares, Peter wants to construct a square, where the edge has length n, and with the following property: The four squares in the corners of an arbitrary subrectangle of the big square, must never have the same colour. How large a square can Peter build? n=6 Comment by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. That the answer is 4 4 is given in the solution of problem #1 of the U.S.A. 1976 Mathematical Olympiad [1]. Reference [1] M.S. Klamkin, U.S.A. Mathematical Olympiads 1972{1986, M.A.A., Washington, D.C. 1988, pp. 93{94. To complete this number we given two comments by Murray Klamkin about earlier solutions. 3. [1994: 279; 1993: 131] The abscissa of a point which moves in the positive part of the axis Ox is given by x(t) = 5(t + 1)2 + a=(t + 1)5, in which a is a positive constant. Find the minimum a such that x(t) 24 for all x 0. Comment by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. A simpler noncalculus solution is given by applying the AM{GM inequality, that is, n 1 o 2 1=7 2a 5(t + 1)2 + 2 (t+1) 5 a Then we have so that
[5 + 2]
4
:
2 1=7 a 7 4 24;
7 min a = 2 24 7 :
5. [1994: 281; 1993: 132] p p For each natural number n, let (1 + 2)2n+1 = an + bn 2 with an and bn integers. (a) Show that an and bn are odd for all n. (b) Show that bn is the hypotenuse of a right triangle with legs an + (,1)n and an , (,1)n : 2 2
164 Comment by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. A more direct solution of part (a) is as follows: p p ( 2 + 1)2n+1 , ( 2 , 1)2n+1
an =
2 2 n + 1 2n + 1sn,1 + + 1; n = 2 + 1 3 p p 2 n +1 + ( 2 , 1)2n+1 bn = ( 2 + 1) 2p 2 2 n + 1 2 n + 1 n n , 1 = 2 + 2 2 + + 2n :
That completes the Corner for this issue. Have a good summer { spend some time solving problems and send me your nice solutions as well as Olympiad Contest materials.
Book wanted! Bruce Shawyer would like to purchase a copy of the outofprint book: On Mathematics and Mathematicians (Memorabilia Mathematica) by Robert Edouard Moritz. Dover Edition published 1942. Originally published as: Memorabilia Mathematica or The Philomath's QuotationBook. Original Edition published 1914. Anyone willing to part with a copy please send him details. Thank you.
165
THE ACADEMY CORNER No. 3
Bruce Shawyer All communications about this column should be sent to Professor Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7 Spring is approaching in parts of Canada, and, as undergraduate students in North America prepare for nal examinations, some nd time to try their skills on some mathematics problems. Memorial University oers some modest prizes for the best students. How do you compare? Please send me your best solutions to these problems.
MEMORIAL UNIVERSITY OF NEWFOUNDLAND UNDERGRADUATE MATHEMATICS COMPETITION March, 1996
n2 + 3n + 1 is an irreducible 1. Prove that if n is a positive integer, then 2 n + 4n + 3 fraction. 2. A jar contains 7 blue balls, 9 red balls and 10 white balls. Balls are drawn at random one by one from the jar until either four balls of the same colour or at least two of each colour have been drawn. What is the largest number of balls that one may have to draw? 3. Find all functions u(x) satisfying u(x) = x +
p
p
Z 1 2
0
u(t)dt.
4. Show that ( 5 + 2) 31 , ( 5 , 2) 31 is a rational number and nd its value. 5. In a quadrilateral ABCD (vertices named in clockwise order), AC and BD intersect in X . You are given that AB k DC , that AB is twice as long as CX and that AC is equal in length to DC . Show that AB and CD are equal in length (and hence ABCD is a parallelogram). 6. Prove that among any thirteen distinct real numbers it is possible to x , y < 2 , p3. choose two, x and y , such that 0 <
1 + xy
166 7. A coast guard boat is hunting a bootlegger in a fog. The fog rises disclosing the bootlegger 4 miles distant and immediately descends. The speed of the boat is 3 times that of the bootlegger, and it is known that the latter will immediately depart at full speed on a straight course of unknown direction. What course should the boat take in order to overtake the bootlegger?
Historical Titbit Taken from a 1950's University Scholarship Paper. A, C , B are collinear points.
Prove that there is one and only one point D such that (ACBD) is harmonic. The following result is stated in a book on geometry: if the pencils O(ACBD) and O0 (ACBD) are harmonic and if A, C , B are collinear, then D lies on the line ACB. Give an example to show that this result is not always true. What alteration is required in order to make it true? Prove the result after making this alteration.
167
BOOK REVIEWS Edited by ANDY LIU All the Math That's Fit to Print, by Keith Devlin, published by the Mathematical Association of America, 1994, ISBN 0883855151, paperbound, 330+ pages, US$29.50. Reviewed by A. Sharma, University of Alberta. In the foreword to his famous book Mathematical Snapshots, Steinhaus says, \My purpose is neither to teach in the usual sense of the word, nor to amuse the reader with some charades. One ne day it happened that I was asked the question: `You claim to be a mathematician, well what does one do all day when one is a mathematician?' " Steinhaus conceived the idea of his book as he tried to explain to his questioner a few geometric problems solved and unsolved while sitting on a bench in a public park. Keith Devlin's book is meant for a much wider audience in a completely dierent milieu. It is also not meant to teach or to amuse the reader with riddles, although it succeeds eminently in doing a bit of both. As he explains in the preface, it is meant for \anyone who regularly reads a serious newspaper and has some interest in matters scienti c, mathematical or is just curious". This very interesting book, with its catching title is a collection of 143 articles written by the author for the Manchester Guardian over a period of eight years from 1983 to 1990. From letters received by him, the author claims that his audience was a mixed bunch: \Students at schools in their early teens, retired people in their nineties, from prison inmates to executives in the computer industry, from truckers to school teachers, both men and women". The articles in this book are chronologically arranged under fancy headings some of which we sample here: Sevenup, A Farey Story, The Vertical Confusion, World's Most Wanted Number, Rabbit Pi, Biblical Fingers Get Stuck Into Pi, Rabbits Do It By Numbers, and so on. Each article can be read and enjoyed independently of the others, although cross references to allied topics are given almost everywhere. It is a book which can be enjoyed at one's leisure and with a pencil and a piece of paper can give hours of delight and education. Devlin follows the style of his mentor Martin Gardner. He is informal and states results when he chooses and sketches a proof if it is short and well within the reach of a layman (see Article 15). Article 19, The measure of all things, begins with the news about the discovery that a long standing conjecture in number theory called Merten's conjecture is false. The author slowly and carefully explains the Mobius function, and tells the reader what Merten's conjecture is, and who proved that it is false, while giving a bird's eye view of the result proved.
168 Article 25 is just half a page, but it tells about the Institute for Pi Research at Emporia University. The Institute, we learn, is campaigning for the value = 3 to be given equal time with the more conventional value in state schools. I personally found the Article 29, Question Time, very interesting, since I spent hours working out some of its problems, specially questions 11, 12, 13 and 15. I was particularly pleased to read in Article 99 about D.R. Kaprekar whom I knew while in India. There were no computers in those days, but I have witnessed his phenomenal abilities at doing long multiplication and other operations without pen and paper almost instantly at several meetings of the Indian Mathematical Society. It would be surprising if a book of almost 320 pages had no misprints. I point out some minor ones which I came across: p. xi, line 2 from bottom  ready should be read; p. 27, column 2, line 6 from top  33 should be 3; p. 46, column 1, line 15 from top  the should be to; p. 71, column 1, line 10 from top  7071 should be 7273; p. 322, column 1, line 3 from bottom  savaged should be salvaged. The range of topics covered in these articles is very impressive  Fermat Primes, Carmichael numbers, Palindromic numbers, information, arti cial intelligence, Hilbert's tenth problem, Bieberbach's conjecture, BanachTarski paradox, cartography and many more. It is a book which should be read piecemeal  an article or two at a time, and as Devlin says, \it is a book for delving". I am sure this book will soon be in most libraries of schools and colleges for the bene t of both teachers and students. Do you know the equation of this curve?
169
PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be typewritten or neatly handwritten, and should be mailed to the EditorinChief, to arrive no later that 1 December 1996. They may also be sent by email to cruxeditor@cms.math.ca. (It would be appreciated if email proposals and solutions were written in LATEX, preferably in LATEX2e). Graphics les should be in epic format, or plain postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication. 2138. Proposed by Christopher J. Bradley, Clifton College, Bristol, UK. ABC is an acute angle triangle with circumcentre O. AO meets the circle BOC again at A0 , BO meets the circle COA again at B 0 , and CO meets the circle AOB again at C 0 . Prove that [A0 B 0 C 0 ] 4[ABC ], where [XY Z ] denotes the area of triangle XY Z . 2139. Proposed by Waldemar Pompe, student, University of Warsaw, Poland. Point P lies inside triangle ABC . Let D, E , F be the orthogonal projections from P onto the lines BC , CA, AB , respectively. Let O0 and R0 denote the circumcentre and circumradius of the triangle DEF , respectively. Prove that p q [ABC ] 3 3R0 R0 2 , (O0P )2; where [XY Z ] denotes the area of triangle XY Z . 2140. Proposed by K.R.S. Sastry, Dodballapur, India. Determine the quartic f (x) = x4 + ax3 + bx2 + cx , c if it shares two distinct integral zeros with its derivative f 0(x) and abc 6= 0.
170
2141. Proposed by Toshio Seimiya, Kawasaki, Japan.
A1A2A3A4 is a quadrilateral. Let B1, B2, B3 and B4 be points on the sides A1 A2 , A2 A3 , A3 A4 and A4 A1 respectively, such that A1B1 : B1A2 = A4B3 : B3A3 and A2B2 : B2A3 = A1B4 : B4A4: Let P1, P2 , P3 and P4 be points on B4 B1 , B1 B2 , B2 B3 and B3 B4 respectively, such that P1P2kA1A2; P2P3kA2A3 and P3P4kA3A4: Prove that P4 P1 kA4 A1 . 2142. Proposed by Victor Oxman, Haifa, Israel. In the plane are given an arbitrary quadrangle and bisectors of three of its angles. Construct, using only an unmarked ruler, the bisector of the fourth angle. 2143. Proposed by B. M???y, Devon, Switzerland. My lucky number, 34117, is equal to 1662 + 812 and also equal to 2 159 + 942, where j166 , 159j = 7 and j81 , 94j = 13; that is,
it can be written as the sum of two squares of positive integers in two ways, where the rst integers occurring in each sum dier by 7 and the second integers dier by 13. What is the smallest positive integer with this property? 2144. Proposed by B. M???y, Devon, Switzerland. My lucky number, 34117, has the interesting property that 34 = 2 17 and 341 = 3 117 , 10, that is, it is a 2N + 1digit number (in base 10) for some N , such that (i) the number formed by the rst N digits is twice the number formed by the last N , and (ii) the number formed by the rst N + 1 digits is three times the number formed by the last N + 1, minus 10. Find another number with this property. 2145. Proposed by Robert Geretschlager, Bundesrealgymnasium, Graz, Austria. n n Y, Y Prove that ak + bk,1 ,ak + bn,k for all a, b > 1. k=1 k=1
171
2146. Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a triangle with AB > AC , and the bisector of \A meets BC at D. Let P be an interior point on the segment AD, and let Q and R be the points of intersection of BP and CP with sides AC and AB respectively. Prove that PB , PC > RB , QC > 0.
2147. Proposed by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore. Let S be the set of all positive integers x such that there exist positive integers y and m satisfying x2 + 2m = y 2. (a) Characterize which positive integers are in S . (b) Find all positive integers x so that both x and x + 1 are in S . 2148. Proposed by Aram A. Yagubyants, Rostov na Donu, Russia.
Suppose that AD, BE and CF are the altitudes of triangle ABC . Suppose that L, M , N are points on BC , CA, AB , respectively, such that BL = DC , CM = EA, AF = NB. Prove that: 1. the perpendiculars to BC , CA, AB at L, M , N , respectively are concurrent; 2. the point of concurrency lies on the Euler line of triangle ABC .
2149. Proposed by JuanBosco Morero Marquez, Universidad de Valladolid, Valladolid, Spain. Let ABCD be a convex quadrilateral and O is the point of the intersection of the diagonals AC and BD. Let A0 B 0 C 0 D0 be the quadrilateral whose vertices, A0 , B 0 , C 0 , D0 , are the feet of the perpendiculars drawn from the point O to the sides BC , CD, DA, AB , respectively. Prove that ABCD is an inscribed (cyclic) quadrilateral if and only if A0 B0C 0D0 is a circumscribing quadrilateral (A0 B0, B0C 0, C 0D0, D0 A0 are tangents to a circle). 2150. Proposed by Sefket Arslanagic, Berlin, Germany. Find all real solutions of the equation p1 , x = 2x2 , 1 + 2xp1 , x2:
172
SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems. 2035. [1995: 130] Proposed by Vaclav Konecny, Ferris State University, Big Rapids, Michigan, USA. If the locus of a point E is an ellipse with xed foci F and G, prove that the locus of the incentre of triangle EFG is another ellipse. Solution by the Science Academy Problem Solvers, Austin, Texas. Let F = (,c; 0) and G = (c; 0) so that the locus of E = (x; y ) satis es GE + EF = 20a (where a is the length of the semimajor axis). Let I be the incentre and I be the point where the incircle is tangent to FG (so that II 0 ? FG). We have
FI 0 = a + c , GE = EF ,2 GE + c [since s = a + c is half the perimeter of EFG, and the tangents to the incircle from the vertices are equal in pairs and sum to 2s]. If O is the origin then FO = c and 2 2 OI 0 = FI 0 , c = EF ,2 GE = EF 4,a EG 2 2 2 2 2 2 = (x + 2cx + c + y ) 4,a (x , 2cx + c + y ) = cx a: Since r = II 0 is the radius of the incircle [and the area of EFG is rs] ) = cy : II 0 = areaa(+EFG c a+c c c y . The mapping (x; y) ! c x; c y is thereThus, I = x; a a+c a a+c fore an ane transformation that maps each point E of the given ellipse to the point I . Ane transformations map ellipses to ellipses so that I traces
out an ellipse. Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Berlin, Germany; FRANCISCO BELLOT Spain; SEFKET ARSLANAGIC, LOPEZ ROSADO, I.B. Emilio Ferrari, and MARIA ASCENSION CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton OCHOA, Logro~no, Spain; College, Bristol, UK; MIGUEL ANGEL CABEZON JORDI DOU, Barcelona, Spain; KEEWAI LAU, Hong Kong; P. PENNING, Delft, the Netherlands; CRISTOBAL SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon, Spain; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.
173
2040. [1995: 130] Proposed by Frederick Stern, San Jose State University, San Jose, California. Let a < b be positive integers, and let a t = 22b ,, 11 :
What is the relative frequency of 1's (versus 0's) in the binary expansion of t? [Ed. My interpretation of the question asked is to nd the ratio of the number of 1's to the number of 0's; most solvers also read it this way. The proposer was the only solver to actually mention that the ratio we are to compute must be done asymptotically. I feel that should have been part of the problem statement.] Solutionby Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USA. Let x = :000 : : : 01, where there are b , 1 zeros (the pattern below the superbar repeats in nitely often). Then (2b , 1)x = :111 : : : 1 = 1, so x = 1=(2b , 1). Therefore, (2a , 1)=(2b , 1) = :00 : : : 01 : : : 11, where there are b , a zeros and a ones in every period, so there are a zeros for every b , a ones when counted from the left. Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; KEITH EKBLAW, Walla Walla, Washington, USA; TOBY GEE, student, the John of Gaunt School, Trowbridge, England; RICHARD I. HESS, Rancho Palos Verdes, California, USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; P. PENNING, Delft, the Netherlands; Science Academy Problem Solvers, Austin, Texas, USA; DAVID STONE, Georgia Southern University, Statesboro, Georgia, USA; PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA; and the proposer. Janous generalizes the problem to show that in base z + 1 we have
(z + 1)a = 000 : : : 0zz : : : z (z + 1)b , 1
where there are b , a zeros and a copies of z when a < b.
2041. [1995: 157] Proposed by Toshio Seimiya, Kawasaki, Japan.
P is an interior point of triangle ABC . AP , BP , CP meet BC , CA, AB at D, E, F respectively. Let M and N be points on segments BF and CE respectively so that BM : MF = EN : NC . Let MN meet BE and CF at X and Y respectively. Prove that MX : Y N = BD : DC .
174 Solution by Jari Lappalainen, Helsinki, Finland. First, we apply Menelaus' theorem to triangles CNY and Y FM with line EB . Dividing the results, we get CE NX Y P ,1 = EN XY PC F B MX Y P = 1 ,1 BM XY PF and substituting CE : EN = FB : BM (which follows directly from BM : MF = EN : NC ), we get NX = PC ; MX PF
or equivalently
FC = MN : PF XM
(1)
In a similar way using Menelaus's theorem for triangles BXM and XNE with line FC , and substituting MF : FB = NC : CE , we get
EB = NM : (2) PE Y N Finally, applying Ceva's theorem to triangle ABC and Menelaus's theorem to triangles CEP and BEA, with lines AB and CF respectively, we nd BD CE AF 1 BD PE FC = 1: DC EA FB = = EB PF BP EC AF CA ,1 (,1) AE BP FC PE CA FB DC EB PF Using (1) and (2)
BD Y N MN = , BD Y N = 1; DC NM XM DC XM or MX : Y N = BD : DC .
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; D.J. SMEENK, Zaltbommel, the Netherlands; HOE TECK WEE, student, Hwa Chong Junior College, Singapore; and the proposer.
175
2042. [1995: 157] Proposed by Jisho Kotani, Akita, Japan, and K.R.S. Sastry, Dodballapur, India. If A and B are threedigit positive integers, let A B denote the sixdigit integer formed by placing them side by side. Find A and B such that A; B; B , A; A B and A B B
are all integer squares. Solution by Kathleen E. Lewis, SUNY Oswego, New York, NY, USA. Since A, B , A and B are all integer squares, their square roots form a Pythagorean triple. Thus, the easiest place to look for numbers meeting the stated conditions is in the multiples of the 3; 4; 5 triple. So, letting A = 9t2 and B = 25t2 for an integer t, we see that B , A = 16t2 , A B = 1000A + B = 9000t2 + 25t2 = 9025t2 = (95t)2 and
A B = (95t)2 = 192: B (5t)2 Any choice of t would yield A and B satisfying all the other conditions, but in order to make A and B three{digit numbers, t would have to be 4, 5 or 6, yielding values of
(144; 400);
(225; 625);
and (324; 900)
for (A; B ). These are in fact the only possibilities. Suppose A = x2 , B , A = y 2 and B = z 2 satisfy the given conditions. Let q = gcd(x; y;z ), x0 = x=q , y0 = y=q and2z0 = 2z=q. 2 Then x0 + y 0 = z 0 and z 0 2 divides (A B , B )=q 2 = 1000x0 2 . Since 0 x and z0 are relatively prime, this means that z02 must divide 1000, so z0 must be 1, 2, 5 or 10. By inspection, none of the others are possible values, so z 0 must be 5. We can also rule out the remaining case of x0 = 4, y 0 = 3 and z 0 = 5 [that is, A = 16t2 and B = 25t2] since (A B )=t2 = 16025 is not a perfect square. Editor's note. As some other solvers point out, the case B , A = 0 should be considered. This is the case y 0 = 0, x0 = z 0 = 1, which is impossible because A B = 1001x2 is not a perfect square. Also solved by CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bris OCHOA, Logro~no, Spain; J.K. FLOYD, tol, UK; MIGUEL ANGEL CABEZON Newnan, Georgia, USA; TOBY GEE, student, the John of Gaunt School, Trowbridge, England; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; DAVID HANKIN, Hunter College Campus Schools, New York, NY, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOHN G.
176 HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHER Y, FerJANOUS, Ursulinengymnasium, Innsbruck, Austria; VACLAV KONECN ris State University, Big Rapids, Michigan, USA; MICHAEL PARMENTER, Memorial University of Newfoundland, St. John's, Newfoundland; P. PENNING, Delft, the Netherlands; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; HOE TECK WEE, student, Hwa Chong Junior College, Singapore; SUSAN SCHWARTZ WILDSTROM, Kensington, Maryland, USA; PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA; and the proposers. Most solvers found all three solutions. The two proposers actually had sent the editor this problem, or something quite similar, independently and at almost the same time.
2043. [1995: 158] Proposed by Aram A. Yagubyants, Rostov na Donu, Russia. What is the locus of a point interior to a xed triangle that moves so that the sum of its distances to the sides of the triangle remains constant? Editor's Comment. The solutionto the problem as stated in the proposal was quite easy using analytic/trigonometric arguments and this was the path chosen by most of the solvers. Some contributors, anxious no doubt to uphold the impeccable standards of CRUX, tried to do something more with the proposal much to the delight of the Editor. One such example is to look at the locus for dierent triangles as portrayed in the featured solution by Hess. Just the observation that all points in the interior is the locus when the given triangle is equilateral can do wonders for the morale of the Editor. The second solution, while restricted to the interior of the triangle, did oer a respite from the analytic. Several solvers made comments of varying substance relating to the case when the point was exterior to the triangle. The third solution by Fritsch did the job nicely and was novel. I. Solution by Richard I. Hess, Rancho Palos Verdes, California, USA. From the gure, where the largest side is the base, we have
h1 = y; h2 = (a , x) sin , y cos ; h3 = x sin , y cos :
p
h3 P h2 h1
a
Note:
P = P (x; y)
177 From h1 + h2 + h3 = k (a constant), we get
y(1 , cos , cos ) = k , a sin + x(sin , sin ):
There are several cases: 1. When = = 60, the whole interior of the triangle has h1 +h2 +h3 = p a 3=2. 2. When cos + cos = 1 but 6= , the locus is x = k , a sin .
3. When = 6= 60 , the locus is y =
sin , sin
k , a sin 1 , 2 cos .
4. Otherwise, the locus is a straight line:
, a sin + x(sin , sin ) . y = 1 ,kcos , cos 1 , cos , cos II. Solution by Toshio Seimiya, Kawasaki, Japan.
Let P be a point interior to triangle ABC , so that the sum of its distances to the sides of triangle ABC is constant, and let D, E, F be the feet of the perpendiculars to BC , CA, AB respectively. We assume that triangle ABC is nonequilateral and that BC is the least side. We put PD + PE + PF = k, where k is a constant.
A
F
G
H P
E
B D C Let G, H be the points on the side AB , AC respectively, such that BG = CH = BC . Then [PBC ] + [PCH ] + [PBG] = 12 a(PD + PE + PF ) = 12 ak; which is constant ([XY Z ] denotes the area of triangle XY Z ). Since the area of quadrilateral BCHG is constant, so also is the area of triangle PGH . Therefore P lies on a xed line `, parallel to GH . Hence the locus of P is the segment of ` contained within the triangle ABC .
III. Solution by Rudolf Fritsch, LudwigMaximiliansUniversitat, Munchen, Germany. Let ABC be a triangle in the real plane; without loss of generality we assume a b c. Then we may choose cartesian coordinates such that
A = (0; h) ; B = (,p; 0) ; C = (q; 0)
178 with h, p, q > 0. The signed distance of a point in the plane from a side of this triangle is taken positive if the point is in the same half plane as the vertex opposite to this side, thus, the signed distance of P = (x; y ) from the line BC is just y . For the other sides of triangle we choose the following equations
AB x sin , y cos + h cos = 0 ; AC ,x sin , y cos + h cos = 0 :
These equations are called the Hessian normal forms of the lines under consideration. The namesake is Otto Ludwig Hesse (born in Konigsberg/East Prussia 1811/4/22, died in Munchen 1874/8/4) although this form has already been used by Gauss in a paper published in 1810. The general idea is to normalize the line equation
g ux + vy + w = 0 by u2 + v 2 = 1, w 0 (or w 0), the line g being unique if it does not pass through the origin. The advantage of this form is that for any point P = (x; y) in the plane the expression d(x; y) = ux + vy + w gives the signed distance of P from g , positive if and only if P is on the same side of g as the origin (origin not on g ). Thus, the sum of the signed distances of P = (x; y ) from the reference triangle is
s(x; y) = x (sin , sin ) + y (1 , cos , cos ) + h (cos + cos ) : Since this expression is linear, the equation s(x; y ) = k describes a line, for any k 2 R. Since the slope is independent of k, all lines obtained in this way form a pencil of parallel lines. A distinguished member of this pencil is obtained by taking k = 0 giving the line connecting the points where the
exterior angle bisectors meet the opposite sides. Taking absolute distances we get buckled lines for the desired locus. If the problem is restricted to the interior points of the triangle ABC , then the result gives parallel line segments. Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; NIELS BEJLEGAARD, Stavanger, Norway; DAVID HANKIN, Hunter College Campus Schools, New York, NY, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, University of Al Y, Ferris State University, Big berta, Edmonton, Alberta; VACLAV KONECN Rapids, Michigan, USA; P. PENNING, Delft, the Netherlands; CRISTOBAL SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon, Spain; ASHISH KR. SINGH, Kanpur, India; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.
179 Klamkin comments: since a1 r1 + a2 r2 + a3 r3 = 2 (where ri is the distance to side ai , and is the area of the triangle), the constant sum of the distances must be bounded by
2 r + r + r 2 ; 1 2 3 a a
3 1 where it is assumed that a1 a2 a3 . Otherwise, there are no points in the locus. This is also true, if, for example, a3 is smaller than the other sides and the constant sum is 2 a3 .
2044. [1995: 158] Proposed by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. Suppose that n m 1 and x y 0 are such that xn+1 + yn+1 xm , ym: Prove that xn + y n 1. I. Solution by Toshio Seimiya, Kawasaki, Japan. There is nothing to prove if x = y since in this case the given inequality implies x = y = 0. Suppose 0 y < x, then xm , y m > 0. From xn+1 xn+1 + yn+1 xm , ym xm and n + 1 > m, it follows that x 1. Using n , m 0 and m 1, we now obtain (xn + yn)(xm , ym) = xn+m , y n+m , (xy )m(xn,m , y n,m) xn+m xn+1 xn+1 + yn+1 xm , ym: Dividing through by xm , y m gives xn + y n 1. II. Solution by the proposer. It is easy to see that 1 x y 0, and thus xm x, xy n y m and n x y y. Therefore, (xn + yn)(x + y ) = xn+1 + yn+1 + xn y + xy n
xm , ym + y + ym x + y: Dividing through by x + y [the case when x + y = 0 being trivial { Ed.] yields xn + y n 1. Equality holds if and only if x = 1, y = 0.
Also solved by NIELS BEJLEGAARD, Stavanger, Norway; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; WALTHER Y, JANOUS, Ursulinengymnasium, Innsbruck, Austria; VACLAV KONECN Ferris State University, Big Rapids, Michigan, USA; and PANOS E. TSAOUSSOGLOU, Athens, Greece.
180 Janous showed that if 1 m n , 1, then the condition can be relaxed to xn+1 + y n+1 xm , y n,1. Flanigan obtained the stronger result that if k = nm+1 , then x(k,1)m + (k , 1)y (k,1)m 1 if mjn + 1 and xkm + kykm 1 if m 6 j n + 1.
2045. [1995: 158] Proposed by Vaclav Konecny, Ferris State University, Big Rapids, Michigan, USA. Show that there are an in nite number of Pythagorean triangles (rightangled triangles with integer sides) whose hypotenuse is an integer of the form 3333 : : : 3. Once again, our readers have been extremely inventive! Most provided a way to construct an in nite sequence (or more than one sequence) of suitable Pythagorean triples (a; b; c), where a2 + b2 = c2 . We summarize the results below. I. Solution by: Niels Bejlegaard, Stavanger, Norway; Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USA; Christopher J. Bradley, Clifton College, Bristol, UK; Miguel Angel Cabez on Ochoa, Logro~no, Spain; Toby Gee, student, the John of Gaunt School, Trowbridge, England; David Hankin, Hunter College Campus Schools,New York, NY, USA; Richard I. Hess, Rancho Palos Verdes, California, USA; Friend H. Kierstead Jr., Cuyahoga Falls, Ohio, USA; Kathleen E. Lewis, SUNY Oswego, NY, USA; Maria Ascension Lopez Chamorro, I.B. Leopoldo Cano, Valladolid, Spain; P. Penning, Delft, the Netherlands; Gottfried Perz, Pestalozzigymnasium, Graz, Austria; David R. Stone, Georgia Southern University, Statesboro, USA; Panos E. Tsaoussoglou, Athens, Greece; and the proposer. First, note that neither 3 nor 33 is the hypotenuse of a Pythagorean triangle. Let (a; b; c) be a Pythagorean triple, where c = 33 : : : 3 is a kdigit integer, k > 2. Then (ma;mb; mc ) is also a Pythagorean triple, for all m = 10k + 1; 102k + 10k + 1; : : : ; Pni=0 10ik; : : : and mc = 33 : : : 3 has 2k; 3k; : : : ; (n + 1)k;: : : digits. Each of the following triples can each be used in this way to generate an in nite sequence of triples: (108; 315; 333) (660; 3267; 3333) (7317; 32520; 33333) (128205; 307692; 333333) (487560; 3297483; 3333333) (25114155; 21917808; 33333333).
181 II. Solution by: HeinzJurgen Seiert, Berlin, Germany.
2 10n(102n , 1); 1 (102n , 1)2; 1 (104n , 1) ; where n 2 N; 3 3 3
gives an in nite sequence of Pythagorean triples with hypotenuse the 4ndigit integer 33 : : : 3. Note that the rst triple, (660; 3267; 3333), is the same as one given above, but the rest of the sequence is dierent. Also solved (in a nonconstructive way) by ASHISH KR. SINGH, student, Kanpur, India; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; and CHRIS WILDHAGEN, Rotterdam, the Netherlands. There was one partial solution.
2047. [1995: 158] Proposed by D.J. Smeenk, Zaltbommel, the Neth
erlands.
ABC is a nonequilateral triangle with circumcentre O and incentre I . D is the foot of the altitude from A to BC . Suppose that the circumradius R equals the radius ra of the excircle to BC . Show that O, I and D are
collinear. Solution by Waldemar Pompe, student, University of Warsaw, Poland. If AB = AC , the I; O lie on AD; hence O; I and D are collinear. Henceforth assume AB 6= AC . Let the angle bisector of the angle \A intersect the circumcircle of ABC at E and let DO and AE meet at J  see the gure. A
O J B
D
C
E OE is perpendicular to BC , so AD and OE are parallel. Therefore, since
R = ra
AJ = AD = ha = ha = 2S = 2ra(s , a) = b + c , a ; JE OE R ra ara ara a
(1)
182 where ha , S , s denote the altitude AD, the area, and the semiperimeter of ABC , respectively. To complete the solution it is enough to show that J is the incentre of ABC . Using Ptolemy's theorem on the quadrilateral ACEB , and the fact that BE = CE , we get (AJ + JE )a = BE(b + c): (2) Since (1) is equivalent to (AJ + JE )a = JE (b + c), comparing it with (2) we obtain JE (b + c) = BE (b + c), which gives JE = BE . Hence \JBA + \BAJ = \BJE = \JBE = \CBJ + \CBE = \CBJ + \EAC: Since \BAJ = \EAC , we get \JBA = \CBJ , which means that BJ bisects \B . And since AJ bisects \A, J is the incentre of ABC , as we wished to prove. Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; P. PENNING, Delft, the Netherlands; TOSHIO SEIMIYA, Kawasaki, Japan; ASHISH KR. SINGH, student, Kanpur, India; and the proposer. Several solvers noted that the result was mentioned in the Editor's comments following the solution to problem 1918 (CRUX, Vol. 21, No. 1).
2046. [1995: 158] Proposed by Stanley Rabinowitz, Westford, Massachusetts, USA. Find integers a and b so that x3 + xy2 + y3 + 3x2 + 2xy + 4y2 + ax + by + 3 factors over the complex numbers. Solution by the proposer. If the cubic is to factor in C [x; y ], then one of the factors must be linear. Without loss of generality, we may assume this factor is of the form x , py , q where p and q are complex numbers. Substituting x = py + q in the original cubic, we get a polynomial in y that must be identically 0. Thus each of its coecients must be 0. This gives us the four equations: 1 + p + p3 = 0 2 4 + 2p + 3p + q + 3p2q = 0 3 + aq + 3q 2 + q 3 = 0 b + ap + 2q + 6pq + 3pq2 = 0: Solving these equations simultaneously, yields a = 4 and b = 5. (The editor being a mere mortal needed Maple to verify this claim!)
183 As a check we note that the resulting polynomial can be written as
(x + y + 2)(y + 1)2 + (x + 1)3. The reducibility of this polynomial will not change if we let x = X , 1 and y = Y , 1. This produces the polynomial X 3 + XY 2 + Y3 3. Letting z = X=Y shows that this polynomial factors over C [x; y ] since z + z + 1 factors over C [z ].
Also solved by RICHARD I. HESS, Rancho Palos Verdes, California, USA. Rabinowitz also remarks that, \except for a few exceptional cases, if f (x;y ) is a cubic polynomial in C [x; y ], there will be unique complex constants a and b such that f (x;y ) + ax + by factors over C [x; y ]." etokrzyski, 2049?. [1995: 158] Proposed by Jan Ciach, Ostrowiec Swi
Poland. Let a tetrahedron ABCD with centroid G be inscribed in a sphere of radius R. The lines AG; BG; CG; DG meet the sphere again at A1 ; B1 ; C1 ; D1 respectively. The edges of the tetrahedron are denoted a; b; c; d; e; f . Prove or disprove that
p
4 1 + 1 + 1 + 1 4 6 1 + 1 + 1 + 1 + 1 + 1 : R GA1 GB1 GC1 GD1 9 a b c d e f
Equality holds if ABCD is regular. (This inequality, if true, would be a threedimensional version of problem 5 of the 1991 Vietnamese Olympiad; see [1994: 41].) Solution to right hand inequality by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. We consider a generalization of the right hand inequality: Let G and O be the centroid and circumcentre of an ndimensional simplex A0 A1 : : : An inscribed in a sphere of radius R. Let the lines Ai G meet the sphere again in points A0i : i = 0; 1; : : : ; n. If the edges are denoted by ej ; j = 1; 2; : : : ; n(n + 1)=2, then s +1)=2 1 n 1 X 8(n + 1) n(nX : 0 n3 i=0 GAi j =0 ej By the Powerofa Point Theorem, we have A01 G Ai G = R2 , OG2 . By the power mean inequality, we have X
AiG X (AiG)2 21 : n+1 n+1
184 So it suces to prove the stronger inequality: s
1 , X (AiG)2(n + 1) 2 8(nn+3 1) R2 , (OG)2 e1 : j It is known that X (AiG)2 = (n + 1) ,R2 , (OG)2 and X e2j : , 2 R , (OG)2 = (n + 1)2 X
(1) (2) (3)
Using (2) and (3), we see that (1) becomes, after raising both sides to the power 2=3, n(n + 1) X e21=3 X 1 2=3 ; j
ej
2
and the result follows immediately from Holder's inequality. There is equality only if the simplex is regular. Comment: Corresponding to the given left hand inequality, the analogous one for the simplex (and not as yet proved) is or, equivalently
n+1 X 1 R A0iG
X
X
AiG (n + 1) ,R2 , (OG)2 = (AiG)2: Even more generally, I conjecture that, for p 1, we have 2R (np + 1) PP(AiG)p+1 : (n + 1) (np,1 + 1) (AiG)p Except for the case p = 1, there is no equality for a regular simplex, but for a degenerate one with n vertices coinciding at one end of a diameter and the R
remaining vertex at the other end of the diameter. No other solutions were received.
2050. [1995: 158] Proposed by Sefket Arslanagic, Berlin, Germany. Find all real numbers x and y satisfying the system of equations p 2x2+y + 2x+y2 = 128; px + py = 2 2:
185 Solution. Essentially identical solutionswere submitted by Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USA; Christopher J. Bradley, Clifton College, Bristol, UK; Toby Gee, student, the John of Gaunt School, Trowbridge, England; David Hankin, Hunter College Campus Schools, New York, NY, USA; Cyrus Hsia, student, University of Toronto, Toronto, Ontario; Walther Janous, Ursulinengymnasium, Innsbruck, Austria; KeeWai Lau, Hong Kong; Waldemar Pompe, student, University of Warsaw, Poland; Panos E. Tsaoussoglou, Athens, Greece; and the proposer. p p p Squaring x + y = 2 2, we obtain that
x + y = 8 , 2pxy 8 , (x + y);
whence Further
x + y 4:
(1)
x2 + y2 12 (x + y)2 8:
(2)
From (1) and (2), we have
x2 +y + 2x+y2 2 64 = 2 2 12 (x2+y+x+y2) 26 = 64: Thus x + y = 4 and x2 + y 2 = 8. These are easily solved to obtain that x = y = 2.
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and LOPEZ MARIA ASCENSION CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; VACLAV KONECNY, Ferris State University, Big Rapids, Michigan, USA; BEATRIZ MARGOLIS, Paris, France; J.A. MCCALLUM, Medicine Hat, Alberta; HEINZ JURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary, Alberta; DAVID R. STONE, MARTHA BELL and JIM BRASELTON; Georgia Southern University, Statesboro, Georgia, USA; STAN WAGON, Macalester College, St. Paul, Minnesota, USA; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and SUSAN SCHWARTZ WILDSTROM, Kensington, Maryland, USA. One other submission was received which assumed that x and y were integers. Janous pointed out that the result can be generalized to: Let x1 ; : : : ; xn (n 2) be nonnegative real numbers such that x1 + : : : + xn = nw, and let b 6= 1 be a positive real number.
186 Let 1; : : : ; n be real numbers, each greater than or equal to 1. Then the equation bx1 1 +x2 2 +:::+xnn + bx2 1 +x3 2+:::+xnn + : : : + bxn1 +x1 2 +:::+xnn,1 = bw 1 +w 2 +:::+w n +logb (n)
has as its only solution x1 = x2 = : : : = xn = w. This can be proved by applications of the AM{GM and power mean inequalities.
2051. [1995: 202] Proposed by Toshio Seimiya, Kawasaki, Japan.
A convex quadrilateral ABCD is inscribed in a circle , with centre O. P is an interior point of ABCD. Let O1, O2, O3, O4 be the circumcentres of triangles PAB , PBC , PCD, PDA respectively. Prove that the midpoints of O1O3 , O2 O4 and OP are collinear.
Combination of solutions by Jordi Dou, Barcelona, Spain and the proposer. The result holds without restriction on the point P . The proof is in two steps. Step 1. The result is trivially true when P is on , (and all the circumcentres coincide with O), so let P be a point o , and let be the conic with foci O and P whose major axis has length R (the radius of ,).
is an ellipse when P is interior to , and a hyperbola when P is exterior. Let Q be the intersection of O1O2 and OB . Because O1 B = O1 P and O2B = O2P , we have that O1O2 is the perpendicular bisector of BP . Therefore BQ = PQ and \BQO2 = \PQO2. When P is interior we conclude that OQ + PQ = OQ + BQ = R; when exterior, OQ , PQ = OQ , BQ = R. Thus, in either case, Q is a point on . Moreover, as O1O2 is a bisector of \OQP; O1O2 is tangent to at Q. Similarly O2O3 ; O3 O4, and O4O1 are tangent to . [Editor's comment by Chris Fisher. Dou refers to , as the focal circle of . I was unable to con rm that terminology in any handy reference, but I did nd the circle mentioned as the basis of a construction of a central conic by folding; see, for example, E.H. Lockwood, A Book of Curves: Draw a circle , on a sheet of paper and mark an arbitrary point P not on ,. For any number of positions B on , fold P onto B and crease the paper. The creases (i.e. the perpendicular bisectors of PB ) envelope a conic .] Step 2. By Newton's theorem the midpoints of O1O3 ; O2 O4, and the centre of (which is the midpoint of OP ) are collinear as desired. As a bonus, also on that line is the midpoint of the segment joining O5 := O1 O4 \ O2O3 to O6 := O1 O2 \ O3O4 . Here is a simple projective proof of this theorem. Consider the pencil of dual conics tangent to the sides of the complete quadrilateral O1 O2O3 O4. As a consequence of the Desargues' involution theorem
187 (see, for example, Dan Pedoe, A Course of Geometry for Colleges and Universities, Theorem II, page 342), the poles of a line l with respect to the conics of the pencil lie on a line. In particular, when l is the line at in nity the line of poles is the line of centres of these conics. The centre of is one such pole. Furthermore, the line of centres passes through the midpoints of the degenerate dual conics of the pencil (consisting of pairs of opposite vertices of the quadrilateral), namely O1 and O3 , O2 and O4 , O5 and O6. LOPEZ Also solved by MARIA ASCENSION CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain.
2053. [1995: 202] Proposed by Jisho Kotani, Akita, Japan.
A gure consisting of two equal and externally tangent circles is inscribed in an ellipse. Find the eccentricity of the ellipse of minimum area. Solution by David Hankin, Hunter College Campus Schools, New York, NY, USA. [Editor's note: By symmetry, we have: (a) the centre of the circumscribing ellipse, E , must be the point of tangency of the two given circles; (b) an axis of E passes through the centres of the two given circles.
6 
All solvers assumed this, most without stating that they had done so.]
}
}
}
} } Let the equations of the circles be (x r)2 + y 2 = r2 , and let the x2 y2 equation of the ellipse be 2 + 2 = 1. a b Solving the rst and third of these equations gives 2 apa2 r2 + b4 , a2 b2 , ar x= : b2 , a2
188 Therefore, at the two points of tangency (in the right halfplane), we have a2 r2 + b4 , a2b2 = 0. From this we get 2 a = pb2b, r2 : Since the area of the ellipse is given by K = ab, we have 3 K = pbb 2 , r2 : dK = b2(2b2 , 3r2) . From this, we obtain that K is minimal Therefore db (b2 , r2 )3=2 3 r when b2 = . At this point a2 = 3b2, and from
2
r
2
2 2 e2 = a a,2 b ;
we obtain that e = . 3 Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; JORDI DOU, Barcelona, Spain; DAVID HANKIN, Hunter College Campus Schools, New York, NY, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; LOPEZ MARIA ASCENSION CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; and the proposer. Two other submissions were received that were almost correct: they used an incorrect formula for the eccentricity.
2054. [1995: 202] Proposed by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. Are there any integral solutions of the Diophantine equation , (x + y + z )3 = 9 x2 y + y 2z + z 2x other than (x; y; z ) = (n;n; n)? I. Solution by Adrian Chan, student, Upper Canada College, Toronto, Ontario. No, there are no integral solutions other than (x; y;z ) = (n; n;n). Without loss of generality, let x y and x z . Let y = x + a and z = x + b, where a and b are nonnegative integers. Then the given equation becomes (3x + a + b)3 = 9 x2 (x + a) + (x + a)2 (x + b) + (x + b)2x :
189 After expanding and simplifying, this is (a + b)3 = 9a2 b, or a3 , 6a2b + 3ab2 + b3 = 0:
(1)
Let a = kb, where k is a rational number. Then (1) becomes b3(k3 , 6k2 + 3k + 1) = 0: By the Rational Root Theorem, k3 , 6k2 + 3k + 1 = 0 does not have any rational roots. So, since k is rational, k3 , 6k2 + 3k + 1 6= 0. Therefore b = 0 and a = 0, so x = y = z. II. Solution by the proposer. Letting y = x + u and z = x + v , the equation reduces to (u + v )3 = 9u2v (2) where u and v are integers. We now show that the only solution to (2) is u = v = 0 so that (x; y;z) = (n;n; n) is the only solution of the given equation. Letting u + v = w, (2) becomes w3 = 9u2(w , u): (3) Hence w = 3w1 where w1 is an integer, and (3) is 3w13 = u2 (3w1 , u). It follows that u = 3u1 for some integer u1 , and we get w13 = 9u21 (w1 , u1): Comparing this equation to (3), we see by in nite descent that the only solution to (3) is u = w = 0, which gives the negative result. Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEEWAI LAU, Hong Kong; and HOE TECK WEE, student, Hwa Chong Junior College, Singapore. There was also one incorrect solution sent in.
2055. [1995: 202] Proposed by Herbert Gulicher, Westfalische WilhelmsUniversitat, Munster, Germany. In triangle ABC let D be the point on the ray from B to C , and E on the ray from C to A, for which BD = CE = AB , and let ` be the line through D that is parallel to AB . If M = ` \ BE and F = CM \ AB , prove that (BA)3 = AE BF CD: Solution by Toshio Seimiya, Kawasaki, Japan; essentially identical solutions were submitted by Jordi Dou, Barcelona, Spain; Mitko Christov Kunchev, Baba Tonka School of Mathematics, Rousse, Bulgaria; and Gottfried Perz, Pestalozzigymnasium, Graz, Austria.
190
C
M E F
D
`
A
B
By Menelaus' Theorem applied to triangle ACF and transverse line BEM , we have
AB FM CE = ,1: BF MC EA BD , giving FM = Since DM kBF , we have MC DC AB BD CE = ,1: BF DC EA Hence we have
AE:BF:CD = ,AB:BD:CE = BA:AB:AB = (BA)3:
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUEL OCHOA, Logro~no, Spain; P. PENNING, Delft, the NetherANGEL CABEZON lands; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer. etokrzyski, 2057?. [1995: 203] Proposed by Jan Ciach, Ostrowiec Swi
Poland. Let P be a point inside an equilateral triangle ABC , and let Ra ; Rb ; Rc and ra ; rb ; rc denote the distances of P from the vertices and edges, respectively, of the triangle. Prove or disprove that r r r a b c 1 + R 1 + R 1 + R 27 8 : a b c
Equality holds if P is the centre of the triangle. Solution by G.P. Henderson, Campbellcroft, Ontario. We will prove that the inequality is true.
191 We have Therefore
sin(PAB ) = Rrc ; sin(PAC ) = Rrb : a
cos(A) = 12 =
a
s
2
1 , Rrb2
a
From this, we have
2
1 , Rrc2 , rRb r2c : a a
Ra = p23 rb2 + rbrc + rc2; and similar expressions for Rb and Rc . q
We can assume, without loss of generality, that P is in the section of the triangle de ned by ra rb rc . Set x = rb =rc and y = ra =rc , so that 0 y x 1. The left side of the given inequality is now ! p3y ! p3x ! p3 1+ p 1+ p : (1) 1+ p 2 2 2 2 x2 + x + 1 2
y
+
y+1
2
x
+
xy + y
We will replace these factors by smaller quantities which do not involve square roots. [Ed. This is a brilliant move. I recommend that the reader draws some graphs to see how eective this is. In fact 76+,3xx is \almost" the p sevPade approximant of p 2 3 Ceby , which is 6:08409 , 1:33375x .] 2
x
We will show that, for 0 x 1,
+
x+1
:
7 + 2 53143
p3 6,x : p 2 2 x + x + 1 7 + 3x
x
This is equivalent to (1 , x)2 ,3 + 36x , 4x2 = (1 , x)2 (4x(1 , x) + 32x + 3) 0: Similarly, we have, for 0 y 1,
p3 6,y : p 2 y 2 + y + 1 7 + 3y
Replacing x by y=x( 1) in (2), we have
p3 6x , y : p 2 2 2 x + xy + y x(7x + 3y )
Using these three expression in (1), it is sucient to prove that
y (6 , x ) x (6 , y ) 6 x , y 1 + 7 + 3x 1 + 7 + 3y 1 + x(7x + 3y ) 27 8:
(2)
192 Now, this is equivalent to
f (x;y) = Py3 + Qy2 + Ry + S 0;
where
P = P (x) Q = Q(x) R = R(x) S = S(x)
= = = =
8(x , 6)(x , 3)(3x , 1); 56x4 , 672x3 + 663x2 + 427x , 504; ,504x4 + 35x3 + 714x2 , 273x , 392; 1008x4 + 423x3 , 3493x2 + 2352x:
We will prove this by showing that f is a decreasing function of y for 0 y x [for xed x] and that f (x; x) 0. We nd that f (x;x) = 10x(1 , x)2 ,8x3 , 124x2 , 35x + 196 0: [Note that 8 + 196 > 124 + 35.] It remains to show that
2 F (x; y) = @f @y = 3Py + 2Qy + R 0:
(3)
First, we note that Q = ,56x3(1 , x) , 616x(1 ,, x)2 , ,569x2 , 1043 x + 504 < 0; R = ,35x(1 , x2) , 238x , 504x2 , 714x + 392 < 0:
[Also, P 0 for 0 x 13 and P 0 for 13 x 1.] In equation (3), when P 0, all three terms are negative, and so F (x;y ) < 0. When P > 0, F (x; y) 0 provided that F (x; 0) 0 and F (x;x) 0; [since @@y2 F2 > 0]. The rst of these terms is R, which is negative. The second is (1 , x) ,,184x4 + 2336x3 , 537x2 , 1673x , 392 = (1 , x) ,,1184x4 , 537x2 (1 , x) , 1673x(1 , x2 ) ,392(1 , x3) , 266x3
0:
Also solved by MARCIN E. KUCZMA, Warszawa, Poland.
193
Ceva meets Pythagoras! K.R.S. Sastry
Dodballapur, Karnataka, India I had assumed that nobody could beat Ptolemy in providing a shortest proof of the Pythagorean theorem. To a rectangle you simply apply his \In a convex cyclic quadrilateral, the product of the diagonals equals the sum of the products of the opposite sides" theorem. In fact we can do better. With the assumed knowledge of ratios of line segments, the area measure of a polygonal region, the de nitions of trigonometrical ratios and the theorems of Ceva and Menelaus, I am going to describe a construction to you about erecting parallelograms on the sides of a triangle. That will enable us to discover a new necessary and sucient condition for the concurrency of three cevians of a triangle. Then, by treating the altitudes of a rightangled triangle as the degenerate case of three concurrent cevians, we deduce the Pythagorean theorem. Intrigued?
PRELIMINARIES.
Suppose D, E , F are points respectively on the sides BC , CA, AB of triangle ABC as in Figure 1. Let BD : DC = 1 : , and CE : EA = 1 : , but AF : FB = : 1. We let a, b, c denote the side lengths BC , CA, AB respectively. You will immediately tell me that 9
BD = 1+a ; DC = 1+a ; > > > = b b CE = 1+ ; EA = 1+ ; > > ; AF = 1+c ; FB = 1+c : >
(1)
A
F B
r
r
D r
E C
Figure 1 We both know from Ceva that the cevians AD, BE , CF concur if and only if = : (2) In the title of our discussion, Menelaus did not meet Pythagoras, but he has his role to play. So let us recall that the points D, E , F will be collinear if and only if = ,.
194
CONSTRUCTION (?).
In addition to the points D, E , F mentioned above, let us construct parallelograms outwardly on the sides BC , CA, AB having widths w1, w2, w3 respectively. These parallelograms, in general, are not similar but mutually equiangular containing an angle , 0 < < 180. Furthermore, we split each parallelogram into two by drawing parallels through the points D, E , F (see Figure 2). We use the notation M [DC; w1] for the area measure of the parallelogram having DC and w1 as side lengths.
w3 F
A E D
q
B
w2 q
q
w1
C
Figure 2 It is now a simple matter for either of us to compute the area measures of the three pairs of parallelograms, each pair sharing exactly one vertex of triangle ABC : 9
bw2 sin M [DC; w1] = aw1+1 sin > ; M [CE; w2 ] = 1+ ; > > = bw sin cw sin 2 3 M [EA; w2] = 1+ ; M [AF; w3] = 1+ ; > > ; aw1 sin : > 3 sin ; M [FB;w3] = cw1+ M [ BD; w ] = 1 1+
WHEN DO CEVIANS AD, BE , CF CONCUR?
(3)
I will provide an answer to that question by relating the area measures of the parallelograms listed in (3). Theorem 1. The cevians AD, BE , CF concur if and only if there exist mutually equiangular parallelograms of appropriate widths w1, w2, w3, depending on the values a, b, c, , , so that the construction (?) yields parallelogram pairs of equal area measure in (3).
195 Proof: Allow me to assume that AD, BE , CF are concurrent cevians. This leaves me with the task of providing consistent expressions for w1 , w2, w3 in terms of a, b, c, , . From my assumption and (2), = follows. So I will substitute = and equate parallelogram area measures of each pair in (3). Observe that the factor sin disappears from our equations:
aw1 = bw2 ; bw2 = cw3 ; 1 + 1 + 1 + 1 + cw3 aw1 1 + = 1 + ; 6= 0; ; 6= ,1:
Indeed we have the consistent equations
aw1 = bw2 = cw3 = k say: 1 + 1 + 1 + These yield, for an appropriate constant k, + ) : w1 = k(1a+ ) ; w2 = k(1 b+ ) ; w3 = k(1 c
(4)
True, we get an in nity of dissimilar parallelograms of varying widths as we vary the value of k. To establish the converse, presently I am obliged to assume that there are parallelograms of appropriate widths w1, w2, w3 so that the construction (?) leads to parallelogram pairs of equal area measure in (3). You will therefore allow me to assume the equations
aw1 = bw2 ; bw2 = cw3 ; cw3 = aw1 : 1+ 1+ 1+ 1+ 1+ 1+
But wait! The above equations imply that
aw1 = bw2 = cw3 = aw1 : 1+ 1+ 1+ 1+
Hence = and we have established the concurrency of the cevians AD, BE, CF by virtue of (2). You will agree with me that, although the proof thus far has proceeded as if , , are all positive, the same goes through should two of them be negative. So, to complete the proof of Theorem 1, I must now answer the question: What if (or ) is 0? As you can see from Figure 3, when = 0, both points D and F coincide with the vertex B . We therefore determine w1 , w2, w3 from the equations
bw2 = cw : 2 aw1 = 1bw ; 3 + 1+
196 This yields
k ; w = k(1 + ) ; w = k w1 = a 2 3 b c for some appropriate constant k. And this springs on us the following delightful surprise!
M [AB; w3] + M [BC; w1] = M [CA; w2]: A
r
F B (D)
( )
(5)
E C
Figure 3 As this equation (5) is an important result, rivalling Pythagoras, let us give it the status of a theorem. Theorem 2. Let E be a point on side AC of triangle ABC distinct from the vertex A or C . Suppose CE : EA = 1 : . Then there exist mutually equiangular parallelograms of appropriate widths w1 , w2, w3, depending on the values a, b, c, , so that the sum of the area measures of the parallelograms on the sides AB , BC equals the area measure of the parallelogram on side AC .
THE PYTHAGOREAN THEOREM.
I know your impatience is growing at an alarming rate to meet and greet Pythagoras. But I am honour bound to show you, in some speci c instance, how I actually compute the w's. For this purpose let us assume that AD, BE, CF are the altitudes of triangle ABC . Then you know BD = c cos B, DC = b cos C , , BD : DC = (c cos B) : (b cos C ) = 1 : , . Hence
cos C ; = c cos A : = bc cos B a cos C
I am very glad to learn that on using (4) you already deduced the nice conclusion w1 = ka; w2 = kb; w3 = kc: (6) For the proof of the Pythagorean theorem, look at Figure 4. If you have the triangle rightangled at B , the altitudes concur at the vertex B . So from Theorem 2, (5) and (6) we have
M [AB; kc] + M [BC; ka] = M [CA; kb]:
197
A E
B
C Figure 4
That is
kc2 sin + ka2 sin = kb2 sin ;
which is more than Pythagoras is said to have said. Speci cally, k = 1, = 90 yields the Pythagorean theorem. You may wish to establish the more general result and its converse: In triangle ABC the cosine law follows by treating its altitudes as three concurrent cevians.
CONCLUSION.
For the sake of completeness, I will state Theorem 3. It can be proved in the manner of Theorem 1. Theorem 3. The points D, E , F on the sides of a triangle ABC are collinear if and only if there exist mutually equiangular parallelograms of appropriate widths w1, w2, w3, depending on the values a, b, c, , , so that the construction (?) yields parallelogram pairs of equal area measure in (3).
ACKNOWLEDGEMENT.
The author thanks the referee for his suggestions.
198
THE SKOLIAD CORNER No. 15
R.E. Woodrow First this issue we pause to correct an error which crept into the solution given for problem 10 of the Eleventh W.J. Blundon Contest, February 23, 1994. Thanks go to Bob Prielipp for noting that we used 6 in place of 5 in the last line on page 155, thus producing a wrong answer! Here is his correction. 10. [1996: 102, 1996: 155] The Eleventh W.J. Blundon Contest. Two numbers are such that the sum of their cubes is 5 and the sum of their squares is 3. Find the sum of the two numbers. Solution by Bob Prielipp, University of Wisconsin{Oshkosh. Let the two numbers be denoted by x and y . Then x3 + y 3 = 5 and 2 x3 + y2 = 3. Since x2 + y2 = 3, (x2 +3 y2)(3x + y) = 3(x + y). Thus (x3 + y )+ xy(x +3y) =33(x + y). Because x + y = 5, xy(x + y) = 3(x + y) , 5. Hence 5 = x + y = (x + y )3 , 3xy (x + y ) = (x + y )3 , 9(x + y )+ 15, so Let x + y = s. Then
(x + y)3 , 9(x + y ) + 10 = 0:
s3 , 9s + 10 = 0 (s , 2)(s2 + 2s , 5) = 0 (s , 2)(s2 + 2s + 1 , 6) = 0 p s = 2 or s + 1 = 6: p Thus s = 2 or s = ,1 6. In this issue, we give the problems of the 14th annual American Invitational Mathematics Examination written March 28, 1996. These problems are copyrighted by the Committee on the American Mathematical Competitions of the Mathematical Association of America and may not be reproduced without permission. Solutions, and additional copies of the problems, may be obtained for a nominal fee from Professor Walter E. Mientka, C.A.M.C. Executive Director, 917 Oldfather Hall, University of Nebraska, Lincoln, NE, USA 68588{0322. As always we welcome your original \nice" solutions and generalizations which dier from the published ocial solutions.
199
14th ANNUAL AMERICAN INVITATIONAL MATHEMATICS EXAMINATION (AIME)
Thursday, March 28, 1996 1. In a magic square, the sum of the three entries in any row, column, or diagonal is the same value. The gure shows four of the entries of a magic square. Find x.
x
19
96
1
2. For each real number x, let bxc denote the greatest integer that does not exceed x. For how many positive integers n is it true that n < 1000 and that blog2 nc is a positive even integer? 3. Find the smallest positive integer n for which the expansion of (xy , 3x +7y , 21)n, after like terms have been collected, has at least 1996 terms. 4. A wooden cube, whose edges are one centimetre long, rests on a horizontal surface. Illuminated by a point source of light that is x centimetres directly above an upper vertex, the cube casts a shadow on the horizontal surface. The area of the shadow, which does not include the area beneath the cube, is 48 square centimetres. Find the greatest integer that does not exceed 1000x. 5. Suppose that the roots of x3 + 3x2 + 4x , 11 = 0 are a, b, and c, 3 and that the roots of x + rx2 + sx + t = 0 are a + b, b + c, and c + a. Find t. 6. In a veteam tournament, each team plays one game with every other team. Each team has a 50% chance of winning any game it plays. (There are no ties.) Let m=n be the probability that the tournament will produce neither an undefeated team nor a winless team, where m and n are relatively prime positive integers. Find m + n. 7. Two of the squares of a 7 7 checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane of the board. How many inequivalent color schemes are possible? 8. The harmonic mean of two positive numbers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs of positive integers (x; y ) with x < y is the harmonic mean of x and y equal to 620 ?
200
9. A bored student walks down a hall that contains a row of closed lockers, numbered 1 to 1024. He opens the locker numbered 1, and then alternates between skippingand opening each closed locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the rst closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens? 10. Find the smallest positive integer solution to 96 + sin96 : tan 19x = cos cos 96 , sin96
11. Let P be the product of those roots of z + z + z + z 6
4
3
2
+1 = 0
that have positive imaginary part, and suppose that P = r(cos + i sin ), where 0 < r and 0 < 360. Find . 12. For each permutation a1; a2; a3; : : : ; a10 of the integers 1, 2, 3, : : : ; 10, form the sum
ja1 , a2j + ja3 , a4j + ja5 , a6j + ja7 , a8j + ja9 , a10j: The average value of all such sums can be written in the form p=q , where p and q are relatively prime positive integers. Find p + q . 13. In triangle ABC , AB = p30, AC = p6, and BC = p15. There is a point D for which AD bisects BC and \ADB is a right angle. The ratio Area (4ADB ) Area (4ABC ) can be written in the form m=n, where m and n are relatively prime positive integers. Find m + n. 14. A 150 324 375 rectangular solid is made by gluing together 1 1 1 cubes. An internal diagonal of this solid passes through the interiors of how many of the 1 1 1 cubes? 15. In parallelogram ABCD, let O be the intersection of diagonals AC and BD. Angles CAB and DBC are each twice as large as angle DBA, and angle ACB is r times as large as angle AOB . Find the greatest integer that does not exceed 1000r.
201 In the last number, we gave the problems of the European \Kangaroo" Mathematical Challenge written 23 March 1995. Here are solutions. 1. B 2. E 3. C 4. E 5. C 6. C 7. C 8. C 9. B 10. D 11. B 12. 27 13. C 14. C 15. D 16. E 17. A 18. E 19. E 20. C 21. D 22. D 23. C 24. E 25. E That completes the Skoliad Corner for this issue. Thanks go to readers who have sent in contest materials and solutions. Please send me more!
Congratulations! Professor Andy Liu We are delighted to congratulate Andy Liu, Department of Mathematical Sciences, University of Alberta, Edmonton, on being awarded, this year, the prestigious David Hilbert International Award at the World Federation of National Mathematics Competitions meeting at ICME8, in Seville. Andy has a long and distinguished record in Mathematics. He has coached both the American and Canadian teams for the IMO, and also participated in the training of teams from Hong Kong and China. He is a well respected problemist. His mathematical interests span discrete mathematics, hypergraph theory, combinatorial geometry, foundations of mathematics, mathematical education and recreational mathematics. He says that the common characteristics of the research problems in which he works are that they are easy to understand, but not so easy to solve. Andy has long been associated with CRUX, and is at present the Editor of the Book Reviews section. Past recipients of the award, who are associated with CRUX, include Ed. Barbeau, Murray Klamkin and Marcin Kuczma.
202
THE OLYMPIAD CORNER No. 175
R.E. Woodrow All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. We begin with the 1996 Canadian Mathematical Olympiad which we reproduce with the permission of the Canadian Mathematical Olympiad Committee of the Canadian Mathematical Society. My thanks go to Daryl Tingley, Chair of the Committee, for forwarding the questions and promising to provide \ocial" and interesting contestant solutions.
1996 CANADIAN MATHEMATICAL OLYMPIAD 1. If , , are the roots of x , x , 1 = 0, compute 3
1+ + 1+ + 1+ : 1, 1, 1,
2. Find all real solutions to the following system of equations. Carefully justify your answer. 8 > > > > > > > > > > < > > > > > > > > > > :
4x2 = y; 1 + 4x2 4y2 = z; 1 + 4y 2 4z 2 = x: 1 + 4z 2
3. We denote an arbitrary permutation of the integers 1; : : : ; n by a1 ; : : : ; an. Let f (n) be the number of these permutations such that (i) a1 = 1; (ii) jai , ai+1 j 2, i = 1; : : : ; n , 1. Determine whether f (1996) is divisible by 3. 4. Let 4ABC be an isosceles triangle with AB = AC . Suppose that the angle bisector of \B meets AC at D and that BC = BD + AD. Determine \A.
203
5. Let numbers P r1 ; r2 ; : : : ; rm be a given set of m positive rational Pm such that m r = 1 . De ne the function f by f ( n ) = n , k=1 k k=1 [rk n] for each positive integer n. Determine the minimum and maximum values of f (n). Here [x] denotes the greatest integer less than or equal to x. The next set of problems are from the twenty fth annual United States of America Mathematical Olympiad written May 2, 1996. These problems are copyrighted by the Committee on the American Mathematical Competitions of the Mathematical Association of America and may not be reproduced without permission. Solutions, and additional copies of the problems, may be obtained for a nominal fee from Professor Walter E. Mientka, C.A.M.C. Executive Director, 917 Oldfather Hall, University of Nebraska, Lincoln, N.E., USA 68588{0322. As always we welcome your original, \nice" solutions and generalizations which dier from the published ocial solutions.
25th UNITED STATES OF AMERICA MATHEMATICAL OLYMPIAD Part I
9 a.m.  12 noon May 2, 1996
1. Prove that the average of the numbers n sin n (n = 2, 4, 6, : : : ; 180) is cot 1 . 2. For any nonempty set S of real numbers, let (S) denote the sum
of the elements of S . Given a set A of n positive integers, consider the collection of all distinct sums (S ) as S ranges over the nonempty subsets of A. Prove that this collection of sums can be partitioned into n classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2. 3. Let ABC be a triangle. Prove that there is a line ` (in the plane of triangle ABC ) such that the intersection of the interior of triangle ABC and the interior of its re ection A0 B 0 C 0 in ` has area more than 2=3 the area of triangle ABC . Part II 1 p.m.  4 p.m. May 2, 1996
4. An nterm sequence (x1; x2 ; : : : ; xn) in which each term is either 0 or 1 is called a binary sequence of length n. Let an be the number of binary sequences of length n containing no three consecutive terms equal to 0, 1, 0 in that order. Let bn be the number of binary sequences of length n that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that bn+1 = 2an for all positive integers n.
204
5. Triangle ABC has the following property: there is an interior point P such that \PAB = 10, \PBA = 20, \PCA = 30, and \PAC = 40 . Prove that triangle ABC is isosceles. 6. Determine (with proof) whether there is a subset X of the integers with the following property: for any integer n there is exactly one solution of a + 2b = n with a; b 2 X . As a third Olympiad for this issue we give the problems of the Italian Mathematical Olympiad written May 6, 1994. My thanks go to Richard Nowakowski who collected this (and many others) when he was Canadian Team Leader to the International Mathematical Olympiad in Hong Kong.
ITALIAN MATHEMATICAL OLYMPIAD May 6, 1994
1. Show that there exists an integer N such that for all n N a square can be represented as a union of n pairwise nonoverlapping squares. 2. Find all integer solutions of the equation y2 = x3 + 16:
3. A journalist wants to report on the island of scoundrels and knights, where all inhabitants are either scoundrels (and they always lie) or knights (and they always tell the truth). The journalist interviews each inhabitant exactly once, and he gets the following answers: A1: on this island there is at least one scoundrel; A2: on this island there are at least two scoundrels; ::: An,1: on this island there are at least n , 1 scoundrels; An: on this island everybody is a scoundrel.
Can the journalist decide whether there are more scoundrels or more knights? 4. Let r be a line in the plane and let ABC be a triangle contained in one of the halfplanes determined by r. Let A0 , B 0 , C 0 be the points symmetric to A, B , C with respect to r; draw the line through A0 parallel to BC , the line through B 0 parallel to AC and the line through C 0 parallel to AB . Show that these three lines have a common point. 5. Let OP be a diagonal of the unit cube. Find the minimum and the maximum value of the area of the intersection of the cube with a plane through OP . 6. On a 10 10 chessboard the squares are labelled with the numbers 1; 2; : : : ; 100 in the following way: the rst row contains the numbers 1; 2; : : : ; 10 in increasing order from left to right, the second row contains the numbers 11; 12;: : : 20 in increasing order from left to right, ... , the last row
205 contains the numbers 91, 92, : : : , 100 in increasing order from left to right. Now change the signs of fty numbers in such a way that each row and each column contains ve positive and ve negative numbers. Show that after this change the sum of all numbers on the chessboard is zero. Before turning to reader's solutions to problems from the 1995 numbers of Crux, I want to give some reactions to material from the Corner. First, comments on the 1995 Canadian Mathematical Olympiad [1995: 190; 1995: 223{226] by Murray S. Klamkin, The University of Alberta. 2. Let a, b, and c be positive real numbers. Prove that
aa bbcc (abc)(a+b+c)=3:
Comment. This same exact problem appeared in the 1974 U.S.A. Mathematical Olympiad [1]. In the solution given, it was also noted that since ln xx is convex for x > 0, it follows more generally and immediately by Jensens's inequality that
aa11 aa22 aan (a1a2 an)(a1+a2++a )=n: 3. Let n be a xed positive integer. Show that for any nonnegative integer k, the Diophantine equation x31 + x32 + + x3n = y3k+2 has in nitely many solutions in positive integers xi and y . n
n
Comment. The following problem, which implies the solution, appeared in the 1985 U.S.A. Mathematical Olympiad [1]: Determine whether or not there are any positive integral solutions of the simultaneous equations
x21 + x22 + + x21985 = y3; x31 + x32 + + x31985 = z2; with distinct integers x1 ; x2 ; : : : ; x1985 . It was shown that there are in nitely many solutions. The solution given easily extends to take care of the case if z 2 is replaced by z 3k+2. For a single equation, we have more generally that
xa11 + xa22 + xan = ya has in nitely many positive solutions if a is relatively prime to all the ai 's. n
We just let
X
xi = uif uai gkp=a where p = i
i
Y
ai so that
X
X
xai = f uai gkp+1: i
i
206 Then we can take y = f uai gl . Finally, since p and a are relatively prime, intgers k and l exist such that kp + 1 = al. Much more general results are to appear in the proceedings of the Belgian Royal Academy. In view of the above comments on previously published problems, the Canadian Olympiad Examination Committee should be more careful about the problems they set in their competitions. Reference [1] M.S. Klamkin, U.S.A. Mathematical Olympiads 1972{1986, Mathematical Association of America, Washington, D.C., 1988, pp. 81, 33{34. P
i
Next we give a comment by Murray S. Klamkin of the University of Alberta about one of the solutions published in the Corner. 2. [1994: 39; 1995: 193{194] 1992 Czechoslovak Mathematical Olympiad, Final Round. Let a, b, c, d, e, f be lengths of edges of a given tetrahedron and S its surface area. Prove that
p S ( 3=6)(a2 + b2 + c2 + d2 + e2 + f 2):
Comment by Murray S. Klamkin, University of Alberta. It is just as easy as the previous solution to obtain the stronger inequality
p S ( 3=4)f(def )2=3 + (abd)2=3 + (bce)2=3 + (caf )2=3g:
This follows from the known inequality
p F ( 3=4)(uvw)2=3 for the area F of a triangle of sides u, v , w, and which is equivalent to
the equilateral triangle being the triangle of maximum area which can be inscribed in a given circle. The rest follows from the AM{GM inequality
(uvw)2=3 3(u2 + v 2 + w2):
207 The next two solutions are in response to challenges to the readership \to ll the gaps" in solutions to problem sets discussed in 1995 and 1996 numbers of the Corner. My thanks to members of the Singapore team to the 35th IMO. 4. [1994: 64{65; 1995: 274] 9th Balkan Mathematical Olympiad (Romania). For every integer n > 3 nd the minimum positive integer f (n) such that every subset of the set A = f1; 2; 3; : : : ; ng which contains f (n) elements contains elements x; y; z 2 A which are pairwise relatively prime. Solution by the joint eorts of Siu Taur Pang and Hoe Teck Wee, silver medallists on the Singapore team to the 35th IMO. Let Tn = ft j t n; 2 j t or 3 j tg. Therefore, among any three elements of Tn , two have a common factor (either 2 or 3). Hence, f (n) jTn j + 1. For 4 n 24 < 52, there are jTn j , 2 composite numbers in the set An = f1; 2; 3; : : : ; ng since every composite number in An is a multiple of 2 or 3. Hence, there exist three irreducible elements (either 1 or a prime) in any (jTn j + 1)element subset of An , which are in pairs relatively prime. Therefore, we have, for 4 n 24, f (n) = jTn j + 1. By the Principle of Inclusion and Exclusion, jTn j = n2 + n3 , n6 ; n n f (n) = + , n + 1:
() Assume that () also holds for some integer k > 3. Consider any (jTk+6 j+1)element subset B of Ak+6 . If at least f (k) elements of B lie in Ak , then there exist three elements in B which are pairwise relatively prime. Otherwise, since jTk+6 j = Tk + 4, at least ve elements of B are from the veelement set C = fk + 1; k + 2; : : : ; k + 6g. Note that the dierence between any two elements of C is at most 5, so the greatest common divisor between any two elements of C is 1; 2; 3; 4 or 5. If any three elements of B \ C are 2
3
6
odd, then there are three elements in B which are pairwise relatively prime, namely these three odd elements since the dierence between any two of these three elements is either 2 or 4 which have no odd factor larger than 1. Otherwise, there are two odd numbers and three even numbers in B \ C . The dierence between any two of these even numbers is either 2 or 4, so at most one of these three numbers is divisible by 3, and at most one is divisible by 5. Hence, there exists an even number in B \ C which is neither divisible by 3 nor 5. Hence, this even number and the two odd numbers are pairwise relatively prime. Thus, f (k + 6) = jTk+6 j + 1, so () holds for k + 6. Therefore, we conclude by the Principle of Mathematical Induction that () holds for all integers n > 3.
208
5. [1994: 129{130; 1996; 24{27] Final Round of the 43rd Mathematical Olympiad in Poland. The regular 2ngon A1 ; A2 ; : : : ; A2n is the base of a regular pyramid with vertex S . A sphere passing through S cuts the lateral edges SAi in the respective points Bi (i = 1; 2; : : : ; 2n). Show that n X
i=1
SB2i,1 =
n X
i=1
SB2i:
Solution by Hoe Teck Wee, silver medallist on the Singapore team at the 35th IMO. It can be proved easily, using vectors that for any regular polygon A1A2 : : : A2n with centre C (where n > 1 is a positive integer) and any point P in space, that n X
j =1
PA j,1 = 2 2
n X
j =1
PA22j ;
by writing PAi PAi = CP CP + CAi CAi , 2CAi CP (i = 1; 2; : : : ; 2n), and using the identity n X
j =1
CA2j,1 =
n X
j =1
CA2j :
Let O denote the center of the sphere passing through S and R denote its radius. Note that we may extend each of the lateral edges so that OAi > R for i = 1; 2; : : : 2n. Therefore, by considering the power of Ai, we have
OA2i , R2 = SAi AiBi = SAi (SAi , SBi) = SA2i , SAi SBi: n X
(SA22j ,1 , SA2j,1 SB2j,1 ) =
j =1
Taking P = S , we have n X
j =1
SA22j,1 =
n X j =1
SA22j )
n X j =1
n X
(SA22j , SA2j SB2j ):
j =1
SB2j,1 =
n X j =1
SB2j :
209 We now turn to reader's solutions to problems from the 1995 numbers of the corner. We begin with the First Stage Exam of the 10th Iranian Mathematical Olympiad [1995: 8{9]. 1. Find all integer solutions of
1 + 1 , 1 = 3: m n mn2 4
Solutions by Mansur Boase, student, St. Paul's School, London, England; Christopher J. Bradley, Clifton College, Bristol, UK; Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany; Cyrus Hsia, student, University of Toronto, Toronto, Ontario; Stewart Metchette, Gardena, California, USA; Michael Selby, University of Windsor, Windsor, Ontario; and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Hsia's writeup, although most solutions were similar. 1 1 3 1 m + n , mn2 = 4 . Note m; n 6= 0. Then
n2 + mn , 1 = 3 mn2 4
giving
2 4(n + 1)(n , 1) : m = n4((3nn ,, 1) = 4) n(3n , 4) Now (n + 1; n) = 1 = (n , 1; n), i.e. n is relatively prime to both n + 1 and n , 1. Since m is an integer, n j 4(n + 1)(n , 1), giving n j 4. Thus n = 1; 2; 4. For n = 1; m = 0 which is impossible ,1)(,3) ; not an integer n = ,2; m = 4( (,2)(,10) n = 2; m = 3 ,3)(,5) ; not an integer n = ,4; m = 4( (,4)(,16)
m = 4(5)(3) 4 8 ; not an integer. Thus the only solution is (n; m) = (2; 3). n = 4;
210
2.
Let X be a set with n elements. Show that the number of pairs
(A; B) such that A, B are subsets of X , A is a subset of B , and A 6= B is equal to: 3n , 2n : Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Mansur Boase, student, St. Paul's School, London, England; Christopher J. Bradley, Clifton College, Bristol, UK; Hans Engelhaupt, Franz{Ludwig{ Gymnasium, Bamberg, Germany; Cyrus Hsia, student, University of Toronto, Toronto, Ontario; Joseph Ling, University of Calgary, Calgary, Alberta; Bob Prielipp, University of Wisconsin{Oshkosh, Wisconsin, USA; Michael Selby, University of Windsor, Windsor, Ontario; and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Boase's argument. The number of subsets of an n element set is 2n . Thus the number of pairs (A; B ) with A = B is 2n . As a result we need to show that the number of pairs (A; B ) with A a subset of B , (possibly equal to B ), is 3n. We do so by induction on n. If n = 1, let X = fa1 g. The pairs (A; B ) are (;; ;), (;; X ), (X;X ). Assume, by induction, that for k elements the number of pairs is 3k . Let X = fa1 ; : : : ; ak ; ak+1 g. By hypothesis there are 3k pairs (A; B ) with A B fa10 ; : :0: ; akg. Each of these can be extended in three ways to yield a pair (A ; B ) with A0 B 0 fa1 ; : : : ; ak ; ak+1 g as follows: (i) A0 = A, B 0 = B (ii) A0 = A [ fak g, B 0 = B [ fak g (iii) A0 = A, B 0 = B [ fak g. Obviously no pairs are repeated, so the number of pairs (A0 ; B 0 ) is 3 3k = 3k+1 . This completes the induction, as required. 4. Let a, b, c, be rational and one of the roots of ax3 + bx + c = 0 be equal to the product of the other two roots. Prove that this root is rational. Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Mansur Boase, St. Paul's School, London, England; Christopher J. Bradley, Clifton College, Bristol, UK; Paul Colucci, student, University of Illinois; Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany; Cyrus Hsia, student, University of Toronto, Toronto, Ontario; Michael Selby, University of Windsor, Windsor, Ontario; and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Amengual Covas's solution (which was similar to the others). Let r1, r2 , r3 , be the roots of the given cubic equation and also let r1 = r2r3. Then r1 + r2 + r3 = 0; (1)
211
r1(r2 + r3) + r1 = r1r2 + r1r3 + r2r3 = ab ;
(2)
r1r2r3 = r12 = , ac :
(3)
(Note that a 6= 0 for the existence of three solutions.) From (1) r2 + r3 = ,r1 and substituting for r2 + r3 in (2) we nd that
,r12 + r1 = ab ;
or equivalently
r1 = ab + r12 : Finally, substituting ,ac for r12 from (3), we obtain r1 = b,a c , which is
rational. 5. Find all primes p such that (2p,1 , 1)=p is a square. Solutions by Christopher J. Bradley, Clifton College, Bristol, UK; Stewart Metchette, Gardena, California, USA; Michael Selby, University of Windsor, Windsor, Ontario; and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Bradley's solution. Lemma 1. The only solution in natural numbers of 2n = x2 , 1 is n = 3, x = 3.n 2 Proof. 2 = x , 1 = (x , 1)(x + 1). The factors of 2n that dier by 2 are 22 and 21 with n = 3 and no others. Lemma 2. The only solutions in natural numbers of 2n = x2 + 1 are x = 0, n = 0, n = 1, x = 1. Proof. The cases mentioned apart, n 2 and the lefthand side is 0 mod 4. Now x2 0; 1 mod 4, so x2 + 1 1; 2 mod 4. Contradiction establishes the result. Let p = 2. Then (2p,1 , 1)=2 = 12 , so p must be odd. Put p = 2k + 1. ,1 2 ,1 Then 2 p ,1 = 22k+1 which is an integer by Fermat's Little Theorem. Now (2 +1)(2 ,1) 22 ,1 = 2k+1 , and since 2k + 1 is prime it must be the case that 2k+1 2k + 1 j 2k + 1 or 2k + 1 ,j 12k , 1. Indeed 2k + 1 and 2k , 1 are coprime, so the only ways in which 2 p ,1 can be a perfect square is for either 22k+1 +1 and 2k , 1 to be perfect squares or 2k + 1 and 22k,+11 to be both perfect squares. Case 1. 2k , 1 a perfect square. By Lemma 2 we have k = 0 or k = 1. Then 22k+1 = 2 (k = 0) or 1 (k = 1). So k = 1, p = 3 gives a solution. +1 Case 2. 2k + 1 a perfect square. By Lemma 1 we have k = 3. Then 2 ,1 7 2k+1 = 7 = 1, so p = 7 gives a solution, and there can be no others. p
k
k
k
k
p
k
k
k
k
212
6. Let O be the intersection of diagonals of the convex quadrilateral ABCD. If P and Q are the centres of the circumcircles of AOB and COD show that PQ AB + CD : 4
Solution by D.J. Smeenk, Zaltbommel, the Netherlands.
C
Q2 Q
D
Q1
B
R1 O
R R2
A Let P1 , Q1 , P2 , Q2 be the midpoints of OB , OD, OA and OC respec
tively. Then
From this, we get
PQ P1Q1 ) PQ 12 BD; PQ P2Q2 ) PQ 12 AC:
1 (BD + AC ) 4 1 (OB + OD + OA + OC ) 4 1 [(OB + OA) + (OC + OD)]: 4 Now OB + OA > AB and OC + OD > CD. Thus PQ 14 (AB + CD).
PQ
That completes the Corner for this issue. Please send me your nice solutions as well as Olympiad contests for use in the future.
213
THE ACADEMY CORNER No. 4 Bruce Shawyer
All communications about this column should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7 In the February 1996 issue, we gave the rst set of problems in the Academy Corner. Here we present solutions to the rst three questions, as sent in by Sefket Arslanagic, Berlin, Germany.
Memorial University Undergraduate Mathematics Competition 1995 1. Find all integer solutions of the equation x4 = y 2 + 71. Solution. From the equation x4 = y 2 + 71, we get
(x2 , y)(x2 + y ) = 71 1:
Since 71 is prime, we have (
x2 , y = 1 I x2 + y = 71
or II
so that
2x2 = 72,
or
(
x2 , y = 71 x2 + y = 1;
2x2 = 72,
giving
x1;2 = 6, y1;2 = 35, x3;4 = 6 y3;4 = ,35. That is,
(x; y) 2 f(6; 35); (,6; 35); (6; ,35); (,6; ,35)g: The other possibility is (x2 , y )(x2 + y ) = (,71) (,1), which gives: x = 6i 2= Z.
Thus we have found all solutions. 2. (a) Show that x2 + y 2 2xy for all real numbers x, y . (b) Show that a2 + b2 + c2 ab + bc + ca for all real numbers a, b, c.
214 Solution. (a) x2 + y 2 2xy , (x , y )2 0 for all x; y 2 R. (b) From (a), we get
that is or
a2 + b2 2ab; a2 + c2 2ac; b2 + c2 2bc;
2(a2 + b2 + c2 ) 2(ab + ac + bc)
a2 + b2 + c2 ab + ac + bc; for all a; b 2 R: Remark: Let a, b, c 2 C . We observe that the equality x3 , 1 = 0; that is, (x , 1)(x2 + x + 1) = 0 p 6 x2 =6 x3 . Let has roots x1 = 1, x2;3 = 12 (,1 i 3), so that x1 = a1 = x1, p b = x21, c = x3p, giving1a + b + c = 0, ab2 + ac2 + bc2 = (,1 + i 3) + 2 (,1 , i 3) + 4 (1 + 3) = 0 and a + b + c = 2 (a + b + c)2 , 2(ab + ac + bc) = 0. (i) Further, in (b) the equality holds because
a2 + b2 + c2 = ab + ac + bc , 12 [(a , b)2 +(a , c)2 +(b , c)2] = 0 if and only if a = b = c 2 R. (ii) Also, (b) holds as well, since a2 + b2 + c2 = ab + ac + bc for a; b; c 2 C and a = 6 b =6 c.
3. Find the sum of the series
1 + 2 + 3 + 4 + : : : + 99 : 2! 3! 4! 5! 100!
Solution. We have
Thus
1, 1 =1, 1 = k : k! (k + 1)! k! (k + 1) (k + 1)!
1 + 2 + 3 + 4 + ::: + n = 1, 1 : 2! 3! 4! 5! (n + 1)! (n + 1)! For n = 99, we get 1 2 3 4 99 1 2! + 3! + 4! + 5! + : : : + 100! = 1 , 100! :
215
BOOK REVIEWS Edited by ANDY LIU She Does Math! reallife problems from women on the job, edited by Marla Parker. Published by the Mathematical Association of America, 1995, paperbound, 272+ pages, ISBN# 0883857022, list price US$24.00. Reviewed by Katherine Heinrich, Simon Fraser University. Thirty eight women working in a variety of careers (engineering, mathematics, computing, biology, management) describe their lives, their feelings about mathematics and the work they currently do. In all cases, competence, con dence and skill in mathematics are essential to their work. Shelley J. Smith avoided math and chose archaeology because it required no math. When she reached graduate school she knew it was needed (to compute shapes of pottery from broken pieces and to do statistical analysis). She overcame her fear and discovered not only that she could do math but that she also enjoyed it. These are \regular" women who, on discovering what they wanted to do, pursued it. They met the usual challenges: of changing goals (Donna McConnaha Sheehy started in commercial art but after a workstudy experience in the Forest Service, moved to civil engineering); of having children (Beth MacConnell, with a new baby, still had to rise at 4 a.m. to track grizzly bears); and of taking care of parents (Nancy G. Roman retired from her job as astronomer at NASA to take care of her mother) to mention a few. Their stories are written for young women of all ages: telling them it is possible to be successful and to have interesting and challenging jobs; and that the rst step is to develop math skills and to have con dence in one's abilities. With each story the author presents problems based on either her work or her other interests. All are mathematical. The problems range from the very simple (formatted stock prices) to the far too obscure and dicult (determining prism diopters, determining a star's perigalactic distance). Many were confusing with important aspects unde ned { they could be answered only after one rst studied the solution. Many were simply substitutions into an unexplained formula  these same problems could often be solved by careful thought without the formula. Some required formulas and ideas from physics which were presented only in the solutions, and then often without explanation. Some were simple arithmetic and others required calculus. On the other hand, there were some wellwritten problems that encouraged the reader to explore and search for understanding: such problems were regrettably in the minority.
216 Perhaps the greatest diculty I had was in trying to understand who the book was written for. I fear that in trying to reach all young women, regardless of math background or feelings about math, it will fail them all. I hope that any future edition will begin by targeting the audience and then ensuring that the problems and their solutions are written speci cally for them.
Mathematical Literacy 1. 2.
3. 4. 5.
Here are the answers to the questions posed in the February 1996 issue. Who thought that the binary system would convince the Emperor of China to abandon Buddhism in favour of Christianity? Leibnitz Who asked which king for one grain of wheat for the rst square of a chess board, two grains for the second square, four grains for the third square, and so on? Grand Vizier Sissa Ben Dahir asked King Shirh^am In which wellknown painting, by whom, does a Magic Square appear? Melancholia by Durer Where was bread cut into \Cones, Cylinders, Parallelograms, and several other Mathematical Figures"? Laputa  (Gulliver's Travels) Which mathematician said: \Philosophers count about twohundred and eighty eight views of the sovereign good"? Pascal
217
PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be typewritten or neatly handwritten, and should be mailed to the EditorinChief, to arrive no later that 1 March 1997. They may also be sent by email to cruxeditor@cms.math.ca. (It would be appreciated if email proposals and solutions were written in LATEX, preferably in LATEX2e). Graphics les should be in epic format, or plain postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication. 2151. Proposed by Toshio Seimiya, Kawasaki, Japan. 4ABC is a triangle with \B = 2\C . Let H be the foot of the perpendicular from A to BC , and let D be the point on the side BC where the excircle touches BC . Prove that AC = 2(HD). 2152. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. n X Let n 2 and 0 x1 : : : xn 2 be such that sin xk = 1. k=1 Consider the set Sn of all sums x1 + : : : + xn . 1. Show Sn is an interval. 2. Let ln be the length of Sn. What is nlim !1 ln ? 2153. Proposed by Sefket Arslanagic, Berlin, Germany.
Suppose that a, b, c 2 R. If, for all x 2 [,1; 1], jax2 + bx + cj 1, prove that
jcx2 + bx + aj 2:
2154. Proposed by K.R.S. Sastry, Dodballapur, India.
218 In a convex pentagon, the medians are concurrent. If the concurrence point sections each median in the same ratio, nd its numerical value. (A median of a pentagon is the line segment between a vertex and the midpoint of the third side from the vertex.) 2155. Proposed by Christopher J. Bradley, Clifton College, Bristol, UK. Prove there is no solution of the equation
1 + 1 = 1 x2 y8 z2
in which y is odd and x; y; z are positive integers with highest common factor 1. Find a solution in which y = 15, and x and z are also positive integers. 2156. Proposed by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore. ABCD is a convex quadrilateral with perpendicular diagonals AC and BD. X and Y are points in the interior of sides BC and AD respectively such that
BX = BD = DY : CX AC AY BC XY : BX AC
Evaluate
2157. Proposed by Sefket Arslanagic, Berlin, Germany. Prove that 2 , 1 is exactly divisible by 1997 . 2158. Proposed by P. Penning, Delft, the Netherlands. 1997 1996
2
Find the smallest integer in base eight for which the square root (also in base eight) has 10 immediately following the `decimal' point. In base ten, the answer would be 199, with sqrt (199) = 14:10673 : : : . 2159. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Let the sequence fxn ; n 1g be given by
xn = n1 (1 + t + : : : + tn,1)
where t > 0 is an arbitrary real number. Show that for all k, l 1, there exists an index m = m(k;l) such that xk xl xm . 2160. Proposed by Toshio Seimiya, Kawasaki, Japan. 4ABC is a triangle with \A < 90. Let P be an interior point of ABC such that \BAP = \ACP and \CAP = \ABP . Let M and N be the
219 incentres of 4ABP and 4ACP respectively, and let R1 be the circumradius of 4AMN . Prove that
1 = 1 + 1 + 1 : R1 AB AC AP
2161. Proposed by JuanBosco Romero Marquez, Universidad de Valladolid, Valladolid, Spain. Evaluate 1 1 : n=1 (2n , 1)(3n , 1) X
2162. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.
In 4ABC , the Cevian lines AD, BE , and CF concur at P . [XY Z ] is the area of 4XY Z . Show that
[DEF ] = PD PE PF 2[ABC ] PA PB PC
2163. Proposed by Theodore Chronis, student, Aristotle University of Thessaloniki, Greece. Prove that if n;m 2 N and n m2 16, then 2n nm. Corrections. 2139. Proposed by Waldemar Pompe, student, University of Warsaw, Poland. Point P lies inside triangle ABC . Let D, E , F be the orthogonal projections from P onto the lines BC , CA, AB , respectively. Let O0 and R0 denote the circumcentre and circumradius of the triangle DEF , respectively. Prove that p q [ABC ] 3 3R0 R0 2 , (O0P )2;
where [XY Z ] denotes the area of triangle XY Z . p p Note: 3 3R0 instead of 3 3R0 , as was printed in the May 1996 issue. Also, problem 2149 was ascribed to JuanBosco Morero Marquez instead of JuanBosco Romero Marquez.
220
SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems. Some solvers are reminded about the instructions given in the Problems section above: : : : please send your proposals and solutions on signed and separate : : : sheets of paper. The emphasis here is on \signed". The secretarial sta may not notice if a submitted solution is not signed, and when it comes time for an editor to read the le, there is no way of knowing who the anonymous person is!
1823. [1993: 77; 1994: 54] Proposed by G. P. Henderson, Campbellcroft, Ontario. A rectangular box is to be decorated with a ribbon that goes across the faces and makes various angles with the edges. If possible, the points where the ribbon crosses the edges are chosen so that the length of the closed path is a local minimum. This will ensure that the ribbon can be tightened and tied without slipping o. Is there always a minimal path that crosses all six faces just once? Correction and editor's comment by Bill Sands. The previous editor's remark on [1994: 55] that Jordi Dou's second example of a minimal path, namely a
b
"" " "" " " "" c
"" " "" " " b "" a
c a
does not result in any further boxes that can be decorated, is false. As Gerd Baron has since pointed out, and Dou and the proposer knew all along, there are boxes that can be decorated in the above way but whose sides a; b; c do not satisfy the triangle inequality. For example, a = 4, b = 2, c = 1 yield
221 with the ribbon as shown. The following is derived from Baron's and the proposer's correct solutions for this problem. From the rst gure above, one can see that such a ribbon exists in this case if and only if
b
b+c < b ; < 2a + c 2a + b + c c
where the middle fraction is the slope of the slanted line. This simpli es to c2 , b2 + 2ac > 0 and b2 , c2 + 2ab > 0: By permuting a; b; c we obtain two other sucient conditions for an a by b by c box to allow a ribbon: a2 , c2 + 2ab > 0 and c2 , a2 + 2bc > 0; and b2 , a2 + 2bc > 0 and a2 , b2 + 2ac > 0: Moreover, these (and the previous solution) exhaust all possible ways for a ribbon to cross all six faces of the box. If we assume that a b c, then an a by b by c box will not possess a ribbon satisfying the condition of the problem exactly if the following three inequalities all hold:
b2 , a2 + 2bc 0;
c2 , b2 + 2ac 0;
b + c a:
For example, the 4 by 3 by 1 box does not have a ribbon. So in retrospect, only Baron, Dou and the proposer sent in complete solutions of this problem. The previous editor (that is, me, Bill Sands) thanks Baron and the proposer for calling attention to my too{hasty reading of the solutions the rst time around.
2044. [1995: 158] Proposed by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. Suppose that n m 1 and x y 0 are such that xn+1 + yn+1 xm , ym:
Prove that xn + y n 1. A typographical error in the LATEX codes used in typesetting resulted in Toshio Seimiya, Kawasaki, Japan being listed as the solver. The solver was indeed HeinzJurgen Seiert, Berlin, Germany. The editor oers his apologies to HeinzJurgen Seiert for any embarrassment caused by this error.
222
2048. [1995: 158] Proposed by Marcin E. Kuczma, Warszawa, Poland.
Find the least integer n so that, for every string of length n composed of the letters a; b; c; d; e; f; g;h; i; j; k (repetitions allowed), one can nd a nonempty block of (consecutive) letters in which no letter appears an odd number of times. Solution by Douglas E. Jackson, Eastern New Mexico University, Portales, New Mexico, USA. Let S be a string of length 211 = 2048 of the 11 letters, a, b, : : : , k. For i = 1; ;, : : : , 2048, let na(i), nb(i), : : : , nk(i) be the number of occurrences mod 2 of the letters a, b, : : : , k respectively in the rst i positions of S . Then, for each position i in S , we have an associated binary string, n(i) = na(i)nb(i) : : : nk(i), for length 11. If, for some i, we have n(i) = 00 : : : 0, then in the rst i positions of S , each letter appears an even number of times. If no position is associated with the zero bit string, then, since there are only 2047 nonzero bit strings of length 11, there must exists i and j such that 1 i j 2048 and n(i) = n(j ). [The Pigeonhole Principle one more time!  Ed.] Then each letter appears an even number of times in positions i + 1 through j of S. Therefore, any string on 11 letters with length at least 2048 will have a block in which no letter appears an odd number of times. In the following sequence of strings, each string is constructed by taking two copies of the previous string and inserting a new letter between these copies. This construction allows a simple inductive argument, that in these strings, every nonempty block contains some letter an odd number of times. Let B be a block in one of the strings of this sequence, other than the rst. If B contains the central (new) letter, then obviously it contains that letter an odd number of times (once). Otherwise, B must be a block of the previous string, and hence is inductively assumed to contain some letter an odd number of times.
a aba aba c aba abacaba d abacaba :::
Clearly string number 11 in this sequence has 11 letters and length 2047. So the minimum value of n which forces the required property is 2048. Also solved by CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; ASHSIH KR. SINGH, student, Kanpur, India; HOE TECK WEE, student, Hwa Chong Junior College, Singapore (all with virtually identical arguments to the one given above); and the proposer. Wee actually considered the general case when the number of letters is k instead of 11. Using the same argument, he showed that the answer is 2k .
223
2052. [1995: 202] Proposed by K.R.S. Sastry, Dodballapur, India.
The in nite arithmetic progression 1 + 3 + 5 + 7 + : : : of odd positive integers has the property that all of its partial sums
1; 1 + 3; 1 + 3 + 5; 1 + 3 + 5 + 7; : : :
are perfect squares. Are there any other in nite arithmetic progressions, all terms positive integers with no common factor, having this same property? Solution by Toby Gee, student, the John of Gaunt School, Trowbridge, England Let the nth term of the arithmetic progression be a + (n , 1)d, and let the sum of the rst n terms be
1)d : Sn = an + n(n , 2 Let p be any prime and consider n = 2p. Then, since S2p is a perfect square, we have S2p = 2ap + p(2p , 1)d = y2 for some integer y . Thus p j y . Let y = px. Then 2a + 2pd , d = px2 0 (mod p) Therefore, p j (2a , d). For all suciently large primes p we have p > 2a , d, implying 2a = d, whence Sn = an + an(n , 1) = an2. Since these numbers have no common factor, we must have a = 1, so there are no other such
progressions. Also solved by CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; JEFFREY K. FLOYD, Newnan, Georgia, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; MICHAEL PARMENTER, Memorial University of Newfoundland, St. John's, Newfoundland; HEINZ JURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary; and the proposer. There were two incorrect solutions submitted. Remark by Christopher J. Bradley, Clifton College, Bristol, UK. The correct generalization of the
1
1+3
1+3+5
1+3+5+7
:::
phenomenon is to be found not in the problem stated, but rather by producing the partial sums of an arithmetic progression such as
15 15 + 33 15 + 33 + 51 15 + 33 + 51 + 69 (a = 15, d = 18) and then adding 1 providing 16 49 100 169 : : : and then tacking on 1 at the beginning giving
12; 42; 72 ; 102; 132 ; : : :
:::
224 The integers whose squares appear are always in arithmetic progression themselves. It always works. For example, if we want to do the reverse for
12; 52; 92 ; 132; 172 ; : : :
we look at 24, 80, 168, 288, etc. and perceive this to be
24; 24 + 56; 24 + 56 + 88; 24 + 56 + 88 + 120; : : : which are the partial sums for a = 24, d = 32. Proofs covering these remarks are straightforward. In the above, d is twice a perfect square and a = 21 d + p 2d. It also works when squares other than 1 (or 12) are tacked on provided
a and d are carefully chosen.
BRADLEY mentions that he has seen this in some magazine before, but does not recall when or where. Perhaps a reader can supply us with the proper reference.
2056. [1995: 203] Proposed by Stanley Rabinowitz, Westford, Massachusetts. Find a polynomial of degree ve whose roots are the tenth powers of the roots of the polynomial x5 , x , 1. I Solution by Christopher J. Bradley, Clifton College, Bristol, UK. If x5 , 1 = x has roots i (i = 1; 2; : : : ; 5), then the substitution y = x5 will provide an equation with roots 5i . Now, (x5 , 1)5 = x5 , so the 5 equation required is (y ,1) = y ; that is, y 5 ,5y 4 +10y 3 ,10y 2 +4y ,1 = 0. If we now put u = y 2, the resulting equation in u will have roots 10 i . 4 2 4 2 Since y (y + 10y + 4) = 5y + 10y + 1, we have, by squaring, that u(u2 + 10u + 4)5 2 = (54 u2 +310u + 1)2 2. Multiplying out and collecting like terms, we get u , 5u + 8y , 30u , 4u , 1 = 0, and the polynomial on the left of this equation is the answer. II Solution by Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USA. Let ri (1 i 5) be the roots of the given polynomial. Since x5 = x + 1, we have ri10 = (ri + 1)2. It is easy to construct a polynomial with roots ri + 1; it is f (x) = (x , 1)5 , (x , 1) , 1 = x5 , 5x4 + 10x3 , 10x2 + 4x , 1:
Now, the polynomial
,f (x)f (,x) = (x5 , 5x4 + 10x3 , 10x2 + 4x , 1) (x5 + 5x4 + 10x3 + 10x2 + 4x + 1) = x10 , 5x8 + 8x6 , 30x4 , 4x2 , 1
225 is even, and has roots ri + 1, ,(ri + 1) (1 i 5). Replacing each x2k term with an xk term, we get a polynomial x5 , 5x4 + 8x3 , 30x2 , 4x , 1, with roots (ri + 1)2 , that is, ri10 . Also solved by HAYO AHLBURG, Benidorm, Spain; MIGUEL ANGEL OCHOA, Logro~no, Spain; THEODORE CHRONIS, student, AristoCABEZON tle University of Thessaloniki, Greece; RONALD HAYNES, student, Memorial University of Newfoundland, St. John's, Newfoundland; RICHARD I. HESS, Rancho Palos Verdes, California, USA; PETER HURTHIG, Columbia College, Burnaby, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, LOPEZ Austria; KEEWAI LAU, Hong Kong; MARIA ASCENSION CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; VEDULA N. MURTY, Andhra University, Visakhapatnam, India; DIGBY SMITH, Mount Royal College, Calgary, Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; MITKO CHRISTOV VINCHEO, Rousse, Bulgaria; and the proposer. One reader submitted a oneline answer only, and another submitted a partially incorrect answer. The solutions given by Hess, Murty and the proposer are very similar to Solution I above.
2058. [1995: 203] Proposed by Christopher J. Bradley, Clifton College, Bristol, UK. Let a; b; c be integers such that a + b + c = 3: b c a
Prove that abc is the cube of an integer. Solution by Michael Parmenter, Memorial University of Newfoundland, St. John's, Newfoundland, modi ed by the editor. Without loss of generality, we may assume that gcd(a; b; c) = 1, since, if d j a; b; c and a0 = ad , b0 = db , c0 = dc , then ab + bc + ac = 3 if and only if a00 + b00 + c00 = 3, and abc is a perfect cube if and only if so is a0 b0 c0 . Rewrite b c a the given equation as a2 c + b2a + c2 b = 3abc: (1) If abc = 1, then we are done. Otherwise, let p be any prime divisor of abc. Then from (1), it is clear that p divides exactly two of a, b and c. By symmetry, we may assume that p j a, p j b and p6 j c. Suppose that pm k a (that is, pm j a, but pm+1 6 j a) and pn k b. Then pm+n j 3abc. We claim that n = 2m. If n < 2m, then n + 1 < 2m, and thus pn+1 j a2 c. Since pn+1 j b2 a, but pn+16 j c2 b, we have pn+1n6 +1 j a2c + b2a + c2b. This is a contradiction since n + 1 m + n implies p j 3abc.
226 If n > 2m, then n 2m + 1, and thus p2m+1 j c2 b. Since p2m+1 j b2 a, but p2m+1 6 j a2 c, we have that p2m+1 6 j a2 c + b2 a + c2 b. This is again a contradiction since 2m + 1 < m + n implies that p2m+1 j 3abc. Hence n = 2m, which implies that pm k abc. It follows immediately that abc is a perfect cube. Also solved by CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; ADRIAN CHAN, student, Upper Canada College, OCHOA, Logro~no, Spain; Toronto, Ontario; MIGUEL ANGEL CABEZON THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; PETER HURTHIG, Columbia College, Burnaby, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer. Six solvers submitted incorrect solutions. All of them assumed inadvertently that a, b, c are positive. (One other reader sent in a partial solution based on this assumption.) In this case, the problem becomes trivial since, from the ArithmeticGeometric Mean inequality, one can deduce immediately that a = b = c. Several other incomplete, or partially incomplete, solutions were also received. Using an argument similar to the one given above, Bosley showed that all solutions are given by a = b = c or a = kr(r + s)2 , b = ksr2 , c = ,k(r + s)s2 for some integers k, r, s, provided that abc 6= 0. Setting k = 1, r = n, s = n + 1 (n 6=2 0; ,1)2 yields the in nite family of solutions (a; b; c), where a = n(2n +1) , b = n (n +1), c = ,(n +1)2(2n +1) [together with all integer multiples of them, and triples obtained by cyclically permuting a, b and c.  Ed.]. Setting k = 1, r = n , 1, s = n + 1, or k = 1, r = n + 1, s = n , 1 (n 6= 0; 1) yields the in nite families of solutions (a; b; c), where a = 4n2(n , d), b = (n2 , 1)(n , d), c = ,2n(n + d)2 , where d = 1. All these families were also found by Penning. Setting k = 1, r = 2, s = 3, one obtains the solution (50; 12; ,45), also found by Hurthig. The special case (4; 1; ,2), obtained by letting k = r = s = 1 was also found by Chronis, Flannigan, Tsoussaglou, and the proposer. Janous asked whether the \n{analogue" of this problem is true (the case n = 2 is trivial); for example, if ab + cb + dc + ad = 4, where a, b, c, d are integers, must abcd be a perfect 4th power? The answer is clearly negative, as shown by the example (a; b; c; d) = (,2; 4; 1; 1), found by this editor.
227 2059. [1995: 203] Proposed by Sefket Arslanagic, Berlin, Germany.
Let A1 A2 : : : An be an ngon with centroid G inscribed in a circle. The lines A1 G; A2 G; : : : ; An G intersect the circle again at B1 ; B2 ; : : : ; Bn . Prove that
A1G + A2G + + AnG = n: GB1 GB2 GBn
Solution by Toshio Seimiya, Kawasaki, Japan. Let O and R be the centre and the radius of the circumscribed circle A1A2 : : : An.
Ai
Ai+1
Ai,1
R O
A2
R An
A1
Since
AiG = AiG2 = AiG2 = AiG2 ; GBi AiG GBi OA2i , OG2 R2 , OG2 for i = 1; 2; : : : ; n, we get n X
n X
AiG2
AiG = i=1 : i=1 GBi R2 , OG2 Since G is the centroid of A1 A2 : : : An , we have n X
i=1
Thus we have
n X
i=1
AiG GO =
AiG = 0:
n X i=1
!
AiG GO = 0 GO = 0:
(1)
228 Hence we have n X
i=1
2
AiO = = =
nR2 = Therefore
n X i=1 n X i=1
n X
i=1 n X i=1 n X
i=1
AiG + GO 2
AiG + 2 2
2
n X i=1
AiG GO + 2
n X i=1
GO
2
AiG + n GO ; that is AiG2 + n GO2:
AiG2 = n(R2 , GO2):
From (1) and (2), we obtain that
(2)
n X
AiG = n. i=1 GBi
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, OCHOA, Logro~no, Spain; JORDI DOU, BarceUK; MIGUEL ANGEL CABEZON lona, Spain; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; LOPEZ MARIA ASCENSION CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; P. PENNING, Delft, the Netherlands; HOE TECK WEE, student, Hwa Chong Junior College, Singapore; and the proposer.
2060. [1995: 203] Proposed by Neven Juric, Zagreb, Croatia. Show that for any positive integers m and n, the integer j
p
m + m2 , 1
n k
is odd (bxc denotes the greatest integer less than or equal to x). I Solution by David Doster, Choate Rosemary Hall, Wallingford, Connecticut, USA. n n p p Let xn = m + m2 , 1 + m , m2 , 1 . Then, using the Binomial Theorem, we get
xn = 2
bn= X2c n k=0
and so xn is an even integer. n p Since 0 < m , m2 , 1 =
n,2k ,m2 , 1k ; m 2k
1 p m + m2 , 1 1, we have that
229 n xjn , 1 m + p mk2 , 1 < xn, and thus that p n which is odd. m + m2 , 1 = xn , 1,
II Solution by P. Penning, Delft, the Netherlands. (modi ed slightly by the editor). p Let u = m + m2 , 1 and
p
n
p
n
Fn = un + u,n = m + m2 , 1 + m , m2 , 1 : Then F1 = 2m. Since 2mFn = F1 Fn = ,u + u,1 ,un + u,n = Fn+1 + Fn,1; we obtain the recurrence relation
Fn+1 = 2m Fn , Fn,1: , Since F2 = 2 2m2 , 1 is even, we conclude that Fn is even for all n. Since un 1, we have 0 < u,n 1, and it follows that bunc = Fn , u,n = Fn , 1, which is odd.
Also solved by HAYO AHLBURG, Benidorm, Spain; SEFKET Berlin, Germany; SEUNGJIN BANG, Seoul, Korea; ARSLANAGIC, FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; TOBY GEE, student, the John of Gaunt School, Trowbridge, England; DAVID HANKIN, Hunter College Campus Schools, New York, NY, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEEWAI LAU, Hong Kong; VEDULA N. MURTY, Andhra University, Visakhapatnam, India; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; CORY PYE; student, Memorial University, St. John's, New foundland; HEINZJURGEN SEIFFERT, Berlin, Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; CHRIS WILDHAGEN, Rotterdam, the Netherlands; PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA; and the proposer. Most submitted solutions are similar to one of the two solutions above. Bellot Rosado pointed out that the problem is a very old one. He located it as Part (a) of Problem 363, Americal Mathematical Monthly, 1912, p. 51. He also gave four similar or related problems.
230
2061. [1995: 234] Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a triangle with centroid G, and P is a variable interior point of ABC . Let D, E, F be points on sides BC , CA, AB respectively such that PD k AG, PE k BG, and PF k CG. Prove that [PAF ]+[PBD]+[PCE] is constant, where [XY Z ] denotes the area of triangle XY Z . I. Solution by Jaosia Jaszunska, student, Warsaw, Poland. In the gure, the line through P :
parallel to AB intersects the sides BC and CA at I and L respectively; parallel to BC intersects the sides CA and AB at K and N respectively; parallel to CA intersects the sides AB and BC at M and J respectively.
A L M F N B
E P
I D J
K C
Since PD and AG are parallel, the dilatation that takes triangle PIJ to ABC , takes D to the midpoint of BC . Thus D is the midpoint of IJ . Similarly, E and F are the midpoints of KL and MN respectively. It is easily seen that the shaded area is equal to the unshaded area. [The vertically shaded pieces such as PID form half of a triangle, while the horizontally shaded pieces such as PBI form half of a parallelogram.] Thus
[PAF ] + [PBD] + [PCE ] = 12 [ABC ];
so that the sum is constant as desired. [Editor's comment: Several solvers mentioned that since the result belongs to ane geometry, one can assume without loss of generality that the given triangle is equilateral. We have therefore used an equilateral triangle in the
gure.]
231 II. Solution by Christopher J. Bradley, Clifton College, Bristol, UK. The result holds for any point P in the plane { there is no need to restrict P to the interior of the given triangle. In areal coordinates, AG = (, 23 ; 31 ; 13 ), and since PDkAG, the point D must have coordinates ( , 2 ; + 3 ; + 3 ), where P is (; ), normalised so that + + = 1, and 3 is chosen so that D lies on BC . Thus = 3=2 and D is (0; + 2 ; + 2 ). From the formula for ratio of areas,
0 0 [PBD] = 1 +
= 1 2 + : 2
2 [ABC ] 0 + 2
Similar formulae hold for the other areas, from which we deduce that
[PAF ] + [PBD ] + [PCE ] 1 2 = [ABC ] 2 + 21 2 + 12 2 + + + = 21 ( + + )2[ABC ] = 12 [ABC ];
which is constant for a given triangle, as P varies. Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; OCHOA, Logro~no, Spain; JORDI DOU, BarceMIGUEL ANGEL CABEZON lona, Spain; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; Y, Ferris State University, Big Rapids, Michigan, USA; VACLAV KONECN GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; WALDEMAR POMPE, student, University of Warsaw, Poland; and the proposer. . [1995: 234] Proposed by K.R.S. Sastry, Dodballapur, India. Find a positive integer n so that both the continued roots 2062
s
r
1995 + n + 1995 + pn +
and
s
r
q
n + 1995 + n + p1995 + q
converge to positive integers. Solution by Miguel Angel Cabezon Ochoa, Logro~no, Spain (loosely translated from the Spanish).
232
p
q
Let a0 = 0, a1 = 1995, and ap+2 = 1995 + n + ap for p 0. q p Similarly let b0 = 0, b1 = pn, and bp+2 = n + 1995 + bp for p 0. We will now show by induction that ap < K for all p 0, where K = maxf1995;ng. This is clearly true for p = 0 and p = 1. Suppose that ap < K for all p, 0 p < t with t 2. Then p
p p p at = 1995 + psn + at,2 < K +spK + K = K + p2K p2K r 1 1 = K K + K 2 = K K + K23 < K
whence ap < K for all p 0. Next we show that fap g is an increasing sequence by induction. It is clear that a0 < a1 < a2 . Suppose that ap < ap+1 for some p 0. Then q
q
ap+2 = 1995 + pn + ap < 1995 + pn + ap+1 = ap+3 which establishes that the sequence is increasing. Similar proofs exist for the sequence fbpg. Thus both fap g and fbp g converge say x and y , respectively. This leads to the equations x = p1995to+limits, y2 and y = pn + x. Let y = (p + 44)2 , 1995; then x = p + 44 and n = y , x. (Editor's comment: this is clearly a solution for all integer values of p 1.) The rst two values for p give the following solutions: (p; n; x; y ) = (1; 855; 45; 30) and (p; n; x; y) = (2; 14595; 46; 121). Also solved by SABIN CAUTIS, student, Earl Haig Secondary School, North York, Ontario; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; TIM CROSS, King Edward's School, Birmingham, England; KEITH EKBLAW, Walla Walla, Washington,USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; ROBERT GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; FRIEND H. KIERSTEAD JR., Cuyahoga Falls, Ohio, USA; JAMSHID KHOLDI, Y, Ferris State University, Big Rapids, New York, NY, USA; VACLAV KONECN Michigan, USA; KEEWAI LAU, Hong Kong; BEATRIZ MARGOLIS, Paris, France; CORY PYE, student, Memorial University, St. John's, Newfound land; CRISTOBAL SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon, Spain; HEINZJURGEN SEIFFERT, Berlin, Germany; ASHSIH KR. SINGH, student, Kanpur, India; PANOS E. TSAOUSSOGLOU, Athens, Greece; CHRIS WILDHAGEN, Rotterdam, the Netherlands; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. There was one incomplete and one incorrect solution submitted.
233 Most solvers found all of the solutions above, but some only found one. Very few solvers directly examined the question of convergence of the continued roots.
2063. [1995: 234] Proposed by Aram A. Yagubyants, Rostov na Donu, Russia. Triangle ABC has a right angle at C . (a) Prove that the three ellipses having foci at two vertices of the given triangle, while passing through the third, all share a common point. (b) Prove that the principal vertices of the ellipses of part (a) (that is the points where an ellipse meets the axis through its foci) form two pairs of collinear triples. Solution by Gottfried Perz, Pestalozzigymnasium, Graz, Austria. (a) Let P be the fourth vertex of the rectangle APBC . Then
PA = BC; PB = AC; PC = AB;
whence
PA + PB = CB + CA; PB + PC = AC + AB; PC + PA = BA + BC; which means that each of the three ellipses passes through P . (b) Let M1 , M2 , M3 be the centres of the three ellipses, and therefore the midpoints of AB , BC and CA respectively. Denote the principal vertices of the ellipses by Xi and Yi , with B lying between A and X1 and between C and X2 , and with C lying between A and X3 . Since M X = AC + BC ; M X = AB + AC ; M X = AB + BC ; 1
1
2
2
2
2
3
3
2
we get with AB = c, BC = a, AC = b (and recalling that a2 + b2 = c2 ),
a + b + c ,a + b + c a , b + c AX1 BX2 CX3 = 2 2 2 a+b,c a+b+c a+b+c BX1 CX2 AX3 2 2 2 2 + 2ab , b2 + c2 , a = a2 + 2ab + b2 , c2 = 22ab ab = 1:
This means, according to Menelaus' Theorem, that X1 , X2 , X3 are collinear.
234 Analogously, Y1 , Y2 , Y3 are collinear. Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and MARIA LOPEZ ASCENSION CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUEL ANGEL OCHOA, Logro~no, Spain; ROBYN M. CARLEY (student) and ANA CABEZON WITT, Austin Peay State University, Clarksville, Tennessee, USA; JORDI DOU, Barcelona, Spain; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; CAMILLA FOX, Toronto, Ontario; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; WALTHER JANOUS, Ursulinengymnasium, Y, Ferris State University, Big Rapids, Innsbruck, Austria; VACLAV KONECN Michigan, USA; WALDEMAR POMPE,student, University of Warsaw, Poland; CRISTOBAL SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon, Spain; TOSHIO SEIMIYA, Kawasaki, Japan; and the proposer. One anonymous solution was received  see note at the start of this section. Bellot Rosado and Lopez Chamorro, Cabezon Ochoa, Dou, Pompe and Seimiya all sent in solutions that are essentially the same as Perz's, whose solution was chosen by the timehonoured method of selecting the superior paper after having thrown the lot down the stairs. The other solvers used coordinates; teachers may wish to note that this problem provides a nonstandard exercise that illustrates eectively several aspects of analytic geometry.
2064. [1995: 234] Proposed by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. Show that 3 max ab + bc + ac ; ab + cb + ac (a + b + c) a1 + 1b + 1c
for arbitrary positive real numbers a; b; c. I. Solution by Sabin Cautis, student, Earl Haig Secondary School, North York, Ontario. Assume without loss of generality that
a+b+c b+c+a: b c a a b c
Using the AM{GM inequality, thus
a + b + c 3r3 abc = 3; b c a bca a b c b c a a b c 3 max b + c + a ; a + b + c = 3 b + c + a
235 ab + cb + ac + ab + bc + ac + ab + bc + ac ab + cb + ac + ab + bc + ac + 3 = (a + b + c) a1 + 1b + 1c :
II. Solution by Chris Wildhagen, Rotterdam, the Netherlands. It is evident that
x + y 2z =) maxfx; yg z for all real numbers x, y and z . Hence it is sucient to show that 3 ab + bc + ac + ab + cb + ac 2(a + b + c) a1 + 1b + 1c ; or
a + b + b + c + c + a 6: b a c b a c This is obviously true, since if p; q > 0 then p + q = p2 + q2 2: q p pq
III. Editorial comments. Note that Solution II says that \max" in the problem statement can be replaced by \average". This stronger result was noticed by several other solvers as well. Here are some other remarks and generalizations made by readers. If a; b; c are the sides of a triangle, then the inequality is true if \max" is replaced by \min", that is, the simpler inequality
a b c 1 1 1 3 b + c + a (a + b + c) a + b + c (1) holds whenever a; b; c are the sides of a triangle. The proposer included
this problem as #2 in his list of six Quickies published in the February 1996 Olympiad Corner (for a solution, see [1996: 59{60]). The proposer also notes that this inequality need not hold otherwise. For example, it fails when a = 8, b = 3 and c = 1. p Bradley proves that in fact (1) holds whenever pa; b; pc are the sides of a triangle. (This happens whenever a; b; c are sides of a triangle and in other cases besides.) He rst proves that if x; y;z are the sides of a triangle and l; m; n are any real numbers, then
x2 l2 + y2m2 + z2n2 , (y2 + z2 , x2)mn ,(z2 + x2 , y2)nl , (x2 + y2 , z2)lm 0;
(2)
236 by completing the square. (If we multiply the left side of (2) by 4x2 , then it can be rewritten in the form
2x2 l , (z2 + x2 , y2)n , (x2 + y 2 , z 2)m2 +(x + y + z )(x + y , z)(y + z , x)(z + x , y )(m , n)2; ,
(3)
with the help of the identity
(x + y + z)(x + y , z )(y + z , x)(z + x , y ) = 4x2 z2 , (z2 + x2 , y 2)2; (4) and (3) is clearly 0 if x; y;z are sides of a triangle.) Bradley then puts 2 = l, c = z 2 = m in (2); the result is equivalent to a = x2 = n, b p= yp p
(1) and holds if a; b; c are the sides of a triangle. Consultation with helpful expert (and current proposer) Murray Klamkin yielded the following alternate way to derive (2). It is known (for example, see Theorem 3, page 306 of Gantmacher's Matrix Theory (The Theory of Matrices), Vol. I, Chelsea, 1959, or any other advanced book on matrix theory) that (2) holds if and only if the matrix 2 6 6 6 6 6 6 6 4
x z2 , x2 , y2 2
2
y ,z ,x 2
2
z2 , x2 , y2
y2 , z2 , x2
y2 x2 , y2 , z2
x2 , y2 , z2
2
2
2
2 2
z2
2
3 7 7 7 7 7 7 7 5
is nonnegative de nite, which is true if and only if the principal minors are nonnegative. Since the minor x2 is nonnegative, and the determinant of the entire matrix is zero (just add all the rows), all we have to prove is that
x z , x2 , y 2 2
z2 , x2 , y2 2
2
y2
2
0:
But this is true when x; y;z are the sides of a triangle because
x
2
z2 , x2 , y2
2 4 z 2 , x2 , y 2 2 y 2 which is 0 by a version of (4).
= 4x2 y2 , (z 2 , x2 , y 2)2
Janous shows that for any n 3 and for all x1 ; : : : ; xn > 0,
n X xk + xl ; (x1 + + xn) x1 + + x1 n , 1 1k<ln xl xk 1 n
237 which generalizes the result of Solution II and can be proved the same way. This can be written as
M1(x1; : : : ; xn) M xk : k 6= l ; 1 M,1 (x1; : : : ; xn) xl where Mt denotes the tth power mean: t t 1=t Mt (x1; : : : ; xn) = x1 + n + xn : He asks for the minimum t > 0 such that M1(x1; : : : ; xn) M xk : k = 6 l t x M (x ; : : : ; x ) ,1
1
n
l
for all x1 ; : : : ; xn > 0. (His conjecture: t = 1.) Janous also writes the inequality of Solution II as
1 1 1 1 a b c b c a M1(a; b; c)M1 a ; b ; c 2 M1 b ; c ; a + M1 a ; b ; c ; and then wonders what the minimum t > 0 is so that M (a; b;c)M 1 ; 1 ; 1 M M a ; b ; c ; M b ; c ; a :
t 1 b c a 1 a b c a b c For an integer n > 3 and positive real numbers a1 ; a2 ; : : : ; an , it is not
1
1
always true that
n max aa1 + aa2 + + aan ; aa2 + aa3 + + aa1 2 3 1 1 2 n 1 1 1 (a1 + a2 + + an) a + a + + a : 1 2 n A counterexample when n = 4 is (a1; a2 ; a3 ; a4 ) = (4; 2; 1; 2), for which the left side of the inequality is 20 and the right side is 81=4.
For readers who still need more problems to solve, here are two more, inspired by the above contributions. (i) Find the smallest t > 0 so that (1) holds whenever at ; bt; ct are the sides of a triangle. Could the answer be t = 1=2? (Klamkin's quickie says t 1, and Bradley proved that t 1=2.) (ii) Find the smallest t > 0 so that
1 (a + b + c) 1 + 1 + 1 t a + b + c + (1 , t) b + c + a 3 a b c b c a a b c whenever a; b; c > 0 satisfy
a+b+c b+c+a: b c a a b c
238 (The original problem says t 1, and Solution II shows that t 1=2.) Berlin, Germany; CARL BOSLEY, Also solved by SEFKET ARSLANAGIC, student, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; HANS ENGELHAUPT, Franz{Ludwig{ Gymnasium, Bamberg, Germany; TOBY GEE, student, the John of Gaunt School, Trowbridge, England; ROBERT GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOE HOWARD, New Mexico Highlands University, Las Vegas, New Mexico, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; DAG Y, Ferris State University, JONSSON, Uppsala, Sweden; VACLAV KONECN Big Rapids, Michigan, USA; SAI C. KWOK, San Diego, California, USA; KEEWAI LAU, Hong Kong; THOMAS LEONG, Staten Island, New York, USA; VEDULA N. MURTY, Andhra University, Visakhapatnam, India; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; WALDEMAR POMPE, student, University of Warsaw, Poland; BOB PRIELIPP, University of Wisconsin{ Oshkosh, Wisconsin, USA; CRISTOBAL SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon, Spain; HEINZJURGEN SEIFFERT, Berlin, Germany; ASHSIH KR. SINGH, student, Kanpur, India; PANOS E. TSAOUSSOGLOU, Athens, Greece; JOHANNES WALDMANN, FriedrichSchillerUniversitat, Jena, Germany (two solutions); EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; and the proposer. There was also one anonymous solution. Many solvers gave solutions similar to I or II. Konecny included a generalization to n numbers which is weaker than Janous's generalization given above.
2065. [1995: 235] Proposed by Stanley Rabinowitz, Westford, Massachusetts. Find a monic polynomial f (x) of lowest degree and with integer coef cients such that f (n) is divisible by 1995 for all integers n. Solutionby Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USA. We have 1995 = 3 5 7 19. It can be shown that the congruence f (x) 0 (mod p) of degree n with integer coecients has at most n solutions for prime p (see, for example, Niven, Zuckerman, and Montgomery's An Introduction to the Theory of Numbers, 5th edition, p. 93). Hence if f (n) 0 (mod 19) for all n and f (x) is monic, f (x) must be of degree 19 or higher. On the other hand, f (x) = x(x + 1)(x + 2) (x + 18); of degree 19, is easily seen to be divisible by 1995 for all integers x.
239 Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; TOBY GEE, student, the John of Gaunt School, Trowbridge, England; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; KEEWAI LAU, Hong Kong; HEINZ JURGEN SEIFFERT, Berlin, Germany; PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA; an anonymous solver; and the proposer. There were ve incorrect solutions submitted. Pompe comments that \this problem has been discussed earlier in the literature; see Problem 17 in Donald J. Newmann's A Problem Seminar". [1995: 235] Proposed by John Magill, Brighton, England. The inhabitants of Rigel III use, in their arithmetic, the same operations of addition, subtraction, multiplication and division, with the same rules of manipulation, as are used by Earth. However, instead of working with base ten, as is common on Earth, the people of Rigel III use a dierent base, greater than two and less than ten. 2066.
BC AB)CBC AB BDC BDC This is the solution to one of their long division problems, which I have copied from a school book. I have substituted letters for the notation originally used. Each of the letters represents a dierent digit, the same digit wherever it appears. As the answer to this puzzle, substitute the correct numbers for the letters and state the base of the arithmetic of Rigel III. Solution by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario. From the long division, note that B AB = AB so B must be 1; CB , AB = BD so D must be zero; since there was no \borrowing" from the C in the previous subtraction, C = A + 1. From C AB = BDC and the rst three conditions, we get (A + 1) A = 10; thus the base must be the product of consecutive numbers. Since the base is greater than two and less than ten, it must be six; thus A = 2 and C = 3. Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; SABIN CAUTIS, student, Earl Haig Secondary School, North York, Ontario; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario;
240 KEITH EKBLAW, Walla Walla, Washington,USA; HANS ENGELHAUPT, Franz{ Ludwig{Gymnasium, Bamberg, Germany; ROBERT GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; FRIEND H. KIERSTEAD JR., Cuyahoga Falls, Ohio, USA; Y, Ferris State University, Big Rapids, Michigan, USA; VACLAV KONECN DAVID E. MANES, State University of New York, Oneonta, NY, USA; J.A. MCCALLUM, Medicine Hat, Alberta; JOHN GRANT MCLOUGHLIN, Okanagan University College, Kelowna, British Columbia; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; CORY PYE, student, Memorial University of Newfoundland, St. John's, Newfoundland; CHRIS WILDHAGEN, Rotterdam, the Netherlands; SUSAN SCHWARTZ WILDSTROM, Kensington, Maryland, USA; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer.
Congratulations! The editors would like to congratulate some regular contributors for making their country's team for the 37th International Mathematical Olympiad, held in July 1996, in Mumbai, India, and on their performances at the IMO. Any student who makes a national team is a champion is his/her own right. Name Carl Bosley Sabin Cautis Adrian Chan Toby Gee Ashish Kr. Singh
Country United States Canada Canada Great Britain India
Award (if any) Gold Medal Bronze Medal Bronze Medal Silver Medal
241
Dissecting Squares into Similar Rectangles ByungKyu Chun (student) Harry Ainlay Composite High School
Andy Liu
University of Alberta and
Daniel van Vliet (student) University of Alberta
1 Introductory Remarks. Karl Scherer and Martin Gardner [1] have proposed the following threepart problem. Cut a square into three similar pieces, where (1) all three are congruent; (2) exactly two are congruent; (3) no two are congruent. Note that (2) really consists of two subproblems, where the congruent pieces are (2a) smaller than the third; (2b) larger than the third. All their solutions to (2) are to (2a). It appears that, alas, (2b) is not to be. Also considered in [1] are related dissections of an equilateral triangle, but they do not concern us here. We generalize the problem of Scherer and Gardner for the square as follows. Given any integer m > 1 and any of its 2m,1 compositions, or ordered partitions, m = a1 + a2 + + an , dissect a square into m similar pieces so that there are a1 congruent pieces of the largest size, a2 congruent pieces of the next largest size, and so on. In the original problem, m = 3 and the compositions are: (1) 3; (2a) 1 + 2; (2b) 2 + 1; (3) 1 + 1 + 1. Our main result is that the dissection problem always admits a solution using rectangular pieces if and only if the composition is not of the form k +1, where k is any positive integer. These solvable cases are covered by two constructions which are only slightly dierent.
242
2 Construction for the case m = a1 + a2 + + an; an > 1.
Suppose the composition is m = a1 + a2 + + an ; an > 1. We start with a rectangle R and divide it into n rectangles R1 ; R2 ; : : : ; Rn as follows. R1 is the right half of R; R2 is the bottom half of R , R1 ; R3 is the right half of R , R1 , R2 ; R4 is the bottom half of R , R1 , R2 , R3 , and so on, except that Rn = R , R1 , R2 , , Rn,1 . The dimensions of these rectangles will be adjusted later. Divide Ri into ai rectangular pieces, using vertical lines if i is odd and horizontal lines if i is even. Let the horizontal and vertical dimensions of each piece in Ri be xi and yi respectively, as shown in Figure 1 for the case 6 = 1 + 3 + 2.
y3 y1
x3
x2
y2 Figure 1
x1
Next, we make all the pieces similar to one another by setting yi = xi t for all i, where t is a given positive number. It follows from the assumption an > 1 that a piece in Ri is larger than a piece in Rj if i > j . We choose xn = 1, so that yn = t. If n is even, yn,1 = an yn = an t and xn,1 = an. If n > 1 is odd, xn,1 = an xn = an and yn,1 = an t. Going backwards step by step on the basis of the recursive formulae yi,1 = aiyi + yi+1 for i even and xi,1 = aixi + xi+1 for i odd, we can compute the dimensions of the pieces in Rn,2 ; Rn,3 ; : : : ; R1 and R. Figure 2 is obtained by applying this process to Figure 1. Note that the xi are always integers while the yi are always integral multiples of t. Finally, we want to nd a t so that R is a square. In other words, t is a solution of y1 = a1 x1 + x2 . Since y1 is linear in t while a1 x1 + x2 is a constant, y1 > a1 x1 + x2 for suciently large t. On the other hand, if t = 1, then the pieces will be square pieces, so that y1 = x1 a1 x1 a1 x1 + x2 . Since m > 1, either a1 > 1 or x2 > 0, so that y1 < a1 x1 + x2 . It follows that there exists a real number t > 1 for which y1 = a1 x1 + x2 .
243
t 1
7t
2t 2
7 Figure 2
From Figure 2, we have 7t = 9 or t = 97 . In general, since t is the solution of a linear equation with integral coecients, it is a rational number. Hence we can apply a uniform magni cation to make all pieces have integral sides. Figure 3 is a magni cation of Figure 2 by a factor of 7.
3 Construction for the case m = a1 + a2 + + an; an = 1 and n 3.
Suppose the composition is m = a1 + a2 + + an ; n 3 and an = 1. The
rst step is identical to the construction in the last Section. We make all the pieces similar to one another by setting yi = xi t, except that xi = yit for the last odd i. We shall prove that the equation y1 = a1 x1 + x2 has a real solution. Note that, as before, we have y1 < a1 x1 + x2 if t = 1. Suppose n 4 is even. Let xn = 1 and yn = t. Then yn,1 = t; xn,1 2= t2; xn,2 = t2 + 1 and yn,32 = t3 + t. By induction, xi is of the form at + b while yi is of the form ct + dt for all i, where a, b, c and d are integers dependent only on i. Thus y1 > a1 x1 + x2 for suciently large t. Suppose n 3 is odd. Let xn = t and yn = 1. Then xn,1 = t; yn,1 = t2; yn,2 = t2 + 1 and xn,22 = t + 1t . By induction, xi is of the form at + bt while yi is of the form ct + d for all i, where a; b;c and d are integers dependent only on i. Thus y1 > a1 x1 + x2 for suciently large t.
244 7
9
63 18 14
49 Figure 3
Figure 4 illustrates the cases 6 = 1 + 2 + 2 + 1 and 6 = 2 + 3 + 1. In many cases covered by this construction, we can also use pieces with integral sides in the same orientation. Figure 5 illustrates one such example, namely, 11 = 4 + 6 + 1.
4 Impossibility proof for the case m = k + 1.
We now consider the remaining cases, compositions of the form m = k + 1, where k is any positive integer. We shall prove that the dissection problem is not solvable with rectangular pieces. Our approach is indirect. We assume that a solution exists, with a \1 by t" piece and k \s by st" pieces, where t 1 and s > 1 are real numbers. Let the side length of the square be `. Consider a horizontal or vertical segment from one side to the other, not running along any side of a piece. Then ` = as + bst for some integers a and b if the segment does not cut the small piece. If it does, we have either ` = cs + dst + 1 or ` = es + fst + t for some integers c, d, e and f . From as + bst = cs + dst = 1, it follows that s is rational if and only if t is. We rst dispose of the case where s and t are rational. Let s = hg and t = ji , where g and h are relatively prime integers, as are i and j . Then agj + bgi = cgj + dgi + hj = egj + fgi + hi. Hence g divides hi and hj . Since g and h are relatively prime, g divides i and j .1 Since i and j are relatively prime, we must have g = 1. However, s = h 1, which is a contradiction.
245
t
1
t2
1
3t2 + 1
4t3 + 3t 2t3 + t
t2 2t2 + 1
4t2 + 3
t
3t + 1t
Figure 4
We now consider the case where s and t are irrational, so that t > 1. Consider a horizontal or vertical segment from one side of the square to the other, not running along any side of a piece and not cutting the small piece. As before, we have ` = as + bst for some integers a and b. We make the important observation that a and b are independent of the choice of such a segment. If we0 have ` = a0 s + b0 st for some integers a0 6= a and b0 6= b, then t = ab,0,ab will be rational. Moreover, since the segment can be horizontal or vertical, we must have a > 0 and b > 0. We now use this to prove that the dissected square does not have a faultline. This is de ned as a segment which divides the square into two rectangles, each containing only complete pieces. Suppose to the contrary that there is a horizontal faultline. Then one of the rectangles is dissected into only large pieces. Divide it into horizontal strips of width s from the top. Each strip contains b \st by s" pieces and a or 2a squares of side s, each of which contains parts of one or two s by st pieces. It follows that the height of the rectangle is equal to ps for some positive integer p. We can rearrange the pieces within this rectangle so that all the horizontal ones are to the left and all the vertical ones are to the right. We then have ps = qst for some positive integer q . However, t will then be rational, which is a contradiction. It follows that the dissected square has no faultlines. We now combine the large pieces into rectangles until the union of any two of them is not a rectangle. The small piece is not part of any rectangle. If this combination is not unique, we choose the one for which the number r of rectangles is minimum. If r 3, there will be a faultline. Hence r 4.
246 28
21 51
68
68
51 21
Figure 5
28
Figure 6
Suppose r = 4. The only possible con guration has the four rectangles at the corners and the small piece in the middle, as shown in Figure 6. Now each of the rectangles is dissected into large pieces all of which are in the same orientation. Since ` = as + bst with constants a > 0 and b > 0, two opposite rectangles must be dissected into a columns and b rows of s by st pieces, with the other two into a rows and b columns of st by s pieces. However, the small piece will then be a square, and t = 1 is a contradiction. Suppose r 5. Then there is a rectangle at a corner of the square which is not adjacent to the small piece. At least one of its two sides within the square is the union of the sides of at least two other rectangles. If the vertical side is not, as illustrated in Figure 7, then the horizontal side must be.
247
Figure 7
Now this corner rectangle and those immediately below it are dissected into large pieces in the same orientation within each. In fact, it must be the same in all of them, as otherwise we can show that t is rational. Now we can expand the corner rectangle until it swallows up a rectangle below it. However, this reduction in r contradicts its minimality assumption.
5 Concluding Remarks.
We conjecture that for the compositions m = k + 1, the dissection problem is not solvable even if nonrectangular polygonal pieces are permitted. This is true for the simplest case, namely, 2 = 1 + 1. We give a proof using an indirect argument. Suppose that a solution exists. Clearly, the smaller polygon P cannot contain two opposite corners of the square S . Hence the larger polygon P 0 contains a side of S . We claim that this is the longest sides of P 0 . Otherwise, the longest side will be inside S , and it separates P 0 from P . Hence it will also be the longest side of P , and the two polygons are in fact congruent. This contradiction justi es the claim. It follows that the longest side of P is shorter than a side of S . Now P and P 0 have the same number of sides inside S , but P 0 has more sides than P on the boundary of S . Hence P and P 0 cannot be similar. On the other hand, Figure 8 shows a \solution" using fractallike pieces. For the cases covered by our two constructions, we have many solutions using polygonal pieces that are not rectangles. Figure 9 illustrates one such example, namely, 7 = 1 + 1 + 1 + 1 + 1 + 1 + 1, based on a solution to a problem in [2].
248
P
P
! P0
! P0
Figure 8
, ,, , , , @ ,, , @@ , @@ ,, , , , , ,, Figure 9
6 References: [1]Martin Gardner, Six Challenging Dissection Tasks, Quantum, Volume 4, Issue 5 (1994) 2627. [2]Peter Taylor, \International Mathematics Tournament of the Towns (19801984)", Australian Mathematics Trust, Canberra (1993) 106.
249
THE SKOLIAD CORNER No. 16
R.E. Woodrow As a contest this issue, we give the Saskatchewan Senior Mathematics Contest. This contest was written Wednesday, February 22, 1995. My thanks go to Garreth Grith, Mathematics Department, The University of Saskatchewan, long time organizer of the contests in Saskatchewan, for permission to use the contest and for the solutions we shall give next issue.
1995 SASKATCHEWAN SENIOR MATHEMATICS CONTEST February 22, 1995  Time: 1.5 hours
1. They sell regular and jumbo sized , orders of sh at Jerry's Fish & Chips Emporium. The jumbo order costs 43 times as much as a regular one and an order of chips costs $1.30. Last Thursday, the Emporium was quite busy over the lunch break (11:30 am  1:30 pm). Exactly thirteen jumbo sized orders of sh along with a quantity of regular orders of sh and chips were sold. $702.52 had been placed in the till. During the period 1:30  4:30, business slackened o. 26 regular orders of sh were sold during this time. Four times as many regular orders had been sold during the lunch break. No jumbo portions were sold and the number of orders of chips declined to one fth the number that had been sold during the lunch break. At 4:30 pm there was $850.46 in Jerry's till. What is the price of a regular sized order of sh at Jerry's? [5 marks]
2.ABCD is a square with side of length s. A circle, centre A and radius r is drawn so that the arc of this circle which lies within the square divides the square into two regions of equal area. Write r as a function of s. [6 marks] 3.(a) Solve the equation 3y = 10y .
(b) Solve the equation 3y = 10. (c) Write t log8 px , 2 log8 y as a single logarithm. 4. Establish the identity 2 cot A = cot A , tan A : 2
2
[3 marks] [3 marks] [4 marks] [5 marks]
250
5. ABC is a triangle, right angled at C . Let a, b, c denote the lengths of the sides opposite angles Ap, B , C respectively. Given that a = 1p, \B = 75 and that tan 75 = 2 + 3, express b and c in the form p + q 3 such p that p = 2 or 2. [8 marks] 6. Determine the function f (x) which satis es all of the following conditions: [8 marks] (i) f (x) is a quadratic function. (ii) f (x + 2) = f (x) + x + 2. (iii) f (2) = 2. 7.Prove that if n is a positive integer (written in base 10) and that if 9 is a factor of n, then 9 is also a factor of the sum of the digits of n.[8 marks]
Last number we gave the problems of the American Invitational Mathematics Examination. These problems and their solutions are copyrighted by the Committee on the American Mathematical Competitions of the Mathematical Association of America and may not be reproduced without permission. Full solutions, and additional copies of the problems, may be obtained for a nominal fee from Professor Walter E. Mientka, C.A.M.C. Executive Director, 917 Oldfather Hall, University of Nebraska, Lincoln, NE, U.S.A. 68588{0322. 1. 200 6. 049 11. 276
Solutions to the 1996 A.I.M.E. 2. 340 7. 300 12. 058
3. 044 8. 799 13. 065
4. 166 9. 342 14. 768
5. 023 10. 159 15. 777
That completes the Skoliad Corner for this issue. Please send me suitable contest materials, your students' nice solutions, and comments or advice for the future of the Corner.
251
THE OLYMPIAD CORNER No. 176
R.E. Woodrow All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. We begin this number with the problems of the 10th Iberoamerican Mathematical Olympiad, written September 26{27, 1995 at Valparaiso (Chile). My thanks go to Professor Francisco Bellot Rosado of Valladolid, Spain for sending me an English translation of the problems.
10th IBEROAMERICAN MATHEMATICAL OLYMPIAD September 26{27, 1995 (Valparaiso, Chile) FIRST DAY  Time: 4.5 hours
1. (Brazil). Determine all the possible values of the sum of the digits of all the perfect squares. 2. (Spain). Let n be an integer bigger than 1. Determine the real numbers x1 ; x2 ; : : : xn 1, and xn+1 > 0, such that the following conditions are simultaneously ful lled: (a) px1 + p3 x2 + + n+1pxn = n pxn+1 x + x + + xn = x . (b) 1 2 n+1 n 3. (Brazil). Let r and s be two orthogonal straight lines, not belonging to the same plane. Let AB be their common perpendicular, with A 2 r and B 2 s. (Note that the plane which contains B and r is perpendicular to s). Consider the sphere with diameter AB . The points M 2 r, and N 2 s, are variable, with the condition that MN is tangent to the sphere at some point T . Find the locus of T . SECOND DAY  Time: 4.5 hours 4. (Argentina). Coins are situated on an m m board. Each coin situated on the board \dominates" all the cells of the row (), the column (6 I ) to which the coin belongs. Note that the coin does R ?) and the diagonal (@ not \dominate" the diagonal (, ). Determine the smallest number of coins which must be placed in order that all the cells of the board be dominated.
252
5. (Spain). The inscribed circumference in the triangle ABC is tangent to BC , CA and AB at D, E and F , respectively. Suppose that this circumference meets AD again at its midpoint X , that is, AX = XD. The lines XB and XC meet the inscribed circumference again at Y and Z , respectively. Show that EY = FZ . 6. (Chile{Brazil). Let N = f1; 2; 3; : : : g. A function f : N ! N is called circular if for each p 2 N there exists n 2 N with n p such that f n(p) = f (f {z (:: : f}(p)) : : : ) = p: n times
The function f has repulse degree k, 0 < k < 1, if for each p 2 N, f i(p) 6= p for all i [k p], in which [x] is the integer part of x. Determine the biggest repulse degree that can be reached by a circular function. The next contest we present in its ocial language. Here is your opportunity to brush up your French. (Of course we will accept your nice solutions in either English or French!) The questions are to the mini and maxi nals of the 19th Belgian Mathematical Olympiad, written in 1994. My thanks go to Richard Nowakowski, Canadian Team Leader at the 35th IMO in Hong Kong, for collecting this contest (and others) and forwarding them to the Olympiad Corner for its use.
DIXNEUVIEME OLYMPIADE MATHEMATIQUE BELGE Mini Finale 1994
1. Combien de nombres enteiers naturels de trois chires non nuls distincts (en base 10) sont primiers avec 10? 2. En prenant comme sommets les points d'intersection des c^otes prolonges d'un hexagone regulier, Jean obtient un nouvel hexagone regulier. Il applique ensuite la m^eme construction a ce nouvel hexagone, et recommence de m^eme : : : Combien de fois Jean doitil eectuer cette construction pour que l'aire du dernier hexagone construit depasse 1994 fois l'aire de l'hexagone initial? 3. Trentehuit lampes numerotees de l a 38 sont disposees en cercle autour d'une lampe centrale numerotee 0. Ces lampes forment des groupes de quatre: f0; 1; 2; 3g; f0; 3; 4; 5g; f0; 5; 6; 7g; f0; 7; 8; 9g; : : : ; f0; 35; 36; 37g; f0; 37; 38; 1g
deux operations seulement sont realisables:
253 () e teindre les quatre lampes d'un m^eme groupe; ( ) changer l'etat de chacune des lampes d'un m^eme groupe (c'estadire, une lampe allumee est e teinte, une e teinte est allumee). Tout e tat initial des 39 lampes estil transformable par une suite de telles operatious en (a) l'etat ou toutes les lampes sont allumees? (b) l'etat ou seule la lampe numero 0 est allumee? 4. Sur un terrain plat et carre de 32 ares (ou 3; 200 metres carres) dont les c^otes sont orientes NOSE et NESO se trouve une villa rectangulaire de 16 metres sur 20 metres, dont les quatre facades font face aux quatre points cardinaux. Le centre de la villa concide avec le centre du terrain. Le reste du terrain est amenage en pelouse. Quelle fraction de la pelouse est constituee de points d'ou sont visibles deux facades de la villa? Maxi Finale 1994
1. Un pentagone plan covexe a deux angles droits non adjacents. Les deux c^otes adjacents au premier angle droit out des longueurs e gales. Les deux c^otes adjacents au second angle droit ont des longueurs e gales. En remplacant par leur point milieu les deux sommets du pentagone situes sur un seul c^ote de ces angles droits, nous formons un quadrilatere. Ce quadrilatere admetil necessairement un angle droit? 2. Des lampes en nombre 2n (avec n 2) et numerotees de 1 a 2n sont disposees en cercle autour d'une lampe centrale numerotee 0. Ces lampes forment des groupes de quatre: f0; 1; 2; 3g; f0; 3; 4; 5g; : : : f0; 2k ,3; 2k ,2; 2k ,1g; f0; 2k ,1; 2k; 2k +1g; f0; 2k + 1; 2k + 2; 2k + 3g; : : : ; f0; 2n , 1; 2n; 1g et deux operations seulement sont realisables: () e teindre les quatre lampes d'un m^eme groupe; ( ) changer l'etat de chacune des lampes d'un m^eme groupe (c'estadire, une lampe allumee est e teinte, une e teinte est allumee). Pour quelles valeurs de n tout e tat initial des 2n + 1 lampes estil transformable par une suite de telles operations en (a) l'etat ou toutes les lampes sont allumees? (b) l'etat ou seule la lampe nuimero 0 est allumee?
3. Existetil une numerotation des ar^etes d'un cube par douze nombres naturels consecutifs telle que (a) la somme des nombres attribues aux ar^etes aboutissant en un sommet soit toujours la m^eme? (b) la somme des nombres attribues aux ar^etes d'une face soit toujours la m^eme?
254
4. Le plan contient il 1994 points (distincts) non tous alignes tels que la distance entre deux quelconques d'entre eux soit un nombre entier? As promised last issue, we now give the \ocial" solutions to problems of the 1996 Canadian Mathematical Olympiad. My thanks go to Daryl Tingley, Chair of the Canadian Mathematical Olympiad Committee of the Canadian Mathematical Society, for forwarding the problems and solutions. Problems with communications during summer break did not allow us to incorporate the novel solutions of some contest participants. Hopefully next year we will be on track earlier and have time to solicit permission to use the submitted solutions.
1996 CANADIAN MATHEMATICAL OLYMPIAD 1. If , , are the roots of x , x , 1 = 0, compute 3
1+ 1+ 1+ + + 1, 1, 1,
:
Solution. If f (x) = x3 , x , 1 = (x , )(x , )(x , ) has roots ; ; standard results about roots of polynomials give + + = 0, + + = ,1, and = 1. Then
+ 1+ + 1+ = N S = 11 + , 1 , 1 , (1 , )(1 , )(1 , )
where the numerator simpli es to
N = 3 , ( + + ) , ( + + ) + 3 = 3 , (0) , (,1) + 3(1) = 7: The denominator is f (1) = ,1 so the required sum is ,7.
2. Find all real solutions to the following system of equations. Carefully justify your answer. 8 > > > > > > > > > > < > > > > > > > > > > :
4x2 =y 1 + 4x2 4y 2 =z 1 + 4y 2
4z 2 =x 1 + 4z 2
255 4t2
Solution. For any t; 0 4t2 < 1 + 4t2 , so 0 < 1. Thus 1 + 4t2 x; y and z must be nonnegative and less than 1. Observe that if one of x y or z is 0, then x = y = z = 0: If two of the variables are equal, say x = y , then the rst equation becomes 2 This has the solution x 1
4x = x: 1 + 4x2 = 0, which gives x =
y = z = 0 and x = 12
which gives x = y = z = . 2 Finally, assume that x; y and z are nonzero and distinct. Without loss of generality we may assume that either 0 < x < y < z < 1 or 0 < x < z < y < 1. The two proofs are similar, so we do only the rst case. 4t2 We will need the fact that f (t) = is increasing on the interval 2 1 + 4t
(0; 1):
To prove this, if 0 < s < t < 1 then
f (t) , f (s) = 1 +4t4t2 , 1 +4s4s2 4t2 , 4s2 = (1 + 4s2 )(1 + 4t2 ) > 0: So 0 < x < y < z ) f (x) = y < f (y ) = z < f (z ) = x, a contradic2
2
tion.
1
Hence x = y = z = 0 and x = y = z = are the only real solutions. 2 Alternate Solution. Notice that x; y and z are nonnegative. Adding the three equations gives 2 2 2 x + y + z = 1 +4z4z2 + 1 +4x4x2 + 1 +4y4y2 :
This can be rearranged to give
x(2x , 1)2 + y(2y , 1)2 + z(2z , 1)2 = 0: 1 + 4x2 1 + 4y 2 1 + 4z 2
Since each term is nonnegative, each term must be 0, and hence each variable is either 0 or 12 . The original equations then show that x = y = z = 0 and x = y = z = 21 are the only two solutions. Alternate Solution. Notice that x, y , and z are nonnegative. Multiply both sides of the inequality
y
1 + 4y 2
0
256 by (2y , 1)2, and rearrange to obtain
2 y , 1 +4y4y2 0;
and hence that y z . Similarly, z x, and x y . Hence, x = y = z and, as in Solution 1, the two solutions follow. Alternate Solution. As for solution 1, note that x = y = z = 0 is a solution and any other solution will have each of x; y and z positive. The arithmeticgeometric mean inequality (or direct computation) 2 1 + 4x2 p shows that 1 4x2 = 2x and hence x 4x 2 = y, with
1 + 4x 1 equality if and only if 1 = 4x , that is, x = . Similarly, y z with equality 2 1 1 if and only if y = and z x with equality if and only if z = . Adding 2 x y; y z and2z x gives x + y + x x + y + z. Thus equality must 1 occur in each inequality, so x = y = z = . 2 3. We denote an arbitrary permutation of the integers 1; : : : ; n by a1 ; : : : ; an. Let f (n) be the number of these permutations such that (i) a1 = 1; (ii) jai , ai+1 j 2, i = 1; : : : ; n , 1. Determine whether f (1996) is divisible by 3. Solution. Let a1 ; a2 ; : : : ; an be a permutation of 1; 2; : : : ; n with prop2
2
erties (i) and (ii). A crucial observation, needed in Case II (b) is the following: If ak and ak+1 are consecutive integers (i.e. ak+1 = ak 1), then the terms to the right of ak+1 (also to the left of ak ) are either all less than both ak and ak+1 or all greater than both ak and ak+1 . Since a1 = 1, by (ii) a2 is either 2 or 3. CASE I: Suppose a2 = 2. Then a3 ; a4 ; : : : ; an is a permutation of 3; 4; : : : ; n. Thus a2 ; a3 ; : : : ; an is a permutation of 2; 3; : : : ; n with a2 = 2 and property (ii). Clearly there are f (n , 1) such permutations. CASE II: Suppose a2 = 3. (a) Suppose a3 = 2. Then a4 ; a5 ; : : : ; an is a permutation of 4; 5; : : : ; n with a4 = 4 and property (ii). There are f (n , 3) such permutations. (b) Suppose a3 4. If ak+1 is the rst even number in the permutation then, because of (ii), a1 ; a2 ; : : : ; ak must be 1; 3; 5; : : : ; 2k,1 (in that order). Then ak+1 is either 2k or 2k , 2, so that ak and ak+1 are consecutive integers. Applying the crucial observation made above, we deduce that ak+2 ; : : : ; an are all either greater than or smaller than ak and ak+1 . But 2 must be to the right of ak+1 . Hence ak+2 ; : : : ; an are the even integers less than ak+1 . The only possibility then, is 1; 3; 5; : : : ; ak,1 ; ak ; : : : ; 6; 4; 2:
257 Cases I and II show that
f (n) = f (n , 1) + f (n , 3) + 1; n 4: (?) Calculating the rst few values of f (n) directly gives f (1) = 1; f (2) = 1; f (3) = 2; f (4) = 4; f (5) = 6: Calculating a few more f (n)'s using (?) and mod 3 arithmetic, f (1) = 1, f (2) = 1, f (3) = 2, f (4) = 1, f (5) = 0, f (6) = 0, f (7) = 2, f (8) = 0, f (9) = 1, f (10) = 1, f (11) = 2. Since f (1) = f (9), f (2) = f (10) and f (3) = f (11) mod 3, (?) shows that f (a) = f (a mod 8); mod 3; a 1: Hence f (1996) f (4) 1 (mod 3) so 3 does not divide f (1996). 4. Let 4ABC be an isosceles triangle with AB = AC . Suppose that the angle bisector of \B meets AC at D and that BC = BD + AD. Determine \A. Solution. Let BE = BD with E on BC , so that AD = EC : A 4x
B
x x
D
2x
4x
2x
C E AB = AD ; so in 4 CED and 4 CAB By a standard theorem, CB DC we have a common angle and
CE = AD = AB = CA : CD CD CB CB Thus, 4CED 4CAB , so that \CDE = \DCE = \ABC = 2x: Hence \BDE = \BED = 4x, whence 9x = 180 so x = 20 : Thus \A = 180 , 4x = 100 : Alternate Solution. Apply the law of sines to 4ABD and 4BDC to get AD = sin x and 1 + AD = BC = sin 3x : BD sin 4x BD BD sin 2x
Now massage the resulting trigonometric equation with standard identities to get sin 2x (sin 4x + sin x) = sin 2x (sin 5x + sin x) :
258 Since 0 < 2x < 90 , we get 5x , 90 = 90 , 4x ;
so that \A = 100 . 5. Let r ; r ; : : : ; rm be a given set of m positive rationalPmnumbers P 1 2 such that m k=1 rk = 1. De ne the function f by f (n) = n , k=1 [rk n] for each positive integer n. Determine the minimum and maximum values of f (n). Here [x] denotes the greatest integer less than or equal to x. Solution. Let
f (n) = n , = =
n
m X
[rk n]
k=1
m X
k=1 m X
rk ,
m X
[rk n]
k=1
frkn , [rkn]g:
k=1
Now 0 x , [x] < 1, and if c is an integer, (c + x) , [c + x] = x , [x]. m X
Hence 0 f (n) < 1 = m. Because f (n) is an integer, 0 f (n) k=1 m , 1. To show that f (n) can achieve these bounds for n > 0, we assume that rk = ak where ak ; bk are integers; ak < bk.
bk
Then, if n = b1 b2 : : : bm ; (rk n) , [rk n] = 0; k = 1; 2; : : : ; m and thus f (n) = 0. Letting n = b1 b2 : : : bn , 1, then
rkn = rk(b1b2 : : : bm , 1) = rk f(b1b2 : : : bm , bk ) + bk , 1)g = integer + rk (bk , 1):
259 This gives
Hence
rk n , [rkn] = rk(bk , 1) , [rk(bk , 1)] a a = k (bk , 1) , k (bk , 1) bk bk a a k k = ak , bk , ak , bk a = ak , k , (ak , 1) bk a = 1 , k = 1 , rk : b k
f (n) =
m X
(1 , rk ) = m , 1:
k=1
Next we give reader solutions to problems of the Second Stage Exam of the 10th Iranian Mathematical Olympiad [1995: 9{10]. 1. In the right triangle ABC (A = 90), let the internal bisectors of B and C intersect each other at I and the opposite sides D and E respectively. Prove that the area of quadrilateral BCDE is twice the area of the triangle BIC . Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Christopher J. Bradley, Clifton College, Bristol, UK; Hans Engelhaupt, Franz{ Ludwig{Gymnasium, Bamberg, Germany; Cyrus Hsia, student, University of Toronto, Toronto, Ontario; Joseph Ling, University of Calgary, Calgary, Alberta; and by Dieter Ruo, Department of Mathematics and Statistics, The University of Regina, Regina, Saskatchewan. We give the solution of Covas. If b and c are the legs, a the hypotenuse, s the semiperimeter and r the inradius of the given right triangle then it is known that r = s , a. The area of such a triangle is bc=2. On the other hand, the area of any triangle is sr. Setting the two expressions equal we have bc = 2sr = 2s(s , a) (1)
260 We see that (see gure 1.1)
C D A
E
B Figure 1.1
area of quadrilateral BCDE = area (4ABC ) , area (4AED): (2) Since AE = bc=(a + b) and AD = bd=(a + c) we have Area (4AED) = 1 bc bc . Substituting, using (2), we have 2 a+b a+c area of quadrilateral BCDE
1 1 bc bc bc , 2 2a+b a+c 1 bc = bc 1 , 2 (a + b)(a + c) abcs : 1 a(a + b + c) = = bc 2 (a + b)(a + c) (a + b)(a + c) =
(3)
Since (a + b)(a + c) = a(a + b + c) + bc = 2as + 2s(s , a) = 2s2 we can write (3) in the form Area of quadrilateral BCDE =
abc : 2s
Finally, we substitute 2rs for bc from (1), simplify, and obtain Area of quadrilateral BCDE = ar = 2(area of 4BIC ), which was to be proved. Editor's Note. Both Ling and Ruo generalized the result proving more. Here is Ruo's generalization. Let 4ABC be a triangle with angles 2, 2 , 2 , D the intersecting point of the angle bisector at B and AC , E the intersecting point of the angle bisector at C and AB , and I the incentre of 4ABC . Then
j4BCI j >=< 12 jBCDEj i >=< 4 :
(1)
261
C
Ia
D 0 I Ib r r 0 A Ic
r
E
B
Figure 1.1a Proof. Let Ia , Ib , Ic be the projections of I to BC , CA, AB respectively and note the LHS of (1) is equivalent to
j4BIIc j + j4CIIc j >=< jBICDEBj; jBICABj , jIIb AIc j >=< jBICABj , j4ADEj
and
j4ADEj >=< jIIb AIcj:
(2)
j4BCI j = 12 jBCDEj
(1)
The essence and heuristic departure point of the following is the proof of respectively
j4ADEj = jIIb AIc j (2) for triangles with right angle at A ( = 4 ). In this case IIc AIc is obviously a square, 4ADE a right triangle, \Ic ID = 0 = , \Ic IE = 0 = , consequently 0 = 4 , 0 , and (2*) becomes for IIb = IIc = r. C
D Ib r A
I
0
0
r
Ic
E
Figure 1.1b
B
262
r2 (1 + tan 0 )(1 + tan( , 0 )) = r2 ; 2
(3*)
4
which holds because of the trigonometric identity (1 + tan x)(1 + tan(
, x)) = 2: 4
Returning to the general case, we rst assume that Ic lies between A and E , and Ib between A and D, in which case (2) amounts to
r2 (cot + tan 0)(cot + tan 0 ) sin2 > r2 cot = 2 <
(3)
respectively to
(1 + tan tan 0 )(1 + tan tan 0 ) cos2 > = 1:
<
1 = 1 + tan2 , we obtain cos2 tan 0 + tan 0 = tan( 0 + 0 ) > = 1 , tan 0 tan 0 < tan :
Multiplying this equation by
(4)
The angle sum in the quadrilateral AEID is and hence
2 + (2 + ) + ( 0 + , 2 + 0 ) + (2 + ) = 2 0 + 0 = , 3( + ) = , 3 2 , 8 < > 4 > = 3 , = for : = 4 : 2< < 4
(5)
(6)
The conditions on the RHS of (6) determine the signs of (4) and one by one those of (3), (2) and (1), q.e.d.
C
Ib D A
0
# I
0
Ic E
Figure 1.1c
B
263 If Ic lies between A and E , but D between A and Ib then, in (3) and in the following, 0 has to be multiplied by ,1, and (4), (5), (6) turn into tan( 0 , 0 ) > = tan ;
and
< 2 + (2 + ) + ( 0 + , 2 , 0 ) + (2 + ) = 2 8
< > 4 0 , 0 = 3 , 2 > for = 4 ; = : < < 4 C Ib D I
A
2
E Ic
(4*) (5*) (6*)
B
Figure 1.1d from which, however, the same conclusions result as before. Since 2 + < , the only de facto applicable condition is < . Also, if E lies between A2 and Ic and D between A and Ib, < 4 is the 4only applicable condition; that the <sign holds in (2) follows directly from the gure. 2. Given the sequence a0 = 1, a1 = 2, an+1 = an + 1+(aann,,11)2 , n > 1, show that 52 < a1371 < 65. Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. Note rst that \n > 1" in the statement should have been \n 1" for the problem to be correct. We show that in general, p2n + 1 a p3n + 2 for all n 0: (1) n
p
p
p
Since n = 1371, 2n + 1 = 2743 52:374 and 3n + 2 = p4115when 64:148, 52 < a1371 < 65 would follow. To establish (1), we
rst show by induction that
an = an,1 + a 1 for all n 1: n,1
(2)
This is clearly true for n = 1 since a1 = 2 = a0 + a10 . Suppose (2) holds for some n 1. Then 2 an = (an,a1) + 1 ) a1 = 1 +a(na,1 )2 n,1 n n,1
264 and thus, from the given recurrence relation, we get an+1 = an + a1n , completing the induction. Since clearly an > 0 for all n, we see from (2) that the sequence fan g is strictly increasing. In particular, a2n1,1 1 for all n 1 and so from a2n = a2n,1 + 2 + a2n1,1 we get
a2n,1 + 2 < a2n a2n,1 + 3 for all n 1:
(3)
Now we use (3) and p induction to establish (1). The case when n = 0 is trivial since a0 = 1 < 2. Suppose (1) holds for some n 0. Then by (3),
an+1 a2n + 3 p3n + 2 + 3 = 3(n + 1) + 2 q
and
p
an+1 > a2n + 2 p2n + 1 + 2 = 2(n + 1) + 1 q
p
and our proof is complete. 3. There is a river with cities on both of its sides. Some boat lines connect these cities in such a way that each line connects a city of one side to a city on the other side, and each city is joined exactly to k cities on the other side. One can travel between every two cities. Prove if one of the boat lines is cancelled, one can travel between every two cities. Solution by Cyrus Hsia, student, University of Toronto, Toronto, Ontario. Originally, one can travel between every two cities. If we consider cities as vertices and boat lines as edges on a graph, then this graph is connected. We must show that if an edge is removed, then the graph is still connected. We interpret cities on the two sides of the river as a bipartite graph since each line connects a city of one side to a city on the other side. As well, each vertex has k edges. If we count the number of edges for each vertex on one side we are counting all the vertices because each edge has exactly one end on that side. Thus, originally we have ak edges (where a is the number of vertices on one of the sides). This shows that we must have a vertices on both sides. Now suppose, on the contrary, that by removing an edge we have two disjoint graphs. Suppose it is divided into parts U , V , X , Y where U [ V and X [ Y are the vertices of the two sides and the remaining edges are between U and X , and V and Y respectively. Further, since one edge was removed, let U and V have kjU j , 1 and kjV j incident edges respectively. If X has kjX j , 1 incident edges, that would mean that originally the whole graph was not connected. Therefore X has kjX j edges remaining and Y has kjY j, 1 edges remaining. This is impossible! kjU j, 1 6= kjV j unless k = 1. If k = 1, a = 1 or else we would have disjoint lines! Now there are cities on both sides, so a > 1, so k > 1, and it is impossible to have two disjoint bipartite graphs by removing one edge.
265
4. Prove that for each natural number t, 18 divides
A = 1t + 2t + + 9t , (1 + 6t + 8t):
Remarks by Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany; and by Stewart Metchette, Gardena, California, USA. There must be a misprint. For t = 1, A = 1 + 2 + 3 + + 9 , (1 + 6 + 8) = 30, and t = 2, A = 1+22 +32 + +92 , (1+62 +82 ) = 285 , 101 = 184. Both are not divisible by 18. 5. In the triangle ABC we have A 90 and B = 2C . Let the internal bisector of C intersect the median AM (M is the midpoint of BC ) at D. Prove that \MDC 45 . What is the condition for \MDC = 45? Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain. It suces to show that tan(\MDC ) 1. If 4ABC has sides a, b, c, in the usual order, then the condition B = 2C is equivalent to the condition b2 = c(c + a) (see this journal [1976: 74] and [1984: 287]). (1) We introduce a Cartesian frame with origin at B and xaxis along BC :
6
A
D B
M
Figure 5
C

The coordinates of A are (c cos B; c sin B ), the coordinates of C are (a; 0), and those of M are (a=2; 0). The internal angle bisector of C has slope
m1 = tan(180 , C=2) = , tan(C=2) and the slope of the median AM is c sin B : m2 = c cos B , a=2 The law of cosines gives
b2 = c2 + a2 , 2ca cos B:
(2)
266 Substituting for b2 from (1), we obtain
c(c + a) = c2 + a2 , 2ca cos B;
and hence
c cos B = (a , c)=2:
Substituting this into (2), we obtain
m2 = ,2 sin B: Since B = 2C and sin 2C = 2 sin C cos C , this equation may be rewritten as m2 = ,4 sin C cos C: Using the formula for the tangent of the angle between two lines, we get
m1 , m2 = , tan(C=2) + 4 sin C cos C : (3) 1 + m1 m2 1 + 4(tan(C=2)) sin C cos C and cos C in terms of tan(C=2). Putting t = tan(C=2),
tan(\MDC ) =
We express sin C we get
sin C =
2t (1 , t2 ) ; cos C = : (1 + t2 ) (1 + t2 )
When these are substituted into (3), it becomes t(1,2t)22 ,t + 8(1+ t)
,t(1 + t2)2 + 8t(1 , t2) 2 ,t2 = (1 + t2 )2 + 8t2 (1 , t2 ) : t 1+ t2 2 We now prove that tan(\MDC ) 1. This holds if and only if tan(\MDC ) =
8 (1 ) (1+ )
8t(1 , t2 ) , t(1 + t2 )2 8t2 (1 , t2 ) + (1 + t2 )2;
or, equivalently,
8t(1 , t2 ) , 8t2 (1 , t2 ) (1 + t2 )2 + t(1 + t2 )2;
or
8t(1 , t2 )(1 , t) (1 + t2 )2(1 + t): Dividing both sides by the positive number 1 + t, we get 8t(1 , t)2 (1 + t2 )2;
equivalent to
8t , 16t2 + 8t3 1 + 2t2 + t4 ;
which is indeed true since
t4 , 8t3 + 18t2 , 8t + 1 = (t2 , 4t + 1)2 0:
267 Equality \MDC = 45 occurs if and only if t2 , 4t + 1 = 0, where p t = tan(pC=2). This is satis ed when t = 2 , 3, i.e. C = 30 + 360 k; or t = 2 + 3, i.e. C = 150 + 360k (k = :: : ; ,2; ,1; 0; 1; 2; : : : ). The only acceptable value for C is 30 . We conclude that \MDC = 45 i A = 90 , B = 60 , C = 30 . 6. Let X be a nonempty nite set and f : X ! X a function such that for all x in X , f p(x) = x, where p is a constant prime. If Y = fx 2 X : f (x) 6= xg, prove that the number of elements of Y is divisible by p. Solution by Cyrus Hsia, student, University of Toronto, Toronto, Ontario; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Wang's solution and remarks. For each x 2 Y consider that orbit, B (x), of x de ned by
B(x) = fx; f (x);f 2(x); : : : ; f p,1(x)g: We claim that all the elements of B (x) are distinct. Suppose not. Then let j be the least positive integer such that f i(x) = f j (x) for some integer i with 0 i < j p , 1. (We de ne f i(x) = x if i = 0.) Then
x = f p(x) = f p,j (f j (x)) = f p,j (f i(x)) = f p,j+i(x) ) f j,i(x) = f j,i(f p,j+i(x)) = f p(x) = x: Since 0 < j , i j , we must have i = 0 and thus f j (x) = x. Now let p j= qj + r, where qjq, r are integers with q > 0 and 0 r < j . Clearly f (x) = x implies f (x) = x and hence f r(x) = f r(f qj (x)) = f p(x) = x: Since r < j , we must have r = 0 and thus p = qj . Since f (x) = 6 x, j > 1. On the other hand,i since j < p , q > 1. Hence p is a composite, a contradiction. Therefore, f (x) = 6 f j (x) for all i = 0; 1; 2; : : : ; p , 1, we see that B (x) Y . Next we show that the orbits of two elements of Y are either disjoint or identical. Let x; y 2 Y and suppose B (x) \ B (y ) = 6 ;. Then f l(x) = f k(y) for some integers l and k, with 0 l k p , 1. Hence y = f k(y) = f p,k(fk(y)) = f p,k(f l(x)) = f p,k+l(x); which show that y 2 B (x). It then follows that B (x) = B (y ). Therefore Y can be partitioned into disjoint orbits each having cardinality p and the result follows. Remarks. (1) Actually, the result still holds even when p = 1 since in this case Y = ; and thus jY j = 0. (2) The result need not hold if p is composite. A counterexample is given by X = f1; 2; 3; 4; 5; 6g, f (1) = 2, f (2) = 1, f (3) = 4, f (4) = 5, f (5) = 6, and f (6) = 3. In this case, p = 4 and Y = X , jY j = 6. That completes the Olympiad Corner for this issue. Send me your contests and nice solutions.
268
THE ACADEMY CORNER No. 5
Bruce Shawyer All communications about this column should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7 In the February 1996 issue, we gave the rst set of problems in the Academy Corner. Here we present solutions to the last three questions, as sent in by Sefket Arslanagic, Berlin, Germany.
Memorial University Undergraduate Mathematics Competition 1995 4. If a, b, c, d are positive integers such that ad = bc, prove that a2 + b2 + c2 + d2 is never a prime number. We will give a generalization: If a, b, c, d, r, s are positive integers such that rad = sbc, prove that r(a2 + d2 ) + s(b2 + c2 ) is never a prime number. Solution. All positive integers greater than or equal to 2 can be written as products of nitely many prime numbers. Therefore rad = sbc = p1 : : : p` (pi; pj not necessarily dierent) are the prime factors of r, a, d and similarly of factors s, b, c. Also, there exist positive integers x11; x12; : : : ; x33 such that
r = x11x12x13 = y1; a = x21x22x23 = y2; d = x31x32x33 = y3; s = x11x21x31 = z1; b = x12x22 x32 = z2; c = x13x23x33 = z3:
We write this as
x11 x21 x31 #
Q
=s
= z1
x12 x22 x32 #
Q
=b
= z2
x13 x23 x33 #
=c = z3
Q
! ! !
Q Q Q
= r = y1
= a = y2 = d = y3
269 Let
xiji
=
px 2 yi px 2 zj k = 1; : : : ; `
Y
px
Now, we get
r(a2 + s2) + s(b2 + c2) = x11 x12 x13 (x221x222 x223 + x231 x232 x233 ) +x11 x21 x31 (x212x222 x232 + x213 x223 x233 = x11 x12 x13 x221 x222 x223 + x11 x12 x13 x231 x232 x233 +x11 x21 x31 x212 x222 x232 + x11 x21 x31 x212 x223 x233 = x11 x12 x21 x222 (x13x21 x223 + x12 x31 x232 ) +x11 x13 x31 x233 (x12x31 x232 + x12 x21 x223 ) = (x11 x12 x21 x222 + x11 x13 x31 x233 ) (x13x21x223 + x12x31x232): Thus, r(a2 + d2 ) + s(b2 + c2 ) is the product of two factors of positive integers, and is greater than or equal to 2, because xij 1 for all i; j . Therefore, r(a2 + d2 ) + s(b2 + c2 ) is never a prime number. For r = s + 1, that is, ad = bc, the sum a2 + b2 + c2 + d2 is never a prime number. 5. Determine all functions f : R ! R which satisfy (x , y )f (x + y ) , (x + y )f (x , y ) = 4xy (x2 , y 2)
for all real numbers x, y . Solution. This has already appeared in CRUX [1993: 41{42]. The problem was used in the XLI Mathematics Olympiad in Poland. 6. Assume that when a snooker ball strikes a cushion, the angle of incidence equals the angle of re ection. For any position of a ball A, a point P on the cushion is determined as shown. Prove that if the ball A is shot at point P , it will go into the pocket B .
270
i
B
ir
ri
i
i
r r
P
A w
r
i
Solution. We will use analytic geometry. We let the axes be Ox OP , Oy0 = OB, Ox ? Oy; and give coordinates: O(0; 0), A(a; b), B(0;c), A (a; 0); we have
B 6y S A O
B0
P A0
x
OA : y = ab xc BA0 : y = , a x +0 c; fSg = OA \ BA ; i,e, S bac+c ; bbc+c and P bac+c ; 0 AP : y = b+a c x , c; that is, B 0 (0; ,c): Therefore jOB j = jOB 0 j = c and OBP ' OB 0 P; that is, \BPO = \B 0 PO: Because \B 0 PO = \APA0 , we obtain \BPO = \APA0 :
271
BOOK REVIEWS Edited by ANDY LIU Experience in Problem Solving  a W. J. Blundon Commemorative, edited by R. H. Eddy and M. M. Parmenter. Published by the Atlantic Provinces Council on the Sciences, 1994, ISBN # 0969896506, 82+ pages, hardbound, $25.00. Reviewed by Murray S. Klamkin, University of Alberta. This book is the culmination of eorts of its editors, whose idea it was to honour the late W. J. Blundon, by collecting his problem solving contributions to the mathematical literature. The preface contains a short bibliography of Blundon. CRUX readers before 1990 may remember some of his contributions. The book consists of all (as far as its editors are aware) the published problems and/or solutions of Blundon and which have appeared in Crux, American Mathematical Monthly, Mathematics Magazine, College Mathematics Journal, Elemente der Mathematik, SIAM Review, and Nieuw Archief voor Wiskunde. The problems and solutions are given under four categories: Geometry, Geometric Inequalitites, Number Theory, and Miscellaneous. Jack Blundon, whom I knew personally from my days on the Canadian Mathematical Olympiad Committee when he was the chairman, was a very personable man as well as mathematician, and was particularly fond of geometric inequalities (as can be seen from the corresponding section of the book). Further evidence of this is his several papers on this topic. In particular, his seminal paper Inequalities associated with the triangle, Canadian Mathematics Bulletin 8 (1963) 615627, was a catalyst for a subsequent series of papers from around the world dealing with the theory for triangle inequalities, on which very little had been done before. He was also, as the editors note and I agree, an ardent disciple of elegance in mathematical exposition, often rewriting a proposal or solution several times in order to meet his exacting standards. As a problem section editor, this is something I would hope my contributors would also do. As a very small sampling of his nice, easily understood proposals from each of the four sections and which did not appear in Crux, we have: If, in triangle ABC , we have sin A + sin B + sin C p = 3; cos A + cos B + cos C prove that at least one angles of the triangle is 60 .
272
For any triangle (other than equilateral), with circumcentre O, incentre I and orthocentre H , let the angles have measures . Prove that (1) 1 < OH IO < 3; IH < 2 ; (2) 0 < OH 3 < 1 if < 60 , IH = IO if = 60 ; (3) 0 < IH IO IH 1 < IO < 2 if > 60 . Find all solutions in integers of the equation
y2 + y = x4 + x3 + x2 + x:
Find necessary and sucient conditions on a, b and c in order that the system of equations
1 x + x1 = a; y + y1 = b; xy + xy =c
has at least one solution. Finally, it is to be noted that the editors, who were younger colleagues of Professor Blundon at Memorial University, have done a good job in compiling this nice collection of eightyone problems and solutions. Not only did it remind me of results that I had forgotten and was looking for, it also suggested to me a number of new problems which I will be submitting to Crux! [Ed.: The book is available, only from: Atlantic Provinces Council on the Sciences Memorial University of Newfoundland P.O. Box 4200, St. John's, Newfoundland Canada A1C 5S7 for CDN $25 plus CDN $3 for shipping and handling.]
273
PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be typewritten or neatly handwritten, and should be mailed to the EditorinChief, to arrive no later that 1 April 1997. They may also be sent by email to cruxeditor@cms.math.ca. (It would be appreciated if email proposals and solutions were written in LATEX, preferably in LATEX2e). Graphics les should be in epic format, or plain postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication. 2164. Proposed by Toshio Seimiya, Kawasaki, Japan. Let D be a point on the side BC of triangle ABC , and let E and F be the incentres of triangles ABD and ACD respectively. Suppose that B , C , E, F are concyclic. Prove that
AD + BD = AB : AD + CD AC
2165. Proposed by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore. Given a triangle ABC , prove that there exists a unique pair of points P and Q such that the triangles ABC , PQC and PBQ are directly similar; that is, \ABC = \PQC = \PBQ and \BAC = \QPC = \BPQ, and the three similar triangles have the same orientation. Find a Euclidean construction for the points P and Q. 2166. Proposed by K.R.S. Sastry, Dodballapur, India. In a rightangled triangle, establish the existence of a unique interior point with the property that the line through the point perpendicular to any side cuts o a triangle of the same area.
274 2167. Proposed by Sefket Arslanagic, Berlin, Germany.
Prove, without the aid the dierential calculus, the inequality, that in a right triangle
a2 (b + c) + b2(a + c) 2 + p2; abc where a and b are the legs and c the hypotenuse of the triangle.
etokrzyski, Poland. 2168. Proposed by Jan Ciach, Ostrowiec Swi
Let P be a point inside a regular tetrahedron ABCD, with circumradius R and let R1 ; R2 ; R3 ; R4 denote the distances of P from vertices of the tetrahedron. Prove or disprove that
R1R2 R3R4 43 R4;
and that the maximum value of R1 R2 R3 R4 is attained. 2169. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. AB is a xed diameter of circle ,1(0;R). P is an arbitrary point of its circumference. Q is the projection onto AB of P . Circle ,2 (P1PQ) intersects ,1 at C and D. CD intersects PQ at E . F is the midpoint of AQ. FG ? CD, where G 2 CD. Show that: 1. EP = EQ = EG, 2. A, G and P are collinear.
2170. Proposed by Tim Cross, King Edward's School, Birmingham, England. Find, with justi cation, the positive integer which comes next in the sequence 1411; 4463; 4464; 1412; 4466; 4467; 1413; 4469; : : : . [Ed.: the answer is NOT 4470.] 2171. Proposed by JuanBosco Romero Marquez, Universidad de Valladolid, Valladolid, Spain. Let P be an arbitrary point taken on an ellipse with foci F1 and F2, and directrices d1 , d2 , respectively. Draw the straight line through P which is parallel to the major axis of the ellipse. This line intersects d1 and d2 at points M and N , respectively. Let P 0 be the point where MF1 intersects NF2. Prove that the quadrilateral PF1P 0 F2 is cyclic. Does the result also hold in the case of a hyperbola? 2172. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Let x; y;z 0 with x + y + z = 1. For xed real numbers a and b, determine the maximum c = c(a; b) such that a + bxyz c(yz + zx + xy ).
275
2173. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Let n 2 and x1 ; : : : ; xn > 0 with x1 + : : : + xn = 1. Consider the terms ln = and where
n X
k=1
s
(1 + xk )
rn = Cn
n Y k=1
1 , xk xk
p11+,xxk
k
Cn = (px , 1)n+1(pn)n=(n + 1)n,1:
1. Show l2 r2 . 2. Prove or disprove: ln rn for n 3.
2174. Proposed by Theodore Chronis, student, Aristotle University of Thessaloniki, Greece. Let A be an n n matrix. Prove that if An+1 = 0 then An = 0. 2175. Proposed by Christopher J. Bradley, Clifton College, Bristol, UK. The fraction 16 can be represented as a dierence in the following ways: 1 1 1 1 1 1 1 1 1 1 1 1 = , ; = , ; = , ; = , : 6 2 3 6 3 6 6 4 2 6 5 30 1 In how many ways can the fraction 2175 be expressed in the form
1 1 1 = , ; 2175 x y
where x and y are positive integers? 2176. Proposed by Sefket Arslanagic, Berlin, Germany. Prove that v u n uY n t (
k=1
ak + bk)
v u n uY n t
where a1 ; a2 ; : : : ; an > 0 and n 2 N.
k=1
ak +
v u n uY n t
k=1
bk
276
SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems.
1987. [1994: 250; 1995: 283285] Proposed by Herbert Gulicher, Westfalische WilhelmsUniversitat, Munster, Germany. A1 In the gure, B2 C1 kA1A2 , B3C2 kA2 A3 and B1 C3 kA3 A1 . Prove that B2C1 , B3 C2 C3 B2 and B1 C3 are concurrent if and only if A1C3 A2C1 A3C2 = 1: C3B3 C1B1 C2B2
B3
C2
A2 A3 C1 B1 II. Solutionby anonymous (spotted on the internet by Waldemar Pompe, student, University of Warsaw, Poland). Just apply Ceva's theorem to 4B1 B2 B3 . Editor's comments by Chris Fisher. 1. Note that if we permit Bi to lie on the side of the triangle extended beyond Ai,1 or Ai+1 (as Bradley did in solution I) then Ceva's theorem fails to apply exactly when B1 ; B2 ; B3 are collinear. Thus the alternative condition in the middle of page 284 (namely u = pq=(p + q , 1) in Bradley's notation) applies if and only if B1 2 B2 B3 . As a consequence, Bradley's extended version of the problem can be restated as
A1A3 A2C1 A3C2 = 1 if and only if either C3B3 C1B1 C2B2 B2C1; B3C2; and B1C3 are concurrent or B1 2 B2B3:
2. It makes an amusing exercise to prove directly that B1 2 B2B3 if and only if C1 2 C2 C3 (in the notation of Gulicher's problem).
277
2067. [1995: 235] Proposed by Moshe Stupel and Victor Oxman, Pedagogical Religious Girls' College \Shanan", Haifa, Israel. Triangle ABC is inscribed in a circle ,. Let AA1 ; BB1 and CC1 be the bisectors of angles A; B and C , with A1 ; B1 and C1 on ,. Prove that the perimeter of the triangle is equal to AA1 cos , A2 + BB1 cos , B2 + CC1 cos , C2 :
Solution by Miguel Angel Cabezon Ochoa, Logro~no, Spain. Let the internal bisector of \BAC meet the circumcircle of 4ABC again in A1 . Let R be the radius of the circumcircle.
A
A 2
B
A 2
A1 From 4BAA1 , we have Thus
Similarly Therefore
C
AA1 = 2R, so that sin B + A2 AA1 = 2R sin ,B + A2 : ,
AA1 cos , A2 = 2R sin ,B + A2 cos , A2 , = R sin(B + A) + sin B c = R sin C + R sin B = b + 2 : BB1 cos , B2 = c +2 a ;
CC1 cos , C2 = a +2 b :
AA1 cos , A2 + BB1 cos , B2 + CC1 cos , C2 = b +2 c + c +2 a + a +2 b :
278 Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Berlin, Germany; FRANCISCO BELLOT Spain; SEFKET ARSLANAGIC, ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; SABIN CAUTIS, student, Earl Haig Secondary School,North York, Ontario; HANS ENGELHAUPT, Franz{Ludwig{ Gymnasium, Bamberg, Germany; HIDETOSI FUKAGAWA, Gifu, Japan; TOBY GEE, student, the John of Gaunt School, Trowbridge, England; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursul Y, Ferris State Uniinengymnasium, Innsbruck, Austria; VACLAV KONECN versity, Big Rapids, Michigan, USA; MITKO CHRISTOV KUNCHEV, Rousse, Bulgaria; VEDULA N. MURTY, Andhra University, Visakhapatnam, India; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; WALDEMAR POMPE, student, University of Warsaw, Poland; CRISTOBAL SANCHEZ{ RUBIO, I.B. Penyagolosa, Castellon, Spain; HEINZJURGEN SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawasaki, Japan; ASHSIH KR. SINGH, student, Kanpur, India; PANOS E. TSAOUSSOGLOU, Athens, Greece; PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA; and the proposers. There was one anonymous solution. Amengual Covas points out that this problem has already appeared in print, in A Treatise on Plane Trigonometry by E.W. Hobson, Cambridge University Press, 2nd Edition, 1897, Example 2 on page 194.
2069. [1995: 235] Proposed by D.J. Smeenk, Zaltbommel, the Neth
erlands.
M is a variable point of side BC of triangle ABC . A line through M intersects the lines AB in K and AC in L so that M is the midpoint of segment KL. Point K 0 is such that ALKK 0 is a parallelogram. Determine the locus of K 0 as M moves on segment BC . I. Essentially the same solution by Gottfried Perz, Pestalozzigymnasium, Graz, Austria; Waldemar Pompe, student, University of Warsaw, Poland; and Toshio Seimiya, Kawasaki, Japan. Suppose that the dilatation with centre A and ratio two transforms the points B , C and M to B 0 , C 0 and N respectively. Since M is the midpoint of both KL and AN , we have that AKNL is a parallelogram. By de nition, 0 is a parallelogram as well, so that K is the midpoint of K 0 N ; also, ALKK 0 K N and AC 0 0are00 parallel. De ne C 00 to be the point such that A is the midpoint of C C . We conclude that, as M varies on BC , N varies on B0 C 0, so that the position of K 0 must vary on the segment B0 C 00. Editor's comments: 1. Note that this solution provides an explicit construction of the points K and L.
279 2. Smeenk points out that K 0 helps in nding the position of M for which the corresponding segment KL has minimum length. (KL = AK ;, and AK 0 has its minimum0 length when K 0 is the foot of the perpendicular 00 from A to the line B C .) II. Essentially the same solution by Tim Cross, King Edward's School, Birmingham, England; Hidetosi Fukagawa, Gifu, Japan; Walther Janous, Ursulinengymnasium, Innsbruck, Austria; Mitko Christov Kunchev, Rousse, Bulgaria; Ashsih Kr. Singh, student, Kanpur, India. Take A to be the origin and set the vectors AB = B , etc. Then M = tB + (1 , t)C, where t varies from 0 to 1 as M moves from C to B. Suppose that K = B and that L = C . Because M is the midpoint of KL, we have
B + C = tB + (1 , t)C; 2
so that = 2t and = 2(1 , t) (since B and C are linearly independent). Since ALKK 0 is a parallelogram, it follows that
K = K ,L = t(2B + (1 , t)(,2C ): Thus the locus of K 0 is the segment joining ,2C (where t = 0) to 2B where t = 1).
Editor's comment: The algebra makes clear that it is not necessary to restrict M to the segment BC ; as t ranges over the real numbers, K 0 moves along its line, while M moves along the line BC . Berlin, Germany; CARL BOSLEY, Also solved by SEFKET ARSLANAGIC, student, Washburn Rural High School, Topeka, Kansas, USA; JORDI DOU, Barcelona, Spain; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bam berg, Germany; CRISTOBAL SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon, Spain; TOSHIO SEIMIYA, Kawasaki, Japan; and the proposer. One anonymous solution was received  see note at the start of this section. There was one incorrect solution received.
2070. [1995: 236] Proposed by Joaqun Gomez Rey, IES Luis Bu~nuel, Alcorcon, Madrid, Spain. For which positive integers n is the Catalan number odd?
1 2n n+1 n
280 I Virtually identical solutions by Toby Gee, student, the John of Gaunt School, Trowbridge, England; Douglas E. Jackson, Eastern New Mexico University, Portales, New Mexico, USA; Thomas Leong, Staten Island, NY, USA; Andy Liu, University of Alberta, Edmonton, Alberta; Waldemar Pompe, student, University of Warsaw, Poland; and the proposer. ,2n 1 It is wellknown that the nth Catalan number, Cn = n+1 n , satis es the recurrence relation
Cn+1 = C0Cn + C1Cn,1 + : : : + CnC0 n 0: We show that Cn is odd if and only if n = 2k , 1 for some integer k 0. We use induction on n. Since C0 = 1, the assertion is true for n = 0. Assume that among the numbers C0, C1, : : : , Cn, only those with n of the form 2k , 1 are odd. If n + 1 is even, then (n,1)=2 X Cn+1 = 2 Ck Cn,k; k=0
which is even. On the other hand, if n + 1 is odd, then
Cn+1 = 2
n, 1)=2 X
(
k=0
2 CkCn,k + Cn= 2;
showing that Cn+1 is odd if and only if Cn=2 is odd. But Cn=2 is odd if and only if n2 = 2k , 1 for some integer k 0. Thus Cn+1 is odd if and only if n + 1 = 2(2k , 1) + 1 = 2k+1 , 1, which completes the proof. II Solution by Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany. ,2n 1 th Let Cn = n+1 n denote the n Catalan number, n = 1, 2, 3, : : : , and let F (n) denote the highest power of 2 that divides n. Then it is well known that
F (n!) = n2 + n4 + n8 + : : : : From this, it is easy to see that F (n!) n , 1, with equality if and only if n = 2k for some integer k 0. [More generally, it is known, and easy to show, that if p is prime and n = ar ar,1 : : : a1 a0 is the basep representation of n, then r X n , ai h = p ,i=01 ; where ph kn! [See, for example, Theorem 2.30 in Elementary Introduction to Number Theory, 3rd edition, by Calvin T. Long  Ed.]
281 Since
n)! = 2n 1 3 5 (2n , 1) ; Cn = (n (2 + 1)! n! (n + 1)! Cn is oddk if and only if F ((n+1)!) = n, which is true if and only if n+1 = 2k or n = 2 , 1 for some integer k 1. We can also allow k = 0 since C0 = 1
is odd. Solutions similar or equivalent to II above were submitted by SEFKET Berlin, Germany; CARL BOSLEY, student, Washburn Rural ARSLANAGIC, High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; SABIN CAUTIS, student, Earl Haig Secondary School, North York, Ontario; EMERIC DEUTSCH, Brooklyn, NY, USA; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; SOLOMON W. GOLOMB, Univ of Southern California, Los Angeles, CA, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOE HOWARD, New Mexico Highlands University, Las Vegas, NM, USA; WALTHER JANOUS, Ursul Y, Ferris State Uniinengymnasium, Innsbruck, Austria; VACLAV KONECN versity, Big Rapids, Michigan, USA; HARRY SEDINGER, St. Bonaventure University, St. Bonaventure, NY, USA; KEEWAI LAU, Hong Kong; HEINZ JURGEN SEIFFERT, Berlin, Germany; SKIDMORE COLLEGE PROBLEM GROUP Saratoga Springs, NY, USA; CHRIS WILDHAGEN, Rotterdam, the Netherlands; KENNETH M. WILKE, Topeka, Kansas, USA; PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA (two solutions). Several solvers out the wellknown fact that the highest power of 2 , pointed that divides 2nn is equal to the number of ones in the binary representation of n. This is, of course, an immediate consequence of the more general fact mentioned in the editor's comment in solution II above. This problem is certainly not new. Deutsch supplied the reference: \The Par ity of the Catalan Numbers via Lattice Paths" by Omer Egecioglu, Fibonacci Quarterly, (21), 1983, 65{66. Godin gave the reference: \Time Travel and other Mathematical Bewilderments" by Martin Gardner, W.H. Freeman and Co. In fact, the result of this problem was also mentioned by Martin Gardner in his article \Mathematical Games; Catalan number: an integer sequence that materializes in unexpected places", Scienti c American, 234(6), 1976, 120{125, but he did not indicate any proof.
2071. [1995: 277] Proposed by Toshio Seimiya, Kawasaki, Japan.
P is an interior point of an equilateral triangle ABC so that PB 6= PC , and BP and CP meet AC and AB at D and E respectively. Suppose that PB : PC = AD : AE. Find angle BPC . Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. We claim that \BPC = 120. Let the side of the triangle be 1. Further, let x = AE; y = AD; = \BCE and ' = \CBD. By the law of
282 cosines p (for 4ABD and 4AEC p respectively) we get (since cos 60 = 1=2): BD = 1 + y2 , y and CE = 1 + x2 , x. Via the law of sines applied to p 1,y 3 2 BD 1+y ,y 2 p3 1,x 1,x 4EBC : sin = CE sin 60 = p1 + x2 , x 2 ' = PC = x [by assumption for the problem!] 4BCP : sin sin PB y 2 x,x2 , that is (with X = x , x2 and Y = y , y 2) y , y Hence, p1+y2 ,y = p1+ x2 ,x
4BCD :
sin ' =
1,y
sin 60 =
p
p1X, X = p1Y, Y :
(1)
Here, since 0 < x; y < 1; 0 < X; Y 1=4: Now, (1) () X 2 (1 , Y ) = Y 2 (1 , X ) () (X , Y )(XY , X , Y ) = 0: Thus, two cases are to be considered. (a) XY , X , Y = 0 () (X , 1)(Y , 1) = 1; which is impossible. (b) X = Y () x , x2 = y , y 2 () (y , x)(x + y , 1) = 0 () y = 1 , x (since x 6= y). Hence 4ABD is congruent to 4BCE ) \ABD = ) + ' = 60 ) \BPC = 180 , ( + ') = 120 : Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and LOPEZ MARIA ASCENSION CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; JORDI DOU, Barcelona, Spain; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; RICHARD I. HESS, Rancho Palos Verdes, California, USA; Y, Ferris State University, Big Rapids, Michigan, USA; VACLAV KONECN P. PENNING, Delft, the Netherlands; D.J. SMEENK, Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; HOE TECK WEE, student, Hwa Chong Junior College, Singapore; one anonymous solver and the proposer. There were two incorrect solutions. etokrzyski, 2073?. [1995: 277] Proposed by Jan Ciach, Ostrowiec Swi
Poland. Let P be an interior point of an equilateral triangle A1 A2 A3 with circumradius R, and let R1 = PA1 , R2 = PA2 , R3 = PA3 . Prove or disprove that
R1R2R3 98 R3:
283 Equality holds if P is the midpoint of a side. [Compare this problem with Crux 1895 [1995: 204].] I Solution by KeeWai Lau, Hong Kong. Without loss of generality p we assume that R = 2, so that the length of each side of the triangle is 2 3. Let \A1 PA2 = ; \A2 PA3 = ; so that \A1 PA3 = 2 , , . Applying the cosine law to A1 PA2; A2 PA3 and A1PA3 , respectively, we obtain cos = (R12 + R2 2 , 12)=(2R1R2 ); cos = (R22 + R3 2 , 12)=(2R2R3 ) and cos( + ) = (R1 2 + R3 2 , 12)=(2R1R3 ): Since [cos cos , cos( + )]2 = (1 , cos2 )(1 , cos2 ), we obtain after
some simpli cation that
R14 + R24 + R34 , R12R22 , R22R32 , R32R12, 12(R1 2 + R2 2 + R3 2 ) + 144 = 0:
Thus
R32 = 21 (z + 24 (12y , 3z2)1=2);
(1)
where y = R1 2 R2 2 ; z = R1 2 + R2 2 , 12 and z 2 4y: By considering R1 = R2 = R3 = 2, we see that we should choose in (1) the , sign from . It follows that
f (y;z) := R12R22R32 = 12 y[z + 24 , (12y , 3z2)1=2]: From df=dz = 0, we obtain z = ,py and f (y; ,py) = y (12 , 2py ) 64: p p Now at the boundary z 2 = 4y , we have jR1 , R2 j = 12 or R1 + R2 = 12, so that ,6 z 0. Hence f = z 2(z +24)=8 81 and equality holds when z = ,6 or P is the midpoint of a side. This proves the required inequality. II Solution by Manuel Benito Munoz and Emilio Fernandez Moral, I.B. Sagasta, Logro~no, Spain The proposed inequality holds (equality only on the midpoint of a side), as a particular case (n = 3) of the following:
R1R2 : : : Rn (1 + cosn n ) Rn when P is any point from a closed regular ngon, A1 A2 : : : An ; with circumradius R and Ri = PAi. (Equality holds only on the midpoint of a side.) Without loss of generality, let us suppose that R = 1 and that the vertices of the ngon are the points 1; 2; : : : ; n of the complex plane, where k = e 2ni (k,1). Let z = ei represent the point P .
284 By the maximum modulus principle applied to the function nk=1 (z , k ); the modulus of that function, i.e., PA1 PA2 : : : PAn = nk=1 jz , k j, assumes its greatest value on the boundary of the ngon. Now, geometrical considerations permit us to avoid much analytical treatment and we conclude that the maximum value is attained on the midpoint of any side Ai Ai+1 : Let P be a point on the side A1 A2 of the ngon; the vertices A3 ; : : : ; An of the ngon are pairwise arranged symmetrically with respect to the mediatrix of the segment A1 A2 (except for the single point A n+1 when n is odd). 2 +1 Let Ak ; Al be one of such pairs of vertices; if M is the midpoint of A1 A2 we have, by the GAmeans inequality and \shortest path principle", that
PA + PAl 2 MAk + MAl 2 k = MAk MAl PAk PAl 2 2
(the latter because MAk = MAl ). When n is even, as PA1 PA2 MA1 MA2 , we are done. When n is odd, let Am be the unpaired vertex and put A1 A2 = 2l, PM = x and MAm = a. Therefore, p
PA1 PA2 PAm = (l2 , x2 ) a2 + x2; and the maximum value of that function for x 2 [0; l] is attained at x = 0; that is, when P = M: (There are no other critical points because a > l:) So that, for every n, we have shown PA1 PA2 : : : PAn MA1 MA2 : : : MAn; where M is the midpoint of a side. Now nally
(PA1 PA2 : : : PAn)2 = nk=1 jz , k j2 = nk=1 (z , k ) nk=1 (z , k ) = (z n , 1) (z n , 1) = 2n , 2n cos n + 1; , and if, P is the midpoint of A1 A2 (, = cos ; = n ); that value is n , , , cos n 2n , 2 cosn n cos + 1 = cosn n + 1 2 , as required.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; WOLFGANG GMEINER, Bundesgymnasium Spittal/Drau, Spittal, Austria (also found the generalization of ); WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta(with a generalization from triangles to tetrahedra); Y, Ferris State University, Big Rapids, Michigan, USA; VACLAV KONECN HEINZJURGEN SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawasaki,
285 Japan; JOHANNES WALDMANN, FriedrichSchillerUniversitat, Jena, Germany; and the proposer.
2074. [1995: 277] Proposed by Stanley Rabinowitz, Westford, Massachusetts. The number 3774 is divisible by 37, 34 and 74 but not by 77. Find another fourdigit integer abcd that is divisible by the twodigit numbers ab, ac, ad, bd and cd but is not divisible by bc. Solution by Paul Yiu, Florida Atlantic University, Boca Raton, Florida, USA. The two possible solutions are 1995 and 2184: Let N = abcd = 100 (ab)+ cd be divisible by ab. Then cd must be divisible by ab as well; say, cd = k (ab) where k < 10. Since N is divisible by cd, so is 100 (ab). It follows that k is a divisor of 100. As such, it must be one of 1, 2, 4, 5. Now, a direct computer search reveals that only 1995, 2184 and 3774 satisfy the requirement. Note: apart from numbers of the form aaaa, 1155 is the only fourdigit number divisible by all twodigit numbers \contained in it". Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; KEITH EKBLAW, Walla Walla, Washington, USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; JEFFREY K. FLOYD, Newnan, Georgia, USA; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHER JANOUS, Y, Ferris State Ursulinengymnasium, Innsbruck, Austria; VACLAV KONECN University, Big Rapids, Michigan, USA; KEEWAI LAU, Hong Kong; J.A. MCCALLUM, Medicine Hat, Alberta; STEWART METCHETTE, Gardena, California, USA; P. PENNING, Delft, the Netherlands; ROBERT P. SEALY, Mount Allison University, Sackville, New Brunswick; HEINZ JURGEN SEIFFERT, Berlin, Germany; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; CHRIS WILDHAGEN, Rotterdam, the Netherlands; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. A bit over half of the solvers found both solutions, often by computer, as there does not appear to be a short way to do the problem by hand. The proposer also used a computer, but it seems to have let him down, as he thought that 1995 was the only answer! Metchette also gave the \solution" 0315, which works, but is not a true fourdigit number. He wonders why all the solutions are multiples of 3; can any reader see an easy explanation?
286
2075. [1995: 278] Proposed by Christopher J. Bradley, Clifton College, Bristol, UK ABC is a triangle with \A < \B < \C , and I is its incentre. BCL, ACM , ABN are the sides of the triangle with L on BC produced, etc., and the points L, M , N chosen so that 1 1 1 \CLI = (\C , \B ); \AMI = (\C , \A); \BNI = (\B , \A): 2 2 2 Prove that L, M , N are collinear. Solution by Gottfried Perz, Pestalozzigymnasium, Graz, Austria. We have \AIN = 180 , (\BAI + \BNI ) \A \B , \A = 180 , + 2 2 \ B = 180 , = \NBI; 2 which implies that triangles 4ANI and 4INB are similar, or
AN : NI : AI = NI : BN : BI Hence it follows (via AN BN = NI 2 ) that AN = NI 2 = AI 2 : BN BN 2 BI 2 Analogously we get
BL = BI 2 ; CM = CI 2 CL CI 2 AM AI 2 and nally AN BL CM = AI 2 BI 2 CI 2 = 1 BN CL AM BI 2 CI 2 AI 2 whence (by the converse of Menelaus' Theorem) M , N and L are collinear. Perz also remarks that, more generally, L, M , N are collinear, if P is a point inside or outside 4ABC and L; M; N are such that \CLP = \BCP , \PBC; \AMP = \CAP , \PCA; \BNI = \ABP , \PAB; where all angles are oriented angles.
287 Also solved by TOBY GEE, student, the John of Gaunt School, Trowbridge, England; CYRUS HSIA, student, University of Toronto, Toronto, On Y, Ferris State University, Big Rapids, Michigan, tario; VACLAV KONECN USA; KEEWAI LAU, Hong Kong; P. PENNING, Delft, the Netherlands; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.
2077. [1995: 278] Proposed by Joseph Zaks, University of Haifa, Israel. The determinant
z1z1 z2z2 z3z3 z4z4
z1 z2 z3 z4
z1 z2 z3 z4
1 1 1 1
equals 0 if and only if the four complex numbers z1 , z2 , z3 , z4 satisfy what simple geometric property? Solutionby Vaclav Konecny, Ferris State University, Big Rapids, Michigan, USA. [modi ed slightly by the editor.] Let D denote the given determinant. If any two (or more) of the four complex numbers are equal, z1 = z2 , say, then D = 0, and the points z2 , z3 , z4 must be collinear (or concyclic). We thus assume that the four numbers are all distinct. It is wellknown that an equation of the circle passing through three noncollinear points (x2 ; y2 ), (x3 ; y3, (x4 ; y4 ), can be written as
x22 + y22 x y 1
x22 + y22 x2 y2 1
= 0: x32 + y32 x3 y3 1
x4 + y4 x4 y4 1
Letting z1 = x + {y and zk = xk + {yk for k = 2; 3, 4, we have
z1 z1 x z1 1
x2 + y 2 x y 1
2 y222 x2 y2 1
: D =
zz23zz23 xx23 zz23 11
= ,2{
xx22 +
3 + y3 x3 y3 1
z4 z4 x4 z4 1
x2 + y 2 x4 y4 1
4 4 Thus D = 0 if and only if z1 lies on the circle passing through the three points z2 , z3 and z4 . There the sought property is that z1 , z2 , z3 , z4 are
either concyclic or collinear (which can be viewed as a degenerate case). [Editor's note: This is certainly a known result. The references supplied by several readers are listed below. The number, if there is one, after a solver's name corresponds to the reference given by that solver.
288 Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Berlin, Germany; FRANCISCO BELLOT Spain; SEFKET ARSLANAGIC, ROSADO, I.B. Emilio Ferrari, Valladolid, Spain [3]; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; WOLFGANG GMEINER, Millstatt, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA [2]; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEEWAI LAU, Hong Kong; HEINZJURGEN SEIFFERT, Berlin, Germany [1]; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer. One anonymous solution [4] and one incomplete solution were also received. References: [1] Conway, J.B., Functionsof One Complex Variable, SpringerVerlag, 2nd ed., 1978, p. 49, Proposition 3.10. [2] Dodge, Clayton W., Complex Numbers, 1993, p. 135 and p. 199. [3] Krzyz, Problems in Complex Variable Theory, Elsevier, PWN, 1971, P. 5 and p. 142. [4] Schwerdtfeger, H., Geometry for Complex Numbers, Dover, NY, 1962.
In the May 1996 issue of CRUX [1996: 168], we asked \Do you know the equation of this curve?"
Here is a hint { it is known as the \butter y"!
289
Letter from the Editor Changes to CRUX
Readers will be interested to learn about some of the changes that are in the works for CRUX. First, CRUX is now available as an online supplement for individual subscribers at http://camel.math.ca/CRUX This new service was instituted at the beginning of September, and is proving to be very popular. A survey of the trac accessing CRUX online shows that many people are sur ng to take a look, and as a result, we have several new subscriptions. Information about each issue will be available to anyone who accesses the CRUX online site. As well as seeing the table of contents, a two page synopsis is also available. This gives a short summary of what is in each published article, as well as some highlights of what is in the various corners. One of the problems is given as a \free sample". This is to try to encourage more problem solvers to become involved. When individualsubscribers renew their subscription for 1997, they will receive information from the CMS Executive Oce on how they can activate their account on CAMEL and gain access to CRUX online. In the future, CRUX online will be an extra free service to individuals with a normal (hard copy) subscription. As well, an online only service (no hard copy) will be available. The rates will be published in the next issue of CRUX. Since CRUX is copyrighted by the Canadian Mathematical Society, there are restrictions on the use of the online version (as there is for the printed version). Here is the statement that appears online: All rights reserved. For private and personal use only. No further distribution of these materials is permitted without the express written permission of the Canadian Mathematical Society (managingeditor@cms.math.ca). Teachers wishing to use these materials are particularly encouraged to contact the CMS for permission. Also, in 1997, CRUX is proposing to increase the quantity of high school level materials in each issue. This will not involve any decrease in the quantity of each of the present sections. We are proposing to increase the issue size to 64 pages (at no increase in the subscription price over that planned for 1997 based on 48 pages per issue). Thus, subscribers, especially those who have connections with high schools, would get a lot more for their money. Watch for a further announcement in the next issue. Bruce Shawyer EditorinChief
290
A Note on the Mean Value Theorem Finbarr Holland
Department of Mathematics, University College, Cork, Ireland Every student of Calculus knows that given two distinct points A; B on the graph of a smooth function there is a point C on the arc of the curve joining A and B where the tangent to the graph is parallel to the chord joining A and B. Moreover, it is easy to construct examples where the number of such points C is either nite, countably in nite or uncountable; see the exercises. The purpose of this note is to show that if the function is convex, then there are only two possibilities: namely, either there is a unique point or uncountably many such points. We recall the de nition of convexity. Geometrically it means that the portion of the graph joining any two points on it must lie below the chord joining them. Analytically, f is convex on an interval [a; b], with a < b, if
, f (s) (x , s); whenever a s x t b. f (x) f (s) + f (t)t , s The result we wish to establish depends on a lemma which we will deal with rst. Lemma 1 Suppose f is convex on [a; b] and dierentiable at c 2 (a; b). Then f (c) + f 0(c)(x , c) f (x); 8 x 2 [a; b]: Proof. To see this, suppose c < t < x b. Then
f (c) (t , c): f (t) f (c) + f (xx) , ,c
Hence
f (t) , f (c) f (x) , f (c) t,c x,c
and so, letting t ! c+ , we deduce that
f (c) ; f 0(c) f (xx) , ,c
291 that is, (x , c)f 0 (c) f (x) , f (c) which proves the desired result for x > c. Next, if a x < t < c, then
Hence
f (x) f (t) f (x) + f (cc) , , x (t , x) f (x) (t , c + c , x) = f (x) + f (cc) , ,x f ( c ) f (x) (t , c) + f (c) , f (x) = f (x) + c , ,x f ( x ) f (c) (t , c): = f (c) + x , ,c f (t) , f (c) f (x) , f (c) t,c x,c
and so, letting t ! c, , we deduce that
f (c) f 0(c) f (xx) , ,c ;
that is, (x , c)f 0 (c) f (x) , f (c) which proves the desired result for x < c. This completes the proof of the lemma since its conclusion certainly holds when x = c. This result means that the graph of y = f (x) lies above the tangent line y = f (c) + f 0(c)(x , c).
Theorem 1 Let ,1 < a < b < 1. Suppose f is continuous and convex on [a; b], and dierentiable on (a; b). Then the equation
f (a) f 0(c) = f (bb) , ,a
has either a unique solution in (a; b) or in nitely many solutions. Proof. The Mean Value Theorem guarantees that the solution set is nonempty. Suppose c1 ; c2 are two distinct solutions in (a; b), so that
f (a) = f 0(c ): f 0(c1) = f (bb) , 2 ,a
We can and do suppose that c1 < c2 . It follows from the lemma that
f (c1) + f 0(c1)(c2 , c1 ) f (c2)
and
f (c2) + f 0(c2)(c1 , c2) f (c1):
292 Hence
f (c1) : f 0(c1) = f 0(c2) = f (cc2) , 2 , c1 But, by convexity, the line joining the two points (c1 ; f (c1)); (c2 ; f (c2)), lies above the graph of f that is, we have f (x) f (c1) + f 0(c1)(x , c1); 8x 2 [c1; c2 ]: Applying the lemma once more, we deduce that
f (x) = f (c1) + f 0(c1)(x , c1); 8 x 2 [c1; c2 ]: In other words, f 0 (x) = (f (b) , f (a))=(b , a) for all x 2 [c1 ; c2 ]. It follows that there are in nitely many points with this property as claimed.
Exercises 1. Concoct examples to show that the number of points on the arc of a smooth curve where the tangent to the curve is parallel to the chord joining the ends of the arc can be (i) nite; (ii) uncountable. 2. Show that, for every nonnegative integer n, the equation
tan x = x +3n
has a unique solution on the interval [0; =2). 3. Consider the function g de ned on [,1; 1] as follows: 3 g(x) = x0; sin(1=x); ifif 0x <= j0x.j 1, Show that g is continuous on [,1; 1] and has a continuous derivative on (,1; 1). Using the previous exercise, or otherwise, show that there is a countably in nite number of points on the graph of y = g (x) where the tangent is parallel to the chord joining the points (,1;g (,1)) and (1; g (1)). 4. Deduce, and/or prove directly, that the function g is not convex on [,1; 1]. 5. Show that f is convex on [a; b] if and only if the line segment joining any two points in the set is a subset of S .
S = f(x; y) : f (x) y;a x bg
293 6. Suppose f is continuous on [a; b], dierentiable on (a; b) and the tangent at any point on the graph of y = f (x) lies below the graph. Show that f is convex on [a; b]. (This is a converse of the lemma.) 7. Suppose f is convex on [a; b] and dierentiable on (a; b). Show that f 0 is increasing on (a; b). 8. Suppose f is twice continuously dierentiable on (a; b) and f 00 0 on (a; b). Show that f is convex on [a; b]. (This provides a simple  and wellknown  test for convexity.) [Hint: if x; c 2 (a; b), then
f (x) = f (c) + f 0(c)(x , c) +
x
Z
c
(x , t)f 00 (t) dt:]
9. Suppose f is twice dierentiable on (a; b) with f 00 > 0 on (a; b). Show that the function
f (x) = sup fxs , f (s)g a<s<b 0 is convex on the range of f . (The function f is called the FenchelLegendre transform of f .)
10. Determine the FenchelLegendre transforms of each of the following functions de ned on (0; 1): ex , x log x , x, 1=x, ,2px, xp=p, xq =q (p > 1, q = p=(p , 1)):
DID YOU KNOW  that the following numbers, containing only the digits 1, 4, 9, are all perfect squares:
1; 4; 9; 49; 144; 441; 1444; 11449; 44944; 991494144; 4914991449; 149991994944 ? Are there in nitely many such numbers?  Neven Juric
294
THE SKOLIAD CORNER No. 17 R.E. Woodrow
As a problem set in this issue, we give the grade 12 level contest from Nova Scotia, written April 26, 1996. My thanks go to Professor Michael Nutt of Acadia University, Wolfville, Nova Scotia, for sending the contest to us for our use.
CANADIAN MATHEMATICAL SOCIETY PRIZE EXAM
Friday, April 26, 1996  Time: 2.5 hours 1. (a) Solve pxp+ 20 , pxp+ 1 = 1. (b) Try to solve 3 x + 20 , 3 x + 1 = 1. 2. Suppose a function is de ned so that f (xy) = f (x) + f (y). (a) Show that if f is de ned at 1, f (1) = 0. (b) Similarly if f is de ned at 0, its value at any x will be 0, so it is a \trivial" function. (c) Show that if f is de ned only for f1; 2; 3; : : : g there are many nontrivial ways to de ne such an f (give an example). 3. Three circles of equal radii r all touch each other to enclose a three cornered concave area A. How big is the area of A? 4. Show that for all real numbers x, (a) x4 4x , 3. (b) x4 is not greater than 3x , 2, even though it appears to be true. 1 1 1 2 3 x4 ,1 0 14 3 2 x 1 0 :004 0:12 :063 1 16 81
4x , 3 ,7 ,3 ,2 ,1:67 ,1 1 5 3x , 2 ,5 ,2 ,1:25 ,1 ,:5 1 4
9 7
5. Two integers are called equivalent, written x y, if they are divisible by the same prime numbers (primes are 2; 3; 5; 7; : : : ) so 2 2 4, 3 27 but 2 6 3. (a) Show that 10 80 but 10 6 90. (b) Prove that if x y , then x2 y 2. 6. We can describe certain fractions in terms of others all with bigger denominators (always in lowest terms). For instance 13 = 14 + 121 and 23 = 1 + 1 + 1 but 2 = 2 + 2 does not work since 2 = 1 and 2 = 1 + 1 does 4 4 6 3 6 6 6 3 3 2 6 not since 2 < 3.
295 (a) Can you write 12 as a sum a1 + 1b for integers 2 < a < b? 1 as 1 + 1 for integers 1996 < a < b. (b) Try to write 1996 a b Last issue we gave the problems of the Saskatchewan Senior Mathematics Competition. Here are the ocial solutions. Many thanks go to Professor Garreth Grith, of the Department of Mathematics of the University of Saskatchewan, and long time contest organizer in Saskatchewan, for providing the problems and solutions.
SASKATCHEWAN SENIOR MATHEMATICS CONTEST Wednesday, February 22, 1995 Time: 1.5 hours
1. They sell regular and jumbo sized , orders of sh at Jerry's Fish & Chips Emporium. The jumbo order costs 43 times as much as a regular one and an order of chips costs $1.30. Last Thursday, the Emporium was quite busy over the lunch break (11:30 am  1:30 pm). Exactly thirteen jumbo sized orders of sh along with a quantity of regular orders of sh and chips were sold. $702.52 had been placed in the till. During the period 1:30  4:30, business slackened o. 26 regular orders of sh were sold during this time. Four times as many regular orders had been sold during the lunch break. No jumbo portions were sold and the number of orders of chips declined to one fth the number that had been sold during the lunch break. At 4:30 pm there was $850.46 in Jerry's till. What is the price of a regular sized order of sh at Jerry's Emporium? [5 marks] Solution. Let x be the price of a regular order of sh at Jerry's. Then, the cost of a jumbo order is 43x . Let y be the number of orders of chips sold at lunchtime. Since 26 regular orders of sh were sold during the afternoon period, 104 were sold at lunch time. Also, y5 orders of chips were sold during the afternoon. The revenue at lunchtime was 4 x (1:30)y + 104x + 13 3 = 702:52
(L)
and the revenue during the afternoon was
(1:30) y5 + 26x = 850:46 , 702:52 = 147:94:
To solve these two equations, consider
4 x 5(PM ) , (L) = 130x , 104x , 13 3 = 37:18
(PM)
296 which is equivalent to
2x = 2:86 or x = 4:29: 3
2. ABCD is a square with side of length s. A circle, centre A and radius r is drawn so that the arc of this circle which lies within the square divides the square into two regions of equal area. Write r as a function of s. [6 marks] Solution. The area of the square is s2 . The area of the (whole) circle is r2 . One quarter of this area lies within the square. This area is r2 =4. Therefore r2 = s2 or r2 = 2s2 so that
4
2
r
r = 2 s: (The sign is unnecessary since r and s are distances.)
3.(a) Solve the equation 3y = 10y .
(b) Solve the equation 3y = 10. (c) Write t log8 px , 2 log8 y as a single logarithm. Solution. (a) Take (common) logs of both sides:
log 3y = log 10y y log 3 = y log 10 = y: 0 = y(1 , log 3) so that the only solution is y = 0. (b) Similarly, y log 3 = 1 so that y = log1 3 : If we use base 3 instead, then y log3 3 = log3 10 or y = log3 10: (The two answers are equal.) (c)
6 log8 px , 2 log8 y = 6 log8 x1=2 , 2 log8 y = log8 (x1=2)6 , log8 y 2 3 = log8 xy 2 :
[3 marks] [3 marks] [4 marks]
297
4. Establish the identity 2 cot A = cot A2 , tan A2 :
Solution.
[5 marks]
cos A=2 , sin A=2 sin A=2 cos A=2 2A 2A = cossin2A,cossinA 2 2 2 2 cos A cos A = 1 sin A = sin A 2 = 2 cot A = the left side:
The right side =
5. ABC is a triangle, right angled at C . Let a, b, c denote the lengths of the sides opposite angles Ap, B , C respectively. Given that a = 1p, \B = 75 and that tan 75 = 2 + 3, express b and c in the form p + q 3 such p that p = 2 or 2. [8 marks] b Solution.pSince ABC is a right angled triangle, tan B = a . Since a = 1, b = 2 + 3. By the theorem of Pythagoras, c2 = a2 + b2 =p1 + b2 p p = 1 + (2 + 3)2 = 8 + 4 3 = r(2 + 3): p
p
Therefore c = 2 2 + 3p . (As p in problem 2, p we dismiss the psign.) The problem now is to write 2 2 + 3 as p + q 3 where p = 2 or 2. Square both sides p p 4(2 + 3) = p2 + 3q 2 + 2pq 3: If p = 2, then 8 = 4 +p3q 2 (1) p and 4 3 = 4q 3: (2) From (2), q =p 1. This is not consistent with (1). If p = 2, then 8 = 2p+ 3qp2 (3) p and 4 3 = 2 2q 3: (4) p From (4), q = 2 which is consistent with (3) so that p p pp p p p = q = 2 and c = 2 + 2 3 or 2(1 + 3): Check! p c2 = 2(1 + 3)2
p
= 2(4 + 2p 3) = 4(2 + 3):
298
6. Determine the function f (x) which satis es all of the following conditions: [8 marks] (i) f (x) is a quadratic function. (ii) f (x + 2) = f (x) + x + 2. (iii) f (2) = 2. Solution. (i) Let f (x) = ax2 + bx + c. (ii) f (x + 2) = a(x + 2)2 + b(x + 2) + c so that a(x + 2)2 + b(x + 2) + c = ax2 + bx + c + x + 2 or
ax2 + 4ax + 4a + bx + 2b + c = ax2 + bx + c + x + 2 so that 4ax + 4a + 2b = x + 2. Since this is an identity,
4a = 1 and 4a + 2b = 2 so that a = 14 and b = 21 :
(iii) f (2) = 4a + 2b + c = 2. Therefore 2 + c = 2 so that c = 0. It follows that x2 x
f (x) = 4 + 2 :
7.Prove that if n is a positive integer (written in base 10) and that if 9 is a factor of n, then 9 is also a factor of the sum of the digits of n.[8 marks] Solution. Let a be the units digit of n; let b be the tens digit of n; let c be the hundreds digit of n and so on. Then n = a + 10b + 100c + = a + (1 + 9)b + (1 + 99)c + = (a + b + c + ) + 9b + 99c + Since 9 divides 9b + 99c + it follows that 9 divides n if and only if 9 divides (a + b + c + ) which is the sum of the digits of n. That completes the Skoliad Corner for this issue. Send me your contests, your suggestions, and your recommendations to improve this feature.
299
THE OLYMPIAD CORNER No. 177
R.E. Woodrow All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4.
We begin this Issue with some of the problems proposed to the jury, but not used at the 36th International Olympiad held at Toronto, Ontario. I welcome your novel, nice solutions that dier from the \ocial" published solutions.
36th INTERNATIONAL MATHEMATICAL OLYMPIAD
Canadian Problems for Consideration by the International Jury Algebra
1. Let a and b be nonnegative integers such that ab c2 , where c
is an integer. Prove that there is a number n and integers x1 ; x2 ; : : : ; xn , y1; y2; : : : yn such that n X
i=1
x2i = a;
n X
i=1
yi2 = b; and
n X
i=1
xiyi = c:
2. Let n be an integer, n 3. Let a1; a2; : : : ; an be real numbers, where 2 ai 3 for i = 1; 2; : : : ; n. If s = a1 + a2 + + an , prove that a21 + a22 , a23 + a22 + a23 , a24 + + a2n + a21 , a22 2s , 2n: a1 + a2 , a3 a2 + a3 , a4 an + a1 , a2
3. Let a, b and c be given positive real numbers. Determine all positive real numbers x, y and z such that x+y+z = a+b+c
and
4xyz , (a2x + b2y + c2z ) = abc:
300 Geometry 4. Let A, B and C be noncollinear points. Prove that there is a unique point X in the plane of ABC such that XA2 + XB 2 + AB 2 = XB2 + XC 2 + BC 2 = XC 2 + XA2 + CA2. 5. The incircle of ABC touches BC , CA and AB at D, E and F respectively. X is a point inside ABC such that the incircle of XBC touches BC at D also, and touches CX and XB at Y and Z , respectively. Prove that EFZY is a cyclic quadrilateral. 6. An acute triangle ABC is given. Points A1 and A2 are taken on the side BC (with A2 between A1 and C ), B1 and B2 on the side AC (with B2 between B1 and A) and C1 and C2 on the side AB (with C2 between C1 and B ) so that \AA1 A2 = \AA2 A1 = \BB1 B2 = \BB2 B1 = \CC1C2 = \CC2C1 : The lines AA1 , BB1 , and CC1 bound a triangle, and the lines AA2 , BB2 and CC2 bound a second triangle. Prove that all six vertices of these two triangles lie on a single circle. Number Theory and Combinatorics 7. Let k be a positive integer. Prove that there are in nitely many perfect squares of the form n2k , 7, where n is a positive integer. 8. Let Z denote the set of all integers. Prove that, for any integers A and B , one can nd an integer C for which M1 = fx2 + Ax + B : x 2 Zg and M2 = f2x2 + 2x + C : x 2 Zg do not intersect. 9. Find all positive integers x and y such that x + y2 + z3 = xyz, where z is the greatest common divisor of x and y . Sequences 10. Does there exist a sequence F (1);F (2);F (3);: : : of nonnegative integers which simultaneously satis es the following three conditions? (a) Each of the integers 0; 1; 2; : : : occurs in the sequence. (b) Each positive integer occurs in the sequence in nitely often. (c) For any n 2, F (F (n163)) = F (F (n)) + F (F (361)):
11. For an integer x 1, let p(x) be the least prime that does not divide x, and de ne q (x) to be the product of all primes less than p(x). In particular, p(1) = 2. For x having p(x) = 2, de ne q (x) = 1. Consider the sequence x0 ; x1 ; x2 ; : : : de ned by x0 = 1 and xn+1 = xqn(px(x)n) n
for n 0. Find all n such that xn = 1995.
301
12. Suppose that x1; x2; x3; : : : are positive real numbers for which xnn =
nX ,1 j =0
xjn
for n = 1; 2; 3; : : : . Prove that for all n,
2 , 2n1,1 xn < 2 , 21n :
Next we turn to the \ocial" results of the 37th IMO which was written in Mumbai, India, July 10{11, 1996. My source this year was Ravi Vakil, former star Olympian and this year's Canadian Team Leader. I hope that I have made no serious errors in compiling the results and transcribing names. This year a total of 426 students from 75 countries took part. This is somewhat up from last year. Sixty ve countries sent teams of six (the number invited to participate in recent years). But there were ten teams of smaller size, two of ve members; ve of four, and one of each of sizes three, two and one. The contest is ocially an individual competition and the six problems were assigned equal weights of seven marks each (the same as the last fteen IMOs for a maximum possible individual score of 42 and a total possible of 252 for a national team of six students). For comparison see the last fteen IMO reports in [1981: 220], [1982: 223], [1983: 205], [1984: 249], [1985: 202], [1986: 169], [1987: 207], [1988: 193], [1989: 193], [1990: 193], [1991: 257], [1992: 263], [1993: 256], [1994: 243], and [1995: 267]. There was only one perfect score. The jury awarded rst prize (Gold) to the thirty ve students who scored 28 or more. Second (Silver) prizes went to the sixtysix students with scores from 20 to 27, and third (Bronze) prizes went to the ninetynine students with scores from 12 to 19. Any student who did not receive a medal, but who scored full marks on at least one problem, was awarded honourable mention. This year there were twentyone honourable mentions awarded. The median score on the examination was 11. Congratulations to the Gold Medalists. Name
Ciprian Manolescu Yu Seek Kong Sug Woo Shin Nikolai Dourov Alexander Harry Saltman Ngo Dac Tuan
Country
Romania Korea Korea Russia U.S.A. Vietnam
Score
42 39 38 37 37 37
302 Name
Peter Burcsi Sachiko Nakajima Serguei Norine Christopher C. Chang Arend Boyer Peter Frenkel Juliy Sannikov Ivan Ivanov David Chkhaidze Constantin Chiscanu David William Bibby Ngo Duc Duy Serguei Chikh Chen Huayi Gunther Vogel Carl Bosley Michael Korn Nguyen Thai Ha He Xuhua Yan Jun Bertram Felgenhauer Gyula Pap Ajay C. Ramdoss Lev Buhovski Adrian Dumitru Corouneanu Senkodan Thevendran Eaman Eftekhary Stefan Radu Niculescu Michael Comyn Ching
Country
Hungary Japan Russia U.S.A. Germany Hungary Ukraine Bulgaria Georgia Romania United Kingdom Vietnam Belarus China Germany U.S.A. U.S.A. Vietnam China China Germany Hungary India Israel Romania Singapore Iran Romania United Kingdom
Score
36 36 36 36 35 35 34 33 33 33 33 33 32 32 31 31 31 31 30 30 30 30 30 30 30 29 28 28 28
Next we give the problems from this year's IMO Competition. Solutions to these problems, along with those of the 1995 USA Mathematical Olympiad will appear in a booklet entitled Mathematical Olympiads 1996 which may be obtained for a small charge from: Dr. W.E. Mientka, Executive Director, MAA Committee on H.S. Contests, 917 Oldfather Hall, University of Nebraska, Lincoln, Nebraska, 68588, USA.
37th INTERNATIONAL MATHEMATICAL OLYMPIAD
July 10{11, 1996 (Mumbai, India) First Day  Time: 4.5 hours
1. Let ABCD be a rectangular board with jABj = 20, jBC j = 12. The board is divided into 20 12 unit squares. Let r be a given positive integer. A coin can be moved from one square to another if and only if the distance between the centres of the two squares is pr . The task is to nd a
303 sequence of moves taking the coin from the square which has A as a vertex to the square which has B as a vertex. (a) Show that the task cannot be done if r is divisible by 2 or 3. (b) Prove that the task can be done if r = 73. (c) Can the task be done when r = 97? 2. Let P be a point inside triangle ABC such that \APB , \ACB = \APC , \ABC:
Let D, E be the incentres of triangles APB , APC respectively. Show that AP , BD and CE meet at a point. 3. Let S = f0; 1; 2; 3; : : : g be the set of nonnegative integers. Find all functions f de ned on S and taking their values in S such that
f (m + f (n)) = f (f (m)) + f (n) for all m; n 2 S: Second Day  Time: 4.5 hours 4. The positive integers a and b are such that the numbers 15a + 16b and 16a , 15b are both squares of positive integers. Find the least possible value that can be taken by the minimum of these two squares. 5. Let ABCDEF be a convex hexagon such that AB is parallel to ED, BC is parallel to FE and CD is parallel to AF . Let RA , RC , RE denote the circumradii of triangles FAB , BCD, DEF respectively, and let p denote the perimeter of the hexagon. Prove that
RA + RC + RE p2 :
6. Let n, p, q be positive integers with n > p + q. Let x0; x1; : : : ; xn be integers satisfying the following conditions: (a) x0 = xn = 0; (b) for each integer i with 1 i n, either xi , xi,1 = p or xi , xi,1 = ,q: Show that there exists a pair (i; j ) of indices with i < j and (i; j ) 6= (0; n) such that xi = xj . Although the IMO is ocially an individual event, the compilation of team scores is unocial, if inevitable. These totals and the prize awards are given in the following table.
304 Rank
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.{12. 11.{12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.{26. 25.{26. 27. 28. 29. 30. 31. 32. 33. 34.{35. 34.{35. 36. 37.{38. 37.{38. 39.{40. 39.{40. 41. 42. 43. 44. 45.{46. 45.{46. 47. 48.{51. 48.{51. 48.{51. 48.{51. 52. 53.{54. 53.{54.
Country
Romania U.S.A. Hungary Russia United Kingdom China Vietnam Korea Iran Germany Bulgaria Japan Poland India Israel Canada Slovakia Ukraine Turkey Taipei Belarus Greece Australia Yugoslavia Italy Singapore Hong Kong Czech Republic Argentina Georgia Belgium Lithuania Latvia Armenia Croatia France New Zealand Norway Colombia Finland Sweden Moldava (Team of 5) Austria Republic of South Africa Mongolia Slovenia Thailand Denmark Macao Former Yugoslav Republic of Macedonia Spain Brazil Mexico Sri Lanka
Score
Gold
Silver
Bronze
Total
44 36 34 34
   
   
   1
   1
187 185 167 162 161 160 155 151 143 137 136 136 122 118 114 111 108 105 104 100 99 95 93 87 86 86 84 83 80 78 75 68 66 63 63 61 60 60 58 58 57 55 54 50 49 49 47 44 44 44
4 4 3 2 2 3 3 2 1 3 1 1  1 1   1   1     1    1                    
2 2 2 3 4 2 1 3 4 1 4 4 3 3 2 3 2  2 2 1 1 2 1 2  1 2 1   1   1 2   1  1  1       
  1 1  1 1  1 1 1  3 1 2 3 4 5 3 3 2 5 3 2 2 3 4 1 3 2 4 2 3 1 1  3 3  2 1 2  2 2 2 1 2 1 2
6 6 6 6 6 6 5 5 6 5 6 5 6 5 5 6 6 6 5 5 4 6 5 3 4 4 5 3 4 3 4 3 3 1 2 2 3 3 1 2 2 2 1 2 2 2 1 2 1 2
305 Rank
55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67.{68. 67.{68. 69. 70. 71. 72.{73. 72.{73. 74. 75.
Country
Estonia Iceland BosniaHerzegovina (Team of 4) Azerbaijan The Netherlands Trinidad & Tobago Ireland Switzerland (Team of 4) Portugal Kazakhstan Morocco Cuba (Team of 1) Albania (Team of 4) Kyrgyztan Cyprus (Team of 5) Indonesia Chile (Team of 2) Malaysia (Team of 4) Turkmenistan (Team of 4) Philippines Kuwait (Team of 3)
Score
Gold
33 31 30 27 26 25 24 23 21 20 19 16 15 15 14 11 10 9 9 8 1
Silver
                    
                    
Bronze
 1 1     1   1 1         
Total
 1 1     1   1 1         
This year the Canadian Team rose to 16th place from 19th last year and 24th the previous year. The Team members were: Richard Hoshino Derek Kisman Saroosh Yazdani Byung Kyu Chun Adrian Chan Sabin Cautis
22 22 22 18 14 13
Silver Silver Silver Bronze Bronze Bronze
This is the rst time every member of the Canadian Team was awarded a medal. The Team Leader was Ravi Vakil, a former Canadian Team Member and now a graduate student; and P.J. Grossman was Deputy Leader, also a former Olympian and current graduate student. The Romanian Team placed rst this year. Its members were: Ciprian Manolescu Constantin Chiscanu Adrian Dumitru Corouneanu Stefan Rodu Niculescu Dragos Ghioca Nicolae Dragos Oprea
Congratulations to the Romanian Team!!
42 33 30 28 27 27
Gold Gold Gold Gold Silver Silver
306 To nish this number of the Corner we discuss solutions sent in by the readers to problems of the 6th Korean Mathematical Olympiad, which we gave in the February 1995 number (6th Korean Mathematical Olympiad, Final Round, April 17{18, 1993, [1995: 45{46]). 1. Let there by a 9 9 array of white squares. Find the largest positive integer n satisfying the following property: There always remains either a 1 4 or a 4 1 array of white squares no matter how you choose n out of 81 white squares and colour them black. Solutions by Mansur Boase, student, St. Paul's School, London, England; by Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany; and by Solomon Golomb, University of Southern California. We give Golomb's solution and comments. The largest positive integer n such that, no matter which n squares of a 9 9 board are blocked, it will still be possible to place a 1 4 rectangle [\straight tetromino"] on the board, either horizontally or vertically, is 19. We prove the following clearly equivalent proposition: \The smallest number of squares which must be blocked on a 99 board so that a straight tetromino will no longer t is 20." First here is a con guration with 20 blocking squares such that no straight tetromino will t:
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
Clearly, a blocking con guration must involve at least two excluded squares in each row and in each column. (One excluded square cannot block a row or column of length nine.) This already shows that 2 9 = 18 excluded squares are required, to block all rows. But consider the two (or more) excluded squares in the bottom row. Their columns each require two additional excluded squares to block those columns. So all the columns have at least two excluded squares, but at least two columns must have at least three excluded squares, for a total of at least 7 2 + 2 3 = 20 excluded squares.
307 Comment. Methods for this type of problem are described in Chapter 3: \Where Pentominoes Will Not Fit," in my book Polyominoes. (Original edition, Charles Scribner's Sons, 1965; expanded, revised edition, Princeton University Press, 1994.) 2. Let ABC be a triangle with BC = a, CA = b, AB = c. Find the point P for which a AP 2 + b BP 2 + c CP 2 is minimal, and nd the minimum. Solutions by Sefket Arslanagic, Berlin, Germany; and by Panos E. Tsaoussoglou, Athens, Greece. We give the solution of Arslanagic. We have (e.g. pages 278 and 280 of [1]): Let P be a point ,! in the plane of,! a triangle ABC and M be an arbitrary P P point in space. Then MP = ( xi MA)=( xi ), where x1 , x2 , x3 are real numbers, and summations are taken cyclically. The following generalization of the wellknown Leibnitz identity is valid: X X X 2 X xi MP 2 = xi xiMA2 , a2x2x3:
For I is the incentre of ABC ), we get (because we can take P P = I , (where xi = 2s, P a2 xixj = abc 2s) X aMA2 = 2sMI 2 + abc; and as a consequence we have the following inequality, X aMA2 abc; and equality holds only for M = I . From this take M = P , aAP 2 + bBP 2 + cCP 2 abc and
min(aAP 2 + bBP 2 + cCP 2 ) = abc
with P = I . Reference [1] D.S. Mitrinovic, Y.E. Pecaric and V. Volenec, Recent Advances in Geometric Inequalities. 3. Find the smallest positive integer x for which 7x25 , 10 is an integer.
83
308 Solutions by Mansur Boase, student, St. Paul's School, London, England; Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany; and by Stewart Metchette, Gardena, California, USA. We give the solution of Boase. We want 7x25 10 mod 83, or 7x25 = 83a + 10. So 83a + 10 0 mod 7 whence a 3 mod 7 and x25 37 mod 83. Now 83 is prime and congruent to 3 mod 5, so there is only one solution to this congruence mod83. (x5 )5 37 mod 83 x5 16 mod 83
x 69 mod 83:
The smallest such x is 69. 4. An integer is called a Pythagorean number if it is the area of a right triangle whose sides are of integral lengths, say x; y;z 2 N such that x2 + y2 = z2. Prove that for each positive integer n (n > 12), there exists a Pythagorean number between n and 2n. Solutions by Mansur Boase, student, St. Paul's School, London, England; and by Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany. We give the solution of Engelhaupt. The smallest Pythagorean numbers are x 3 6 5 9 8 y 4 8 12 12 15 z 5 10 13 15 17 Pythagorean Number 6 24 30 54 60 Thus for n = 13 to 23 we have 24 in the interval [n; 2n], for n = 24 to 29 we have 30 in the interval [n; 2n], and for n = 30 to 53 we have 54 in the interval [n; 2n]. Now the numbers 6k2 , with k 2 N are Pythagorean numbers (x = 3k, 2 y = 4k). Observe that 6(k6+k21) < 2 if 2k + 1 < k2, or k > 2. Henceforth, k 3. For n = 6k2 to 6(k + 1)2 , 1 we have 6(k + 1)2 lies in the interval [n; 2n], for example for n = 54 to 95, 96 lies in [n; 2n]. for n = 96 to 149, 150 lies in [n; 2n]; for n = 150 to 215, 216 lies in [n; 2n]; and so on. 5. Let n be a given natural number. Find all the continuous functions f (x) satisfying: !
n 0
!
f (x) +
n 1
f
, 2 x +
!
n 2
f x
22
+
+
n
n,1
!
f x
2n,1
!
+
n 2n f x n
309 Solution by the editor. We prove by induction that f (x) 0 if f is a continuous function satisfying the condition (*). With n = 0 there is nothing to prove as the condition becomes f (x) = 0 for all x. For n = 1, assume that f (x) + f (x2) = 0 for all x with f continuous. Now f (,x) = ,f (x2) = f (x) so f is an even function and it suces to prove the result for x 0. From f (0) = ,f (0) and f (1) = ,f (1) we get f (0) = 0 = f (1). So consider a xed value of x > 0. Now limk!1 x1=k = m1 from which we obtain limm!1 x1=2m = 0. It is easy tom see that f (x1=2 ) = (,1)mf (x) for m 0. However, limm!1 f (x1=2 ) = f (1) = 0. From this it follows that f (x) = 0, establishing the result for n = 1. Now assume f is a continuous function satisfying nX +1 n + 1 2j ) = 0 for all x: f ( x j j =0 Set n n X g(x) = f (x2l ): l l=0 Now n n n n X X l 2 2 2)2l ) g(x) + g(x ) = f ( x ) + f (( x l=0 l l=0 l n n X n 2l ) + X n f (x2l+1 ) = f ( x l=0 l l=0 l nX +1 n + 1 2j ) = 0 = f ( x j j =0 , ,n , n since n+1 j = j + j ,1 . But then g (x) is a continuous function satisfying g (x) + g (x2) = 0 for all x. By the case n = 1, g (x) = 0 for all x, i.e. n X
n f (x2l) = 0 for all x: l=0 l By the induction hypothesis f (x) = 0 for all x and the induction is complete. 6. Let ABC be a triangle with BC = a, CA = b and AB = c. Let D be the midpoint of the side BC , and let E be the point on BC for which the line segment AE is the bisector of angle A. Let the circle passing through A, D, E intersect with the sides CA, AB at F , G respectively. Finally let H be the point on AB for which BG = GH , i.e. BH = 2BG. Prove that the triangles EBH and ABC are similar and then nd the ratio EBH ABC of these g(x) =
areas.
310 Solution by Mansur Boase, St. Paul's School, London, England.
A H
q
G B
q
F
q
D E q
q
C
Now \EAF = \EGF = \FDE and \GAD = \GFD = \GED. Thus GED ABD since two angles are the same. Thus
BE BH HE BE BG GE AB = BD = AD and AB = BC = AC : Thus BHE ABC ., 2 BE BHE The ratio ABC = C . By the sine rule
BE = AE sin A2 sin B a , BE = AE : sin C sin A2 Therefore BE sin B = (a , BE ) sin C , and so C = a 2cR BE = sin aBsin + sin C 2bR + 2cR = b ac + c;
and
BE 2 : 1 = ac 2 : 1 = a2 : c c(b + c) (b + c)2
That completes the solutions we have and the Olympiad Corner for this issue. Send me your contests and nice solutions!
311
THE ACADEMY CORNER No. 5
Bruce Shawyer All communications about this column should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7 The Mathematics Archives on the World Wide Web. Everyone knows that there is a plethora of information on the World Wide Web. I was directed recently to the Mathematics Archives: Lessons, Tutorials and Lecture Notes at http://archives.math.utk.edu/tutorials.html
This contains over 140 links to mathematical assistance. The whole is too large to mention here, but I mention a few of the links that could be of use to CRUX readers. 1. http://www.lance.colostate.edu/auth/lessons/1/ Aristotle University of Thessaloniki Lesson on the NewtonRaphson method 2. http://www.math.uga.edu/andrew/Binomial/index.html Arithmetic Properties of Binomial Coecients Binomial Coecients is an online dynamic survey, available on the World Wide Web, and accessible from various mathematical sites and ejournals. It is an edited document, in HTML, and can and will be edited as new developments arise. Content is at the discretion of the editors. Editors are Andrew Granville and Richard Witt. 3. http://www2.ncsu.edu/math/Projects/MA141Manual/Contents.html CALCULUS I WITH MAPLE V, North Carolina State University 4. http://www.math.ilstu.edu/day/305S96.html Combinatorics: Topics for K8 Teachers Information and materials from a course taught by Roger Day at Illinois State University. 5. ftp://ftp.utirc.utoronto.ca/pub/ednet/maths/model_html/ExpMath.html Experiments in Mathematics Using Maple A Computer Lab Book For High Schools and Home Computers by C. T. J. Dodson and E. A. Gonzalez, University of Toronto.
312 6.
http://millbrook.lib.rmit.edu.au/fractals/exploring.html
7.
http://math.holycross.edu/
8.
http://williamking.www.drexel.edu/top/class/gametoc.html
9.
http://www.geom.umn.edu/docs/education/institute91/
Exploring Chaos and Fractals Exploring Chaos and Fractals is an electronic textbook which includes full text, work sheets, sound, video and animation. Parts of the material have been placed on a Web server as an experiment in electronic publishing of hypertext based material. The Fibonacci Numbers
davids/fibonacci/fibonacci.html
Game Theory: An Introductory Sketch
Geometry and the Imagination Notes and handouts for an innovative geometry course developed at Princeton and the Geometry Center by John Conway, Peter Doyle, Jane Gilman and Bill Thurston.
10.
http://www.utm.edu:80/departments/math/graph/
11.
http://www.intergalact.com/threedoor/threedoor.html
Graph Theory : Tutorials This is the home page for a series of short interactive tutorials by Chris K. Caldwell introducing the basic concepts of graph theory. They are designed with the needs of future high school teachers in mind. The Three Door Puzzle
12.
http://www.wam.umd.edu/ krc/numtheory/prime.html
13.
http://www.math.ilstu.edu/
Prime Numbers These pages were written by Kevin Coombes as an experiment to aid students to better understand proofs of theorems. From an email message: \It seems to me that one of the main diculties that students of (higher) mathematics encounter stems from the insistence on getting all the logical prerequisites set up before trying to explain anything interesting. (Of course, this style of presentation goes back at least as far as Euclid.)"
day/326S96.html
Technology Tools for Secondary School Mathematics Materials developed by Roger Day at Illinois State University. Acknowledgment: Many of the descriptions were taken from and/or modi ed from the documents from the listed site.
313
BOOK REVIEWS Edited by ANDY LIU Five Hundred Mathematical Challenges by Edward J. Barbeau, Murray S. Klamkin and William O.J. Moser. Published by The Mathematical Association of America. softcover, 227+ pages, ISBN # 0883856194, US $ 29.50. Reviewed by Marcin E. Kuczma, University of Warsaw. The authors' names alone are a sucient recommendation of this book, which is a must for your bookshelf  unless you have already the previously published collection of booklets, some 20 years ago, by the same authors, containing the same problems. Actually, even in that case, it is still worth having the new book, since it is a revised and expanded version. Solutions are presented with greater care and elegance. Multiple solutions are often supplied, with extensions and remarks providing profound insight into the nature of the problems. There are also new bibliographical references. I could have stopped the review right here. However, I should perhaps say some more about the book than just express my enchantment with it. The title is a little bit misleading. Why so? The word \Challenges" calls for re ection. The readers may be relieved to learn that some of the problems presented are no more dicult than very simple \puzzlers" often found in nonmathematical journals. Here is an example. Problem 471 Ma and Pa and brother and me, The sum of our ages is eightythree, Six times Pa's is seven times Ma's, And Ma's is three times meno fractions please. Another frequent type is: Without using a calculator tell which one of the two numbers (expressed in terms of horrifying root combinations) is greater. I am sure the readers will welcome with much pleasure such exercises, which provide healthy entertainment. They bring a refreshing eect as they appear, from time to time, amidst heavier artillery: the more or less typical competition problems. As mentioned, the book arose from a series of earlier publications. It must be kept in mind that in those years, publications of that type were rare, at least in America. Problem solving competitions had not yet expanded into a \branch of industry" as they are today. Students were not acquainted with any ocurricular techniques. That explains the presence of many problems which are immediately reducible to direct applications of, say, Jensen's Inequality, Menelaus' Theorem, the Factor Theorem, the Pigeonhole Principle, congruence considerations, and so on. Nowadays, those techniques are taught at every training course. Any student trying to tackle the olympiad
314 or other serious mathematics competitions must be familiar with them, and this takes the challenge somewhat away. A few problems can be quite \challenging" to a high school student, while being standard at the rstyear college level. Examples are onesided or twosided estimates of the partial sums of some typical series or of expressions involving factorials and binomial coecients; rate of growth: polynomial versus logarithmic or exponential; and maximization of functions by multivariable calculus methods. I would hesitate to call any one of those questions a \challenge". What has been said so far should by no means be construed as criticism. Even if we agree that some trick has become \standard", this fact is not necessarily re ected in the standard problem literature. The book under review is a treasury of such training material. About a quarter of the book consists of problems at the olympiad level true challenges. I must confess to having real diculties with many of them, such as the following. Problem 432 Show that ve or more great circles on a sphere, no three of which are concurrent, determine at least one spherical polygon having 5 or more sides. With some of the problems I had no trouble only because they were wellknown to me. Again, this is no criticism! Problems do circulate in the literature. I am quite sure that several problems were actually born and rst published many years ago, in one of those booklets by the authors. They may have since been repeatedly reused in other publications and contests, to reappear in the present edition. Several problems have been reprinted, with reference given, from other journals. Special mention must be made about those taken from the issues of Math. Gazette in the previous century(!)a precious gift to every problem collectioner. The authors write in the Preface: \ : : : we make no claim for the originality of most of the problems; we acknowledge our debt to the unsung creators : : : ". Too modest! Many of the problems have without any doubt been created by the authors of the book. The majority of those truly challenging and beautiful problems come from these domains: algebra of polynomials, in one or more variables; sophisticated inequalities, often involving symmetric forms; and smart combinatorial reasonings. Each one of the authors has left his own imprint. The correlation between \challenging" and \beautiful" is wellknown to every problemist; nice problems are usually the dicult ones. Now, it is a pleasant feature of this book that it also contains relatively many problems of medium or even less than medium diculty, which nevertheless have more than average charm, sometimes revealed only after arriving at the solution. Here are some examples.
315 Problem 28 A boy lives in each of n houses on a straight line. At what point should the n boys meet so that the sum of the distances that they walk from their houses is as small as possible? (Two solutions are provided, contrasting a common sense argument against an analytic one.) Problem 292 Let f (x) be a nondecreasing function of a real variable. Let c be any real number. Solve the equation x = c , f (x + f (c)). Problem 327 Let three concentric circles be given such that the radius of the largest is less than the sum of the radii of the two smaller. Construct an equilateral triangle whose vertices lie one on each circle. Problem 445 Prove that if the top 26 cards of an ordinary shued deck contain more red cards than there are black cards in the bottom 26, then there are in the deck at least three consecutive cards of the same colour. This sample re ects my taste; another reader may point out other problems she/he considers particularly nice. While it is impossible that all 500 problems should be equally appealing to everybody, it is rather certain that everybody shall nd in this book enough items matching her/his taste, needs and \sense of pleasure" accurately. With a certain dose of pride, I can say that I was able to solve most of the problems I tried. And with a certain dose of shame I must say that, in most cases, I failed to nd solutions as elegant as those included in the book. For many problems, dierent approaches are shown. The neatness of presentation must be emphasised. The solutions are clear and concise. Occasionally, a rigorous argument is preceded by a heuristic one. The solution to Problem 327 (see above) ends with a \rider": Given equilateral triangle ABC such that PA = 3, PB = 4, PC = 5 for an interior point P , nd AB . Anyone who solved the problem (or has read and understood the solution) is invited to tackle this new one, employing the similar lines of reasoning, and yet revealing some new and unexpected features. Such \riders" follow the solutions to many problems. Here is another example. Problem 380 Prove that the function f (x;y ) = 12 (x + y )(x + y + 1) + x is a onetoone map from the set of lattice points (x; y ) with x; y 0, other than (0; 0), onto the set N of positive integers. Too easy? or wellknown? Try the \rider", then: For each n 2 N establish the existence of a polynomial Fn (x1 ; : : : ; xn ) such that Fn : N n ! N is a bijection.
316 Problem 447 If m and n are positive integers, show that m,1=n + n,1=m > 1: Known? Perhaps, but consider the intriguing \rider" which I very much recommend to every reader: Can you give a noncalculus proof that xy + y x > 1 for x; y > 0 ? These \riders" show extensions (sometimes farreaching), teach analogies, and provide further training to learn the techniques from the given solutions. Their value cannot be overstated. Five hundred is a huge number (in fact, there are altogether some six hundred problems, since the \riders" might be counted separately). In such a voluminous text, small mistakes are unavoidable. The solution to Problem 177 (a functional equation) misses certain discontinuous functions. According to the statement of Problem 365, the father's age has to be a perfect square and a prime number, at the same time. A mistake in sign occurs in the statement of the InclusionExclusion Principle. Minor typos are not entirely absent, but they are rare enough not to aect the impression that the book has been prepared and edited with great care. The book ends with a Tool Chest section. It is a handy help for the readers. For authors of problem collections, this is a model example of how such a section ought to be written. In conclusion, I repeat that this book is a must. (!)
Rider: To buy or not to buy?
(Answer on page ??.)
317
PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be typewritten or neatly handwritten, and should be mailed to the EditorinChief, to arrive no later that 1 May 1997. They may also be sent by email to cruxeditor@cms.math.ca. (It would be appreciated if email proposals and solutions were written in LATEX). Graphics les should be in epic format, or plain postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication.
Correction to 2137 2137. [1996: 124] Proposed by Aram A. Yagubyants, Rostov na Donu, Russia. Three circles of (equal) radius t pass through a point T , and are each inside triangle ABC and tangent to two of its sides. Prove that: rR , (ii) T lies on the line segment joining the cen(i) t = R+2 of the circumcircle and the incircle of [NB: r instead of 2] tres ABC . 2177. Proposed by Toshio Seimiya, Kawasaki, Japan.
ABCD is a convex quadrilateral, with P the intersection of its diagonals and M the midpoint of AD. MP meets BC at E . Suppose that BE : EC = (AB)2 : (CD)2. Characterize quadrilateral ABCD.
318 UK.
2178.
Proposed by Christopher J. Bradley, Clifton College, Bristol,
If A; B; C are the angles of a triangle, prove that sin A sin B sin C 8 ,sin3 A cos B cos C + sin3 B cos C cos A + sin3 C cos A cos B 3p3 ,cos2 A + cos2 B + cos2 C :
2179. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. For real numbers pn(na+a) ,1, we consider the sequence 1 ; n 1g. F (a) := f(1 + n ) Determine the sets D, respectively I , of all a, such that F (a) strictly decreases, respectively increases. 2180. Proposed by JuanBosco Romero Marquez, Universidad de Valladolid, Valladolid, Spain. Prove that if a > 0; x > y > z > 0; n 0 (natural), then 1. ax (yz )n(y , z ) + ay (xz )n(z , x) + az (xy )n(x , y ) 0; 2. ax cosh x(y , z ) + ay cosh y (z , x) + az cosh z (x , y ) 0. 2181. Proposed by Sefket Arslanagic, Berlin, Germany.
Prove that the product of eight consecutive positive integers cannot be the fourth power of any positive integer. 2182. Proposed by Robert Geretschlager, Bundesrealgymnasium, Graz, Austria. Many CRUX readers are familiar with the card game \Crazy Eights", of which there are many variations. We de ne the game of \Solo Crazy Eights" in the following manner: We are given a standard deck of 52 cards, and are dealt k of these at random, 1 k 52. We then attempt to arrange these k cards according to three rules: 1. Any card can be chosen as the rst card of a sequence; 2. A card can be succeeded by any card of the same suit, or the same number, or by any eight; 3. Anytime in the sequence that an eight appears, any suit can be \called", and the succeeding card must be either of the called suit, or another eight. (This means that, in eect, any card can follow an eight). The game is won if all dealt cards can be ordered into a sequence according to rules 1{3. If no such sequence is possible, the game is lost. What is the largest value of k for which it is possible to lose the game?
319
2183. Proposed by Vaclav Konecny, Ferris State University, Big Rapids, Michigan, USA. Suppose that A, B , C are the angles of a triangle and that k, l, m 1. Show that: 0 < sink A sinl B sinm B kkllmmS S2 (Sk2 + P ), k2 (Sl2 + P ), 2l (Sm2 + P ), m2 ; where S = k + l + m and P = klm. 2184. Proposed by Joaqun Gomez Rey, IES Luis Bu~nuel, Alcorcon, Madrid, Spain. Let n be a positive integer and let an denote the sum bn= X2c n , k k=0
(,1)k
k
:
Prove that the sequence fan : n 0g is periodic. 2185. Proposed by Bill Sands, University of Calgary, Calgary, Alberta. Notice that 22 + 42 + 62 + 82 + 102 = 4 5 + 5 6 + 6 7 + 7 8 + 8 9; that is, the sum of the rst n (in this case 5) even positive squares is equal to the sum of some n consecutive products of consecutive pairs of positive integers. Find another value of n for which this happens. (NOTE: this problem was suggested by a nal exam that I marked recently.) 2186. Proposed by Vedula N. Murty, Andhra University, Visakhapatnam, India. Let a, b, c respectively denote the lengths of the sides BC , CA, AB of triangle ABC . Let G denote the centroid, let I denote the incentre, let R denote the circumradius, r denote the inradius, and let s denote the semiperimeter. Prove that 1 (a , b)(a , c)(b + c , a) GI 2 =
9(a + b + c)
+ (b , c)(b , a)(c + a , b) + (c , a)(c , b)(a + b , c) :
320 Deduce the (known) result
GI 2 = 19 s2 + 5r2 , 16Rr :
2187. Proposed by Syd BulmanFleming and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. It is easy to show that the maximum number of bishops that can be placed on an 8 8 chessboard, so that no two of them attack each other, is 14. (a) Prove or disprove that in any con guration of 14 nonattacking bishops, all the bishops must be on the boundary of the board. (b) Describe all of the con gurations with 14 nonattacking bishops. 2188. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.
Suppose that a, b, c are the sides of a triangle with semiperimeter s and area . Prove that
1+1+1< s: a b c
DID YOU KNOW  that the nine consecutive integers
2; 3; 4; 5; 6; 7; 8; 9; 10; partitioned into three groups of three as shown, can be permuted inside each group to form three squares 324 = 182; 576 = 242 ; 1089 = 332 ? Is there another sequence of nine consecutive positive integers with this property?  K. R. S. Sastry
321
SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems.
1940. [1994: 108; 1995: 107; 1995: 206] Proposed by Ji Chen, Ningbo University, China. Show that if x; y;z > 0, (xy + yz + zx) (x +1 y )2 + (y +1 z )2 + (z +1 x)2 94 :
In CRUX [1995: 206], the editor asked `if anyone nds a \nice" solution to this problem [CRUX [1994:108; 1995: 107]], the editor would be interested to see it.' Comment by Vedula N. Murty, Andhra University, Visakhapatnam, India. Without loss of generality, assume that 0 < x y z , and let a = x+2 y , b = x+2 z , and c = y+2 z . Then we have 0 < a b c; (1) and the inequality proposed in CRUX [1994: 108] is equivalent to (2bc + 2ca + 2ab , a2 , b2 , c2 ) , a12 + b12 + c12 9: (2) We now proceed to establish (2). For this, we note that a, b and c represent the side lengths of a triangle. The left side of (2) is identical to , , , 9 + (b , c)2 2 , 12 + (c , a)2 2 , 12 + (a , b)2 2 , 12 0: bc
a
ca
b
ab
Hence, (2) is established by proving bc(b , c)2(2a2 , bc) + ca(c , a)2(2b2 , ca) + ab(a , b)2(2c2 , ab) 0: To prove (3), we note that (1) implies that either 0 2a2 , bc 2b2 , ca 2c2 , ab or
2a2 , bc < 0 < 2b2 , ca 2c2 , ab:
c
(3) (4) (5)
If (4) holds, then (3) is true, and we are done. If (5) holds, we note that the sum of the rst two terms of (3) is nonnegative, and hence that (3) is true.
322 2068. [1995: 235] Proposed by Sefket Arslanagic, Berlin, Germany. Find all real solutions of the equation p p 17 + 8x , 2x2 + 4 + 12x , 3x2 = x2 , 4x + 13:
Solution by Toby Gee, student, the John of Gaunt School, Trowbridge, England. We have p p 17 +p8x , 2x2 + 4 + 12 x , 3x2 p 2 = p 25 , p 2(x , 2) + 16 , 3(x , 2)2 25 + 16 = 9 9 + (x , 2)2 = x2 , 4x + 13
with equality throughout if and only if x , 2 = 0. Thus x = 2. Also solved by SEUNGJIN BANG, Seoul, Korea; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; PAUL BRACKEN, University of Waterloo; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; SABIN CAUTIS, student, Earl Haig Secondary School, North York, Ontario; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; TIM CROSS, King Edward's School, Birmingham, England; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; JEFFREY K. FLOYD, Newnan, Georgia, USA; ROBERT GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOE HOWARD, New Mexico Highlands University; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; FRIEND H. KIERSTEAD JR., Cuya Y, Ferris State University, Big Rapids, hoga Falls, Ohio, USA; VACLAV KONECN Michigan, USA; MITKO CHRISTOV KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; SAI CHONG KWOK, San Diego, CA, USA; KEEWAI LAU, Hong Kong; THOMAS LEONG, Staten Island, NY, USA; DAVID E. MANES, State University of New York, Oneonta, NY, USA; BEATRIZ MARGOLIS, Paris, France; J.A. MCCALLUM, Medicine Hat, Alberta; VEDULA N. MURTY, Andhra University, Visakhapatnam, India; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; BOB PRIELIPP, University of Wisconsin{Oshkosh,Wisconsin, USA; CORY PYE, student, Memorial Univer sity, St. John's, Newfoundland; JUANBOSCO ROMERO MARQUEZ, Univer sidad de Valladolid, Valladolid, Spain; CRISTOBAL SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon, Spain; K.R.S. SASTRY, Dodballapur, India; HEINZ JURGEN SEIFFERT, Berlin, Germany; ASHSIH KR. SINGH, student, Kanpur, India; DIGBY SMITH, Mount Royal College, Calgary, Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; CHRIS WILDHAGEN, Rotterdam, the Netherlands; SUSAN SCHWARTZ WILDSTROM, Kensington, Maryland, USA;
323 KENNETH M. WILKE, Topeka, Kansas, USA; PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA; an anonymous solver; and the proposer.
2072. [1995: 277] Proposed by K.R.S. Sastry, Dodballapur, India. Find positive integers x, y , u, v such that x2 + y2 = u2 and x2 , xy + y2 = v2:
(Equivalently, nd a rightangled triangle with integral sides x, y surrounding the right angle and a triangle with sides x, y surrounding a 60 angle, and with the third side an integer in both cases.) Solution by Paul Yiu, Florida Atlantic University, Boca Raton, Florida, USA. We need only consider x and y relatively prime, in which case x2 + y 2 and x2 , xy + y 2 are also relatively prime. These are squares if and only if their product is a square. It is convenient to consider the possibility that the second equation admit negative integral values of x as well (in that case jxj and y surround a 120 angle instead). Writing s = xy , we consider the rational points of the quartic curve t2 = (s2 + 1)(s2 , s + 1) = s4 , s3 + 2s2 , s + 1: This corresponds to the elliptic curve v2 = u3 + 2u2 , 3u , 6: The quartic curve has an obvious rational point (0; 1), which corresponds to a rational point P = (2; 2) on the elliptic curve. By standard procedures, we determine the multiples nP , 2 n 8. These are rational points on elliptic curves with corresponding rational points on the quartic curve. For n = 2; 3; : : : we have rational points on the quartic curve with
1768 ; : : : s = 15 ; , 8 2415
with corresponding values of x, y , u, and v given below. We separate those with dierent signs of x. The two solutions corresponding to n = 2; 3 can be found in Dickson's history of the Theory of Numbers, vol.2, p. 481. n x y u v
2 4 6 7 15 8109409 101477031226926255 4676030077060796052820312 8 10130640 422390893185635192 1382348542917116969367345 17 12976609 434409546986238833 4876078831979879983187537 13 9286489 381901401745295077 4160798170065530232858973
Corresponding to n = 3; 5; 8 we have solutions with negative x (corresponding to triangles with a 120 angle):
324 n
3
5
8
,x 1768 498993199440 227124445985970945806894399956799 y u v
2415 136318711969 654056791401866496244333771257120 2993 517278459169 692369699180530962038852086430401 3637 579309170089 792419135769606228834850391429041
Remark. It is, however, not possible to nd two triangles with integral sides, one with sides x, y surrounding a 60 angle, another with sides x, y surrounding a 120 angle, and with the third side an integer in both cases. This is because the quadratic forms x2 + xy + y 2 and x2 , xy + y 2 cannot be simultaneously made squares. See Dickson, ibid. Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; JEFFREY FLOYD, Newnan, Georgia; ROBERT GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; VACLAV Y, Ferris State University, Big Rapids, Michigan, USA; KEEWAI KONECN LAU, Hong Kong; P. PENNING, Delft, the Netherlands; HEINZJURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia; PANOS E. TSAOUSSOGLOU, Athens, Greece; CHRIS WILDHAGEN, Rotterdam, the Netherlands; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. There was one incorrect solution. Most solvers simply gave the solution corresponding to n = 2 in YIU's solution above; several found sucient conditions rst. 2078?. [1995: 278] Proposed by Sefket Arslanagic, Berlin, Germany. Prove or disprove that
pa , 1 + pb , 1 + pc , 1 pc(ab + 1)
for a; b;c 1. Solution by Theodore Chronis, student, Aristotle University of Thessaloniki, Greece. Let a , 1 = x2 , b , 1 = y 2, c , 1 = z 2, where x; y;z 0. Then the inequality to be proved becomes p x + y + z (z2 + 1)[(x2 + 1)(y2 + 1) + 1]: From the Cauchy{Schwarz inequality,
p
x + y = x 1 + 1 y (x2 + 1)(y2 + 1)
325 and
p
p
p
z + (x2 + 1)(y2 + 1) z2 + 1 (x2 + 1)(y2 + 1) + 1;
and we are done. ~ and EMILIO FERNANDEZ Also solved by MANUEL BENITO MUNOZ MORAL, Logro~no, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; WOLFGANG GMEINER, Millstatt, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOE HOWARD, New Mexico Highlands University, Las Vegas, New Mexico, USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; PETER HURTHIG, Columbia College, Burnaby, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; DAG JONSSON, Uppsala, Sweden; MURRAY S. KLAMKIN, University of Alberta, Y, Ferris State University, Big Rapids, Edmonton, Alberta; VACLAV KONECN Michigan, USA; KEEWAI LAU, Hong Kong; JUANBOSCO ROMERO MARQUEZ, Universidad de Valladolid, Valladolid, Spain; HEINZJURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary, Alberta; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; and the proposer. As several solvers pointed out, and as can be seen from the above proof, equality holds in the original inequality if and only if
a = 1 + t;
b = 1 + 1t ;
for some positive real number t.
1 c = 1 + ab
2079. [1995: 278] Proposed by Cristobal Sanchez{Rubio, I. B. Penyagolosa, Castellon, Spain. An ellipse is inscribed in a rectangle. Prove that the contact points of the ellipse with the sides of the rectangle lie on the rectangular hyperbola which passes through the foci of the ellipse and whose asymptotes are parallel to the sides of the rectangle. Solution by P. Penning, Delft, the Netherlands. , x 2 , y 2 Let the ellipse be a + b = 1 with a > b. Suppose that m and m0 are the slopes of the sides of the rectangle, so that mm0 = ,1. By symmetry, the centre of the hyperbola will coincide with the origin. The square of one half of the distance between the foci is a2 , b2. Thus, the equation of the hyperbola is (mx , y)(m0x , y ) = a2 , b2: Let x = a = cos(), y = b sin() be any of the four points of contact of the rectangle with the ellipse. The tangent to the ellipse at this point is
a cos() + y sin() = 1: a b
326 Thus we have
,b ; m0 = a tan() ; m = a tan( ) b cos2 () , b sin() = ,b ; mx , y = ,bsin( ) sin() 2 2 m0x , y = ,(a , bb ) sin() :
Substitution into the equation of the hyperbola shows that this point of contact lies on the hyperbola. [Ed: This argument fails if the rectangle has sides parallel to the axes. Only Konecny pointed out that in this case, the hyperbola is degenerate and is, in fact, given by the axes.] Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; JORDI DOU, Barcelona, Spain; ROBERT GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; WALTHER JANOUS, Ursulinengymnasium, Y, Ferris State University, Big Rapids, Innsbruck, Austria; VACLAV KONECN Michigan, USA; KEEWAI LAU, Hong Kong; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.
2080. [1995: 278] Proposed by Marcin E. Kuczma, Warszawa, Poland.
Let S = f1; 2; 3; 4; 5; 6; 7g. Find the number of maps f from S to S such that f 2080(x) = x for every x 2 S . (Here the superscript denotes iteration: f 1 (x) = f (x) and f n(x) = f (f n,1(X )) for all n > 1.) Solution by Robert Geretschlager, Bundesrealgymnasium, Graz, Austria. Perhaps not too surprisingly, knowing Marcin, there are 2080 such maps. We can see this in the following manner. First of all, each f must be onetoone in S , since there would otherwise exist an x 2 S such that there is no y 2 S with f (y ) = x, which would contradict f (f 2079(x)) = f 2080(x) = x. f must therefore be a permutation of the elements of S . We know that every permutation of a nite set can be expressed uniquely as the product of cyclic permutations with no common elements. The problem is therefore equivalent to nding the number of such products of cycles, such that the 2080th iteration is the identity. In order for this to be the case, the length of each cycle must be a divisor of 2080. The lengths of such cycles can therefore be either 1, 2, 4 or 5. We now consider three cases. (a) There exists a cycle of length ve. , There are 75 ways to select the ve numbers in the cycle, and 4! ways to build a cycle of ve given numbers. Since the remaining numbers can either
327 , build a cycle of length two, or two of length one, there are 75 4! 2 = 1008 such maps. (b) There exists a cycle of length four. , There are 74 ways to select the four numbers in the cycle, and 3! ways to build a cycle of four given numbers. The remaining three numbers can either each build a cycle of length one, or there are three ways in which to choose two to build a cycle of length , two, leaving the other to build a cycle of length one. There are therefore 74 3! 4 = 840 such maps. (c) The only other cycles that exist are of lengths one and two. is one such map with no cycle of length two (the identity). There , There are, 72 , = 21 such maps with precisely one cycle of length two. There are 1 7 5 = 105 such maps with precisely two cycles of length two (choosing 2 2 2 2 from 7, then 2 from the remaining 5, and then dividing of , by,the number , repetitions of the chosen cycles). Finally there are 3!1 72 52 32 = 105 such maps with precisely three cycles of length two (for the analogous reason as for two such cycles). Adding up, we have
1008 + 840 + 1 + 21 + 105 + 105 = 2080 maps with the desired property. Also solved by HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEEWAI LAU, Hong Kong; ROBERT P. SEALY, Mount Allison University, Sackville, New Brunswick; HOE TECK WEE, student, Hwa Chong Junior College, Singapore; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer. There were also six incorrect solutions sent in. Sealy nds all positive integers n such that there are exactly n maps f : S ,! S (with S = f1; : : : ; 7g) satisfying f n(x) = x for all x 2 S. He gets:
n = 1; 351; 505; 721; 1072; 1225; 1575; 2080; 2800; 3312; 4320; 5040: As Sealy remarks, \You now have proposed problems 2800, 3312, 4320 and 5040."! This list contains two obvious members, 1 and 5040 = 7!, and the analogous numbers would appear in the corresponding list if we were to use S = f1; : : : ; N g for an arbitrary positive integer N . How many numbers are in the list for N ? What are the second{smallest and second{largest numbers in the list?
328
2081. [1995: 306] Proposed by K.R.S. Sastry, Dodballapur, India.
In base ten, if we write down the rst doubledigit integer (10) followed by the last two singledigit integers (8 and 9) we form a fourdigit number (1089) which is a perfect square (332). What other bases exhibit this same property? Solution by HeinzJurgen Seiert, Berlin, Germany. In base b 2, Nb has the four digits 10(b , 2)(b , 1), so Nb = b3 + (b , 2)b + (b , 1) = (b , 1)(b + 1)2; which shows that Nb is a perfect square if and only if b , 1 is a perfect square. (Editor's note: a number of solvers ignored base b = 2, where 10012 = 910 = (310)2 = (112)2:) Benito Mu~ ager pnoz, Fern~andez Moral, and Geretschl all noted that if b , 1 = m2 , then Nb is the base b twodigit number with p both digits equal to m; Vella remarked that in base 10pthis Nb is just the product of m and b +1. For example, if b = 5 = 22 +1, N5 = 225 = 1210.) Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, ~ and EMILIO FERNANDEZ Spain; MANUEL BENITO MUNOZ MORAL, I.B. Sagasta, Logro~no, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; TIM CROSS, King Edward's School, Birmingham, England; LUIS V. DIEULEFAIT, IMPA, Rio de Janeiro, Brazil; HANS ENGELHAUPT, Franz{ Ludwig{Gymnasium, Bamberg, Germany; JEFFREY K. FLOYD, Newnan, Georgia, USA; TOBY GEE, student, the John of Gaunt School, Trowbridge, England; ROBERT GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KATHLEEN E. LEWIS, SUNY Oswego. Oswego, NY, USA; DAVID LINDSEY, Austin Peay State University, TN, USA; DAVID E. MANES, State University of New York, Oneonta, NY, USA; J.A. MCCALLUM, Medicine Hat, Alberta; P. PENNING, Delft, the Netherlands; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; CORY PYE, student, Memorial University of Newfoundland, St. John's, Newfoundland; CRIST OBAL SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon, Spain; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; DAVID C. VELLA, Skidmore College, Saratoga Springs, NY, USA; CHRIS WILDHAGEN, Rotterdam, the Netherlands; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. There was one incorrect solution.
2082. [1995: 306] Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a triangle with \A > 90 , and AD, BE and CF are its altitudes (with D on BC , etc.). Let E 0 and F 0 be the feet of the perpendiculars from E and F to BC . Suppose that 2E 0 F 0 = 2AD + BC . Find \A.
329 Solution by KeeWai Lau, Hong Kong. As usual, let a = BC , b = CA and c = AB . We have AD = b sin C; and E0F 0 = a(1 , sin2 B , sin2 C ); where the latter equation comes from E 0 F 0 = a , CF 0 , BE 0 , with CF 0 = BC0 , BF 0 2= a , BF cos B = a , a cos2 B = a sin2 B, and, similarly, BE = a sin C . Since we are given 2E 0 F 0 = 2AD + BC , by the sine law, we have 2 sin A(1 , sin2 B , sin2 C ) = 2 sin B sin C + sin A: Hence, and so
sin A(1 , 2 sin2 B , 2 sin2 C ) = 2 sin B sin C;
sin A 2 cos(B , C ) cos(B + C ) , 1 = cos(B , C ) , cos(B + C ); or (since cos A = , cos(B + C ))
, cos(B , C )(1 + 2 sin A cos A) = cos A + sin A;
or
(cos A + sin A) 1 + (cos A + sin A) cos(B , C ) = 0:
Since \A is obtuse, we have
p
1 + (cos A + sin A) cos(B , C ) = 1 + 2 cos( A, 45) cos(B , C ) p ,1 = 0: > 1+ 2 p 2 Thus, cos A + sin A = 0, and therefore \A = 135 .
Berlin, Germany; MANUEL Also solved by SEFKET ARSLANAGIC, ~ BENITO MUNOZ and EMILIO FERNANDEZ MORAL, I.B. Sagasta, Logro~no, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; RICHARD Y, I. HESS, Rancho Palos Verdes, California, USA; VACLAV KONECN Ferris State University, Big Rapids, Michigan, USA; P. PENNING, Delft, the Netherlands; D.J. SMEENK, Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer.
2083. [1995: 306] Proposed by Stanley Rabinowitz, Westford, Massachusetts. The numerical identity cos2 14 , cos 7 cos 21 = sin2 7 is a special case of the more general identity cos2 2x , cos x cos 3x = sin2 x. In a similar manner, nd a generalization for each of the following numerical identities:
330 (a) tan 55 , tan 35 = 2 tan 20 ; (b) tan 70 = tan 20 + 2 tan 40 + 4 tan 10 ; (c)? csc 10 , 4 sin 70 = 2. [Ed: it was not easy to decide which submitted solution to highlight here, since the answers to the problem are not unique. We decided to give the proposer's solution to (a) and (b), and one of several submitted for (c).] Solutions: (a) and (b) by the proposer, (c)? by several solvers. (a) tan x , tan(90 , x) = 2 tan(2x , 90 ), (b) tan(x + 60 ) = tan(30 , x) + 2 tan(60 , 2x) + 4 tan(4x , 30 ), (c)? csc(2x) , 4 sin(15 , x) sin(75 , x) csc(2x) = 2.
[Ed: all solvers listed below solved parts (a) and (b). A ? before a solver's name indicates that part (c) was also solved, while a y before a solver's name indicates that a restricted case of part (c) was also solved.] Berlin, Germany; ? Manuel Benito Solved by ? SEFKET ARSLANAGIC, Mu~noz and EMILIO FERNANDEZ MORAL, I.B. Sagasta, Logro~no, Spain; ? CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; ? CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; ? THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; CURTIS COOPER, Central Missouri State University, Warrensburg, Missouri, USA; ? TIM CROSS, King Edward's School,Birmingham, England; ? DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut,USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; ? RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus Y, Ferris State University, Big Rapids, Michigan, tria; yVACLAV KONECN USA; yDAVID E. MANES, State University of New York, Oneonta, NY, USA; yP. PENNING, Delft, the Netherlands; ?BOB PRIELIPP, University of Wis consin{Oshkosh, Wisconsin, USA; HEINZJURGEN SEIFFERT, Berlin, Ger? many; TOSHIO SEIMIYA, Kawasaki, Japan; yDIGBY SMITH, Mount Royal College, Calgary, Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer.
2084. [1995: 306] Proposed by Murray S. Klamkin, University of Alberta. Prove that cos B2 cos C2 + cos C2 cos A2 + cos A2 cos B2 1 , 2 cos A2 cos B2 cos C2 ; where A, B , C are the angles of a triangle.
331
to
Solution by the proposer. Let x = cos A2 , y = cos B2 , z = cos C2 . The inequality is now equivalent
x 1,y 1,z 1 11 , + x + 1 + y + 1 + z;
or
1 tan2 A4 + tan2 B4 + tan2 C4 : Since tan2 x4 is convex for x 2 [0; ], the latter inequality follows by the majorization inequality. Note that
(; 0; 0) (A; B; C ) , 3 ; 3 ; 3 :
So there is equality only from the degenerate triangle of angles , 0, 0. It also follows that p tan2 A4 + tan2 B4 + tan2 C4 3 tan2 12 = 21 , 12 3: and with equality if and only if the triangle is equilateral. Berlin, Germany; MANUEL Also solved by SEFKET ARSLANAGIC, ~ BENITO MUNOZ and EMILIO FERNANDEZ MORAL, I.B. Sagasta, Logro~no, Spain; CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOE HOWARD, New Mexico Highlands University, Las Vegas, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; DAVID E. MANES, State University of New York, Oneonta, NY, USA; HEINZJURGEN SEIFFERT, Berlin, Germany; and PANOS E. TSAOUSSOGLOU, Athens, Greece.
2085. [1995: 307] Proposed by Iliya Bluskov, student, Simon Fraser University, Burnaby, BC, and Gary MacGillivray, University of Victoria, B. C. Find a closedform expression for the n by n determinant
n ,1 ,1 ,1 ,1 ,1 ,1
,1 3 ,1 0 0 0 0
,1 ,1 3 ,1 0 0 0
,1 0 ,1 3 0 0 0
.. .
,1 ,1 ,1
.. .
0 0 0
.. .
0 0 0
.. . . . .
.. .
.. .
..
: .
0 3 ,1 0 0 ,1 3 ,1 0 0 ,1 3
Combination of solutions by Theodore Chronis, student, Aristotle University of Thessaloniki, Greece, and HeinzJurgen Seiert, Berlin, Germany.
332 that
Let Dn denote the determinant under consideration. We shall prove
Dn = F2n,1 + F2n+1 , 2
for all positive integers n; where (Fk) is the sequence of the Fibonacci numbers de ned by F1 = F2 = 1 and Fk = Fk,1 + Fk,2 for k 2. Adding all rows to the rst row and then adding all columns to the rst column, we get that Dn equals the n by n determinant
3 1 0 0 1 3 ,1 0 0 ,1 3 ,1 0 0 ,1 3 .. .
.. .
0 0 1
.. .
0 0 0
0 0 0
0 0 0 0
.. . . . .
0 0 0 0
.. .
1 0 0 0
.. : .
.. .
0 3 ,1 0 0 ,1 3 ,1 0 0 ,1 3
By expanding this determinant in minors along the rst row, we therefore have Dn = 3An,1 , ,A,n,2 + (,1)n(,1)n,2 +(,1)n+1 (,1)n,2 + (,1)nAn,2 = 3An,1 , 2An,2 , 2 for n 3, where An is the n by n determinant
3 ,1 0 ,1 3 ,1 0 ,1 3 .. .
0 0 0
.. .
0 0 0
.. . . . .
0 0 0 .. .
0 0 0 .. .
0 0 0
.. : .
0 3 ,1 0 0 ,1 3 ,1 0 0 ,1 3
By expanding this along the rst row, we nd the recurrence relation An = 3An,1 , An,2; n 3: Since A1 = 3 = F4 and A2 = 8 = F6 , from the known identity Fn = 3Fn,2 , Fn,4 it follows that An = F2n+2 for all n 1. So the original determinant is equal to Dn = 3F2n , 2F2n,2 , 2 = 2F2n + (F2n+1 , F2n,1) , 2(F2n , F2n,1 ) , 2 = F2n+1 + F2n,1 , 2
333 for n 3. The cases n = 1 and n = 2 are done by direct computation [D1 = 1 = F3 + F1 , 2, D2 = 5 = F5 + F3 , 2]. ~ and EMILIO FERNANDEZ Also solved by MANUEL BENITO MUNOZ MORAL, I.B. Sagasta, Logro~no, Spain; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; and the proposers. Benito and Fernandez give the answer in the form Dn = L2n , 2, where Ln is the nth Lucas number, de ned by L1 = 1, L2 = 3, Ln = Ln,1 + Ln,2 . They also point out that the answer could be written in yet another interesting form: ( 2 2 4; n even; Dn = 5LF2 n==5LFn2 , n n , 4; n odd. In his article \The sequence 1 5 16 45 121 320 : : : in combinatorics", in the Fibonacci Quarterly Vol. 13 (1975) pp. 51{55, Kenneth Rebman considers the same sequence of integers de ned by the determinant in this problem, and gives examples involving graphs and matrices where this sequence arises. He also evaluates a determinant which can easily be shown to be equal to the rst determinant in the above solution.
2086. Proposed by Aram A. Yagubyants, Rostov na Donu, Russia.
If the side AC of the spherical triangle ABC has length 120 (that is, it subtends an angle of 120 at the centre), prove that the median from B (that is, the arc of the great circle from B to the midpoint of AC ) is bisected by the other two medians. Solution by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. This is a special case of a known result: If G is the concurrency point of the three medians and A0 is the sin AG0 = 2 cos a , etc. midpoint of BC then sin GA 2 0 So, if a = 120 , then AG = GA . (The other possibility AG + GA0 = 180 is ruled out since A and A0 would then be antipodal, forcing triangle ABC to be degenerate.) Proof. Let A~ , B~ , C~ denote unit vectors from the centre of the sphere to the vertices A, B , C , respectively. It now follows easily that
~ ~ ~ G =
A~ + B~ + C~
A + B + C
lies on each of the medians and hence is the point of concurrency. Also, ~ ~ A~0 =
B + C
: Then
B~ + C~
334
A~ A~ + B~ + C~
A~ B~ + C~
sin AG =
~ ~ ~
=
~ ~ ~
;
A + B + C
A + B + C
~ ~ + C~
A~ + B~ + C~ B~ + C~
A B 0 sin GA =
~ ~ ~
~ ~
=
~ ~ ~
~ ~
;
A + B + C B + C
A + B + C B + C
so that
sin AG =
B~ + C~
= 2 cos a : sin GA0 2
~ and EMILIO FERNANDEZ Also solved by MANUEL BENITO MUNOZ MORAL, Logro~no, Spain; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; P. PENNING, Delft, the Netherlands; and the proposer. Benito and Ferna~ndez found the general result as an exercise in a nineteenth century text by J.A. Serret; they show it to be an easy application of spherical trigonometry. 2088. [1995: 307] Proposed by Sefket Arslanagic, Berlin, Germany. Determine all real numbers x satisfying the equation
2x + 1 + 4x + 5 = 3x , 1 ; 3 6 2
where bxc denotes the greatest integer x. Solution by Luis Dieulefait, IMPA, Rio de Janeiro, Brazil. Let Zdenote the integers. Then, since b(2x + 1)=3c + b(4x + 5)=6c = (3x , 1)=2 must be an integer,
x = 2n3+ 1 for some n 2 Z:
(1)
2x + 1 + 1 = 4x + 5 : 3 2 6
(2)
We will also use the fact that
For every real number we have:
1 1 () + 2 , + 2 21 [that is, the fractional part of is < 1=2 if and only if the fractional part of + 1=2 is 1=2], so that 1 1 bc > , 2 () + 2 :
, bc < 12
335 From this, using by c y we obtain
bc + + 12 2;
and using by c > y , 1 we obtain
(3)
bc + + 12 > 2 , 1; (4) both for every real number . [Editor's remark: for example, to get (3) note that for any either 1 + 2 and bc ;
or
bc , 21
and
1 + 2 + 12 ;
and adding gives (3) in either case.] Putting inequalities (3) and (4) together and using (2) gives, for = (2x + 1)=3:
2(2x + 1) , 1 < 2x + 1 + 4x + 5 2(2x + 1) : 3 3 6 3
Using the original equation, this is
4x , 1 < 3x , 1 4x + 2 : (5) 3 2 3 From this we obtain 1 < x 7. Using (1) the only possibilities are x = 35 ; 73 ; 93 ; : : : ; 21 3 :
So this is exactly the set of all solutions. Editorial comment. Alternatively, equations (3) and (4) imply that 1 bc + + 2 = b2c
for all , so the original equation becomes
4x + 2 = 3x , 1 ; 3 2
which is equivalent to (5) and (1) together. Benito and Fernandez, Wildhagen and the proposer also gave solutions along these lines. Also solved by CLAUDIO ARCONCHER, Jundia, Brazil; MANUEL ~ and EMILIO FERNANDEZ BENITO MUNOZ MORAL, I.B. Sagasta, Logro~no,
336 Spain; CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; TIM CROSS, King Edward's School, Birmingham, England; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut,USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; JEFFREY K. FLOYD, Newnan, Georgia, USA; TOBY GEE, student, the John of Gaunt School, Trow bridge, England; ROBERT GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER Y, FerJANOUS, Ursulinengymnasium, Innsbruck, Austria; VACLAV KONECN ris State University, Big Rapids, Michigan, USA; KEEWAI LAU, Hong Kong; DAVID E. MANES, State University of New York, Oneonta, NY, USA; P. PENNING, Delft, the Netherlands; CORY PYE, student, Memorial Uni versity of Newfoundland, St. John's; CRISTOBAL SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon, Spain; HRISTOS SARAGHIOTES, student, Aristotle University of Thessaloniki, Greece; ROBERT P. SEALY, Mount Allison Uni versity, Sackville, New Brunswick; HEINZJURGEN SEIFFERT, Berlin, Germany; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer. As well, there were ten incorrect solutions sent in, most of which assumed that x had to be an integer. In the May 1996 issue of CRUX [1996: 168], we asked: \Do you know the equation of this curve?"
In the September 1996 issue of CRUX [1996: 288], we gave the hint: \it is known as the \butter y"!" The answer is r = ecos() , 2 cos(4) + sin5 (=12). If you have any other \nice" curves, please send them to the editor. (Answer to Rider on page ??:) To buy or not to buy, that is not the question.
Piglet.
337
Letter from the Editors As announced in the November issue, 1997 will see an increase in the quantity of material devoted to high school mathematics, without a decrease in any of the present composition of CRUX. We are pleased to announce that this will be eected through an amalgamation of CRUX MATHEMATICORUM with Mathematical Mayhem. For those of you not familiar with it, Mayhem was started by a group of Canadian IMO Alumni, and aimed to be a \journal of high school and college level mathematics, written by and for students". It has run continuously for eight years, and has a circulation of about 130. Its previous editors have included the distinguished problemists Ravi Vakil (cofounder with Patrick Surry) and J.P. Grossman, who were the Deputy Leader and Leader of the Canadian delegation of the 1996 IMO. The present editors are Naoki Sato and Cyrus Hsia, who will be joining the CRUX with MAYHEM Editorial Board in January. Mayhem has published items such as expository articles, hints of problem solving, IMO problems and solutions, and three levels of problems  high school, advanced and challenge board  (set by the editors or contributors). By keeping communication alive amongst the Canadian IMO alumni, team members and \wanabees", Mayhem has played an important role in keeping the IMO spirit alive and well in Canada. This amalgamation will give CRUX readers additional high school level material (if you have forgotten CRUX's mandate, please reread the back inside cover), as well as giving Mayhem a wider and international exposure. CRUX with MAYHEM will have 64 pages per issue and will have problems to challenge all readers from interested high school students to senior undergraduate students. We hope that high school readers who will be new to CRUX with MAYHEM will nd the range of problems useful in the classroom and stimulating. If the issue that you habitually read is a library copy, please ensure that your librarian is aware of this change. Since CRUX is the older journal, and taken by more libraries, the Mayhem editors have agreed that CRUX with MAYHEM will continue the volume numbering of CRUX (next year will see volume 23) as well as maintain the general external appearance. However, the table of contents will be moved to the outside back cover since it will be enlarged, and also to make space for the words Mathematical Mayhem on the front cover. With all best wishes, Bruce Shawyer Graham Wright EditorinChief Managing Editor
338
Lettre des redacteurs
Comme nous l'avons annonce dans le numero de novembre, le CRUX sera davantage axe sur les mathematiques de niveau secondaire en 1997, sans pour autant que l'on touche a son contenu actuel. Il nous fait plaisir de vous annoncer que nous reussirons ce tour de force en fusionnant le CRUX MATHEMATICORUM et le Mathematical Mayhem. Pour ceux d'entre vous qui ne conna^itraient pas le Mayhem, sachez qu'il a e te fonde par d'anciens membres de l'equipe canadienne a l'OIM s'etant donne pour mission de produire un periodique sur les mathematiques de niveau secondaire et collegial, concu par et pour des e tudiants. Ce magazine, qui para^it regulierement depuis huit ans, est tire a environ 130 exemplaires. Parmi ses anciens redacteurs, mentionnons les excellents problemistes Ravi Vakil (cofondateur avec Patrick Surry) et J.P. Grossman, respectivement chef d'equipeadjoint et chef d'equipe de la delegation canadienne a l'OIM 1996. Les redacteurs actuels, Naoki Sato et Cyrus Hsia, se joindront a l'equipe de redaction du CRUX with Mayhem en janvier. Le Mayhem a entre autres publie des articles de fond, des trucs pour faciliter la resolution de problemes et certains problemes poses aux dernieres OIM accompagnes. Les problemes presentes sont classes selon trois niveaux de diculte, determines par les redacteurs ou les collaborateurs : secondaire, avance et challenge board . En permettant les e changes entre les anciens participants canadiens a l'OIM, les membres de l'equipe canadienne actuelle et les aspirants a ce titre, le Mayhem a grandement contribue a preserver l'esprit de l'OIM au Canada. La fusion de ces deux magazines permettra aux lecteurs du CRUX d'avoir acces a une plus grande quantite d'articles sur les mathematiques de niveau secondaire (pour un rappel du mandat du CRUX, relire la troisieme de couverture) et procurera au Mayhem une visibilite accrue et internationale. Chaque numero du CRUX with Mayhem comptera 64 pages et presentera des problemes qui sauront susciter l'inter^et de tous les lecteurs, des e leves du secondaire aux e tudiants en n de bac. Nous esperons que les lecteurs du secondaire qui decouvriront le CRUX with Mayhem y trouveront des problemes stimulants et utiles pour leurs e tudes. Si vous lisez habituellement un exemplaire provenant d'une bibliotheque, veri ez si votre bibliothecaire est au courant de cette fusion. Puisque le CRUX est le plus ancien des deux magazines et le plus commun dans les bibliotheques, les redacteurs du Mayhem ont accepte que le CRUX with Mayhem suive la numerotation par volumes du CRUX (l'an prochain : volume no 23) et conserve en gros sa presentation exterieure. On deplacera toutefois la table des matieres en quatrieme de couverture, puisqu'elle sera agrandie et qu'il faudra faire de la place aux mots Mathematical Mayhem sur la pagetitre. Meilleurs voeux a tous! Bruce Shawyer Graham Wright Redacteur en chef Directeur de redaction
339
In memoriam  Pal Erdo} s
Pal Erd}os  19130326 to 19960920 In appreciation of the \Prince of Problem Posers", Pal Erdos, } who, sadly for those of us who had the privilege of knowing him, died while he was in Warsaw, attending a mathematics meeting. It always seemed that Erdos } was eternal; it's hard to realize that we won't hear his delightful language again. He in uenced hundreds, probably thousands, of mathematicians, and each in an individual way. Anything one can say must be personal. A major part of Erdos's } genius was asking innumerable questions, and, most importantly, asking them of the right person. He seemed to know, better than you yourself, what problems you could solve. He gave the con dence that many of us needed to embark on research. Erdos } was the problem proposer par excellence. Almost anyone can ask questions that are impossibly dicult or are trivially easy. To achieve the balance between these extremes is given to few of us. Erdos's } questions were always just right. Many have remained as outstanding, but important problems, but most have been attacked and partially, if not completely solved. But he didn't only pose problems; he wrote more than 1500 papers. Those who didn't know him well thought that he just threw out ideas and got others to write for him. Certainly he had a phenomenal number of coauthors, but a good fraction of his papers were solo eorts, and, far from depending on his coauthors, he often wrote the paper himself and added on the other names, even on occasions when their contributions were comparatively minor. He opened up new areas of research in number theory, combinatorics and geometry. Notable are probabilistic number theory with Renyi & Turan and the partition calculus with Rado & Hajnal.
340 Perhaps `bosses', `slaves', `captured', `epsilons', `noise', `poison', `the SF', `The Book', `the Sam & Joe show', `vot vos this ven it vos alive?', `is your brain open?' will remain in the vocabulary of a few of his many friends, but the words won't sound the same now that it's not Erdos } saying them. R.K. Guy
For the information of those readers who may not know who Pal Erdos } was, here is some further information. Erdos } was highly regarded in the mathematical community as one of the most brilliant and eccentric mathematicians of the 20th century. His sole interest was in mathematics, resulting in him being only known to the mathematical community. He lived a very simple life, eschewing personal comfort and possessions. He considered money to be an encumbrance, and gave away, as prizes for solving mathematical problems, whatever he earned that was over and above his basic simple needs. Erdos } published over 1,500 papers, most of them jointly with other mathematicians. He gave generously of his ideas to everyone with whom he came into contact. He would ascertain your mathematical interests, and then proceed with \I have this problem, : : : ". Over 4,000 mathematicians have published jointly with Erdos. } Pal was born in Budapest into a HungarianJewish family, the son of two mathematics teachers. While his father was a prisoner of war in Siberia for six years, his mother taught him at home. He took his doctorate in 1934 at the University of Budapest, and then went for further study in Manchester, England. The conditions in Europe made it unsafe for him to return to Hungary, and he moved to the United States. In the 1950's, the United States denied him reentry, as he was suspected of being a Soviet sympathiser. He then spent much time in Israel. He was allowed reentry to the United States in the 1960's, and his mother, then in her eighties, began to travel with him. Erdos } never saw the need to restrict himself to one institution, so he crisscrossed the world, inspiring mathematicians wherever he went. After his mother died in 1971, he immersed himself in his work with great enthusiasm, often spending 18 hours daily on mathematical problems. He was still following this active lifestyle when the great reaper called on him in Warsaw.
341
Algebraic Integers and Tensor Products of Matrices. Shaun M. Fallat
Dept. of Mathematics and Statistics University of Victoria Victoria B.C., Canada A complex number is called an algebraic integer if it is a zero of a monic (i.e., the coecient of the term with the highest exponent is one) polynomial with integer coecients. Let C denote the complex numbers. A set S C is called a subring if 1 2 S , and for every a; b 2 S , a + b 2 S and a b 2 S . It is wellknown that the set of algebraic integers forms a subring of the complex numbers (see [3]). The aim of the present note is to show that this result follows easily from basic facts about tensor products of matrices. First we will need some notation. The set of n n matrices over a subring R is denoted by Mn (R). The notation A = [aij ] means the (i; j )th entry of A is aij . The set of polynomials with integer coecients is denoted by Z [x], where x is an indeterminate. Let det A denote the determinant of an n n matrix A. If A 2 Mn (R), then pA (x) = det(xIn , A) is the characteristic polynomial of A, where In denotes the n n identity matrix. Since the operations involved in computing the determinant of A are multiplication and addition, it is readily veri ed that if A 2 Mn (Z ), then pA (x) is a monic polynomial in Z [x]. Recall that if is a zero of pA (x), then is called an eigenvalue of A. We begin with an example for completeness. p p Example 1. It is easy to check that 2, 3 are algebraic integers. For examp ple, 2 is p a zeropof x2 , 2. Thereforepthere p mustpexist a polynomial in Z [x], for which 2 3 is a root. Since 2 3 = 6,pit is easy to verify that x2 , 6 is an example of a polynomial in Z [x] with p 6pas a root. Similarly, there must exist a polynomial in Z [x], for which 2+ 3 is a root. Consider
p p p p f (z) = (x , ( 2 + p 3))(px , ( 2 , p3)) p (x , (, 2 + 3))(x , (, 2 , 3)) = x4 , 10x2 + 1: p p Hence x4 , 10x2 + 1 is an example of a polynomial in Z [x] with 2 + 3 as a root. The following result will establish the relationship between algebraic integers and matrices.
342 Lemma 2. Let be a complex number. Then is an algebraic integer if and only if is an eigenvalue of a square matrix with integer coecients. Proof. If is an eigenvalue of A 2 Mn (Z ), then pA (x) is a monic polynomial in Z [x], and pA () = 0. Thus is an algebraic integer. To prove the converse, assume that f () = 0, where f (x) is a monic polynomial in Z [x], say f (x) = xn + c1 xn,1 + + cn,1 x + cn. Let A be the companion matrix of f (x), i.e. 2 3
66 A = 666 64
0 0
1 0
0 1
0 0 77
77 : 77 0 0 0 1 5 ,cn ,cn,1 ,c2 ,c1 .. .
.. .
... ...
.. .
Expanding pA (x) by the rst column and using a simple induction argument, it follows that pA (x) = f (x), and this fact can be found in [1; pp 230231]. Therefore is an eigenvalue of A and A 2 Mn (Z ). De nition 3. If A = [aij ] is an m n matrix and B = [bij ] is a p q matrix then the Kronecker or tensor product of A and B , denoted A B , is the mp nq matrix de ned as follows:
2 a B a B 3 1n 66 11. . .. 77 : A B = 4 .. .. . 5 am1B amnB Note that if A and B have integer entries, then so does A B . From the above de nition it is easy to verify that if A; B; C; D 2 Mn (R), then (A B)(C D) = AC BD: Vectors in C n may be identi ed with n 1 matrices over C, so the above de nition may be used to de ne tensor products x y with x 2 C n and y 2 C m. The following proposition establishes a part of the wellknown
results that identify the eigenvalues of tensor products see, e.g. [2; pp 242245]. We provide an easy proof for completeness. Proposition 4. Let A 2 Mn (Z ), B 2 Mm (Z ), let be an eigenvalue of A and let be an eigenvalue of B . Then (a) is an eigenvalue of A B , (b) + is an eigenvalue of (A Im ) + (In B ). Proof. There exist nonzero vectors x 2 C n and y 2 C m such that Ax = x and By = y . It follows immediately that
343
(A B )(x y ) = (x y ); and
((A Im ) + (In B ))(x y ) = ( + )(x y ):
This proves the propostion. Notice that if A 2 Mn (Z ), B 2 Mm (Z ), then A B 2 Mmn (Z ). Hence, if two algebraic integers ; are given, the tensor product is a useful tool for determining the existence of an integral matrix, for which or + is an eigenvalue. We summarize this in the following theorem. Theorem 5. The set of algebraic integers forms a subring of the ring of complex numbers. The proof follows directly from Lemma 2 and Proposition 4.
References [1] K. Homan and R. Kunze, Linear Algebra, PrenticeHall, New Jersey, 1971. [2] R.A. Horn and C.R. Johnson, Topics in Matrix Analysis, Cambridge University Press, New York, 1991. [3] H.L. Montgomery, I. Niven and H.S. Zuckerman, An Introduction to the Theory of Numbers, John Wiley and Sons, Inc. New York, 1991.
PHOTO PROBLEM
Can you identify this regular CRUX contributor?
344
THE SKOLIAD CORNER No. 18
R.E. Woodrow Last issue we gave the problems of the Canadian Mathematical Society Prize Exam for Nova Scotia. Here are solutions.
CANADIAN MATHEMATICAL SOCIETY PRIZE EXAM Friday, April 26, 1996  Time: 2.5 hours
1. (a) Solve px + 20 , px + 1 = 1. p
p
(b) Try to solve 3 x + 20 , 3 x + 1 = 1. p p Solution. (a) Set A = x + 20 and B = x + 1. We get A2 , B 2 = A + B as A , B = 1. So x + 20 , (x + 1) = 19 = A + B. Now 2A = 20, A = 10 and x = 80. p p (b) Let A = 3 x + 20 and B = 3 x + 1 so A , B = 1. Multiplying by A22+AB+B2 gives A23,B3 = A2+AB+B2 or A2+AB+B2 = 19. As 2 A , 2AB + B = 1 = 13 we3 obtain3 3AB = 19 , 1 = 18 or AB = 6. So (x + 20)(x + 1) = A B = 6 = 216. This gives the quadratic equation x2 + 21x , 196 = 0, or (x + 28)(x , 7) = 0 and x = ,28 or x = 7. 2. Suppose a function is de ned so that f (xy) = f (x) + f (y). (a) Show that if f is de ned at 1, f (1) = 0. (b) Similarly if f is de ned at 0, its value at any x will be 0, so it is a \trivial" function. (c) Show that if f is de ned only for f1; 2; 3; : : : g there are many nontrivial ways to de ne such an f (give an example). Solution. (a) With x = 1 = y we obtain f (1 1) = f (1) + f (1), so f (1) = 2f (1) giving f (1) = 0. (b) Suppose that f is de ned at 0. Then with y = 0 and x arbitrary we have f (0) = f (x 0) = f (x) + f (0) and f (x) = 0. (c) To see there are many such f with domain the natural numbers consider the natural logarithm, logarithms to base 10, and the following family of examples: Let S be a set of prime Q numbers. A given natural number can be written uniquely in the form Pim where P1 ; P2 ; : : : ; lists the prime numbers and mi is a nonnegative integer. P(Of course only nitely many of the mi are nonzero. De ne fs (n) = mi . i
345
3. Three circles of equal radii r all touch each other to enclose a three cornered concave area A. How big is the area of A? Solution.
D
q
q
E
F q
Let the centres of the circles be labelled D, E , Fp. Then DEF is an equilateral triangle with side length 2r and area 3 r2 . The area A is the area of the triangle less the three sectors, each subtending an angle of 60 in a circle of radius r, so the area of A is
p3 r2 , 3 1 r2 = p3 , r2: 6
2
4. Show that for all real numbers x,
(a) x4 4x , 3. (b) x4 is not greater than 3x , 2, even though it appears to be true. 1 1 x ,1 0 14 1 3 2 x4 1 0 :004 0:12 :063 1 4x , 3 ,7 ,3 ,2 ,1:67 ,1 1 3x , 2 ,5 ,2 ,1:25 ,1 ,:5 1
2 16 5 4
Solution. (a) x4 4x , 3 is the same as x4 , 4x + 3 0. Set f (x) = x4 , 4x + 3. Now
f (x) = f ((x , 1) + 1) = [(x , 1) + 1]4 , 4[(x , 1) + 1] + 3 = (x , 1)4 + 4(x , 1)3 + 6(x , 1)2 +4(x , 1) + 1 , 4(x , 1) , 4 + 3 = (x , 1)4 + 4(x , 1)3 + 6(x , 1)2 = (x , 1)2[(x , 1)2 + 4(x , 1) + 6]:
3 81 9 7
346 Now (x , 1)2 0 and the discriminant of X 2 +4X +6 is 42 , 4 1 6 < 0 so both factors of f (x) are nonnegative. The minimum value is 0 at x = 1, as required. (b) Consider x = 0:9. Now x4 = 0:6561 and 3x , 2 = 0:7, so that x4 6 3x , 2. 5. Two integers are called equivalent, written x y if they are divisible by the same prime numbers (primes are 2; 3; 5; 7; : : : ) so 2 2 4, 3 27 but 2 6 3. (a) Show that 10 80 but 10 6 90. (b) Prove that if x y , then x2 y 2. Solution. (a) 10 = 2 5, 80 = 24 5 so 10 and 80 are divisible by the same primes, namely 2 and 5. (b) Suppose x y . Consider any xed prime p. If p divides x2 then p must2 also divide p. But then if2 p divides x2 it divides x and y, and thus y . Similarly if p divides y it also divides x2 . Thus x2 y 2, as required. 6. We can describe certain fractions in terms of others all1 with1 bigger denominators (always in lowest terms). For instance 3 = 4 + 121 and 2 = 14 + 41 + 16 but 23 = 26 + 62 doesn't work since 26 = 13 and 23 = 12 + 16 3 doesn't since 2 < 3. (a) Can you write 12 as a sum a1 + 1b for integers 2 < a < b? 1 (b) Try to write 1996 as a1 + 1b for integers 1996 < a < b.
1 1 1 2 3 6 1 1 1 = 1 1 + 1 = 1 + 1 : (b) = 1996 998 2 998 3 6 2994 5998
Solution. (a) = + and 2 < 3 < 6.
That completes the Skoliad Corner for this issue. Send me your contests, suggestions, and recommendations to improve this feature.
347
THE OLYMPIAD CORNER No. 178 R.E. Woodrow
All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. We begin this Issue with the remaining problems proposed to the jury, but not used at the 36th International Olympiad held at Toronto, Ontario in July 1995. As always, I welcome your novel, nice solutions that dier from the \ocial" published solutions.
36th INTERNATIONAL MATHEMATICAL OLYMPIAD
Problems proposed to the jury but not used Algebra
1. Let R be the set of real numbers. Does there exist a function f : R ! R which simultaneously satis es the following three conditions? (a) There is a positive number M such that ,M f (x) M for all x. (b) f (1) = 1. (c) If x 6= 0, then 2 f x + 1 = f (x) + f 1 : x2
x 2. Let n be an integer, n 3. Let x1; x2 ; : : : ; xn be real numbers such that xi < xi+1 for 1 i n , 1. Prove that 1 ! 0X ,1 n n(n , 1) X x x > nX (n , i)xi @ (j , 1)xj A : i j 2 i<j
i=1
j =2
Geometry 3.0 Let A1A2A3A4 be a tetrahedron, G its centroid, and A01, A02, A03 and A4 the points where the circumsphere of A1 A2 A3 A4 intersects GA1 , GA2, GA3 and GA4 respectively. Prove that GA1 GA2 GA3 GA4 GA01 GA02 GA03 GA04
and
1
1
1
1
1
1
1
1
GA01 + GA02 + GA03 + GA04 GA1 + GA2 + GA3 + GA4 :
348
4. O is a point inside a convex quadrilateral ABCD of area S. K , L, M and N are interior points of the sides AB, BC , CD and p DA prespectively. p If OKBL and OMDN are parallelograms, prove that S S1 + S2, where S1 and S2 are the areas of ONAK and OLCM respectively. 5. Let ABC be a triangle. A circle passing through B and C intersects the sides AB and AC again at C 0 and B 0 , respectively. Prove that BB 0 , CC 0 and HH 0 are concurrent, where H and H 0 are the orthocentres of triangles ABC and AB0C 0 respectively.
Number Theory and Combinatorics 6. At a meeting of 12k people, each person exchanges greetings with exactly 3k + 6 others. For any two people, the number who exchange greetings with both is the same. How many people are at the meeting? 7. Does there exist an integer n > 1 which satis es the following condition? The set of positive integers can be partitioned into n nonempty subsets, such that an arbitrary sum of n , 1 integers, one taken from each of any n , 1 of the subsets, lies in the remaining subset. 8. Let p be an odd prime. Determine positive integers x and y for which x y and p2p , px , py is nonnegative and as small as possible. Sequences 9. For positive integers n, the numbers f (n) are de ned inductively as follows: f (1) = 1, and for every positive integer n, f (n + 1) is the greatest integer m such that there is an arithmetic progression of positive integers a1 < a2 < < am = n and
f (a1) = f (a2) = = f (am): Prove that there are positive integers a and b such that f (an + b) = n + 2 for every positive integer n.
10. Let N denote the set of all positive integers. Prove that there exists a unique function f : N ! N satisfying f (m + f (n)) = n + f (m + 95) P f (k)? for all m and n in N. What is the value of 19 k=1
We now turn to readers' solutions to problems of the contest given in the Corner as the Turkish Mathematical Olympiad Committee Final Selection Test of April 4, 1993. This problem set was given in the April 1995 number of the corner [1995: 117118].
349
TURKISH MATHEMATICAL OLYMPIAD COMMITTEE FINAL SELECTION TEST April 4, 1993  Part I (Time: 3 hours)
1. Show that there is an in nite sequence of positive integers such that the rst term is 16, the number of distinct positive divisors of each term is divisible by 5, and the terms of the sequence form an arithmetic progression. Of all such sequences, nd the one with the smallest possible common dierence between consecutive terms. Solutions by Cyrus Hsia, student, University of Toronto, Toronto, Ontario; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Wang's solution. Clearly the question should have been reworded as \of all..., nd the one with the smallest nonzero common dierence between consecutive terms." For otherwise the constant sequence 16; 16; 16; : : : is clearly the answer. Let d denote the common dierence. If d < 0 then the terms of the sequence will eventually become negative. Hence assume that d > 0. For natural numbers n, let (n) denote the number of positive divisors of n. Then it is wellknown that if n = PQ11 P22 Pk denotes the prime power decomposition of n, then (n) = ki=1(i + 1). Furthermore, is a multiplicative function, i.e., if gcd (n;m) = 1 then (n m) = (n) (m). Therefore if 16 + d = 24 (2q + 1), where q = 0; 1; 2; : : : , then 5j (16+ d). To minimize d, or 16 + d, we take q = 1 since q = 0 implies d = 0. Then d = 3 24 , 24 = 25. To see that this value of d would yield an arithmetic progression with the desired property, note rst that for all m = 0; 1; 2; : : : , (16+ md) = (24 +25 m) = (24(2m +1)) 0 mod 5. Finally, straightforward checking reveals that 56 j (16 + t) for t = 1; 2; : : : ; 31. Therefore, the required sequence is indeed f16 + 32mg, m = 0; 1; 2; : : : with d = 32. 2. Let M be the circumcentre of an acuteangled triangle ABC , and assume the circle (BMA) intersects the segment [BC ] at P , and the segment [AC ] at Q. Show that the line CM is perpendicular to the line PQ. Solution by Cyrus Hsia, student, University of Toronto, Toronto, Ontario. Let triangle ABC have angles A, B , and C . \AMC = 2\B in circle (ABC ). Thus k
180 , 2B (in isosceles triangle AMC ) 2 = 90 , B; \PQC = \ABP (in circle (AMB )) = B:
\MCA =
350
A
q
q
B
N
q
P
q q
Q
M
q
q
C
Thus, extending CM to meet PQ in N we have \QCN = \MCA = 90 , B and \NQC = \PQC = B . It follows that \QNC = 180 , (\QCN + \NQC )
= 180 , (90 , B + B ) = 90:
That is, CM ? PQ. 3. Let fbng be sequence of positive real numbers such that
b2
b2
b2
for each n 1; b2n+1 13 + 23 + + n3 : 1 2 n Show that there is a natural number K such that K X
bn+1 > 1993 : b + b + + b n 1000 n=1 1 2
Solution by Cyrus Hsia, student, University of Toronto, Toronto, Ontario.
(13 + 23 + + n3)(b2n+1) 2 2 2 (13 + 23 + + n3) 1b13 + 2b23 + + nbn3 (b1 + b2 + + bn)2
by the Cauchy{Schwartz inequality. Thus
2 2 b2n+1 1 = (b1 + b2 + + bn )2 13 + 23 + + n3 n(n + 1) :
It follows that
bn+1 2 b1 + b2 + + bn n(n + 1)
351 since the sequence fbn g has only positive real terms. Thus K X
K X bn+1 2 n=1 b1 + b2 + + bn n=1 n(n + 1) K X = 2 n(n1+ 1) n=1 K 1 X 1 , = 2 n=1 n n + 1 = 2 1 , K 1+ 1 = K2K + 1:
By setting K = 999 we have 999 X
bn+1 2(999) = 1998 > 1993 ; n=1 b1 + + bn 999 + 1 1000 1000 as required.
April 4, 1993  Part II (Time: 3 hours)
1. Some towns are connected to each other by some roads with at most one road between any pair of towns. Let v denote the number of towns, and e denote the number of roads. Show that (a) if e < v , 1, then there are at least two towns such that it is impossible to travel from one to the other, (b) if 2e > (v , 1)(v , 2), then travelling between any pair of towns is possible. Solution by Cyrus Hsia, student, University of Toronto, Toronto, Ontario. (a) The minimum value that e could be such that the graph with v towns is connected is e = v , 1. We prove this by induction. For v = 3, we must have e at least 2 since e = 1 would only connect two towns. (And e = 2 is possible, as shown ,@). Suppose that the minimum for v towns is e = v , 1. Consider a connected situation with v + 1 towns. We can assume there is at least one town with at most one road connecting it to other towns otherwise 2e 2(v + 1) and e v + 1. Removing that town (and its road to another town) must r
r
r
352 leave a connected situation with v towns. So e , 1 v , 1 and e v as required. It is easy to see that one can construct a connected example with v + 1 towns and v + 1 , 1 = e edges, completing the induction. (b) The statement is proved by induction. For v = 3, we have 3 > (v,1)(2 v,2) = 1, so that e 2. ties.
With e = 2 we have
r
,@ r
r
and e = 3
r
,@
r
r
as the essential possibili
Suppose travelling between any pair of towns is possible for v towns if
2e > (v , 1)(v , 2). Now consider v + 1 towns joined by 2e > (v )(v , 1) roads. Consider a xed town. If it is connected to the v remaining towns, we are done. Otherwise, let it be joined to x v , 1 towns. If the town and its roads are removed we are left with v towns and
e , x > v(v 2, 1) , x v(v 2, 1) , (v , 1) = (v , 1)(v , 2) 2
roads, that is, the remaining v towns have more than (v,1)(2 v,2) roads and are connected. If x > 0, the (v + 1)st town is joined to some one of the remaining towns, and hence connected ,to all of them. If x = 0 then the v towns can have a maximum number of v2 = v(v2,1) roads. But e > v(v2,1) , so x 6= 0. Thus for v towns, e > (v,1)(2 v,2) implies all towns are connected and one can travel between any pair of towns. 2. On a semicircle with diameter AB and centre O points E and C are marked in such a way that OE is perpendicular to AB , and the chord AC intersects the segment OE at a point D which is interior to the semicircle. Find all values of the angle \CAB such that a circle can be inscribed into the quadrilateral OBCD. Solutions by Cyrus Hsia, student, University of Toronto, Toronto, Ontario; and by D.J. Smeenk, Zaltbommel, the Netherlands. We give Smeenk's solution. E q
C
A A A A A A A q
D
A q
q
O q
B q
353 We set OA = OB = 1 and denote \BAC = . Then OD = tan , OB = 1 and BC = 2 sin , CD = 2 cos , cos1 . For a circle to be inscribed in quadrilateral OBCD, we have OB + CD = OD + BC 1 + 2 cos , cos1 = tan + 2 sin cos + 2 cos2 , 1 = sin + 2 sin cos cos + cos 2 = sin + sin 2 2 cos 32 cos 2 = 2 sin 32 cos 2 : Since cos 6= 0, we have
2
tan 32 = 1 3 = 2 4 = 6:
3. Let Q+ denote the set of all positive rational numbers. Find all functions f : Q+ ! Q+ such that
for every x; y 2 Q+; f x +
y = f (x) + f (y) + 2y: x f (x)
Solutions by Cyrus Hsia, student, University of Toronto, Toronto, Ontario; and by Beatriz Margolis, Paris, France. We give the solution by Margolis. We show that the only solution is f (x) = x2 , x 2 Q+. Take x = 1, this gives
y) + 2y: f (1 + y) = f (1) + ff ((1)
(1)
f (y + 1) = f (y) + 1 + 2y:
(2)
Now taking x = y we get
From (1), and (2) we obtain
y) = 0: (f (1) , 1) 1 , ff ((1)
Now f can not be a constant function, so we have that
f (1) = 1:
(3)
354 By induction, from (2) and (3) we get
k 2 N+:
f (k) = k2
Therefore, if n, k 2 N+, using (4), we obtain that
f k + nk = f (k) + ff ((nk)) + 2n n 2 2 n 2 = k + k2 + 2n = k + k :
(4)
(5)
Let x = mk 2 Q+, and N 2 N+. By (2), we have
f (x + N ) , f (x) = =
NX ,1
[f (x + j + 1) , f (x + j )]
j =0 NX ,1 j =0
[1 + 2(x + j )]
= N (2x + 1) + (N , 1)N = (x + N )2 , x2 : Therefore f (x + N ) , (x + N )2 = f (x) , x2 , N 2 N+, x 2 Q+. In particular, using (5), we obtain that
n n 2 n n 2 0=f k +k , k +k = f k , k = f (x) , x2; where x = nk , with n, k 2 N+. Thus f (x) = x2 for x 2 Q+. That completes the solutions we have on le and the Olympiad Corner for this issue. Send me your nice problem sets and solutions!
355
THE ACADEMY CORNER No. 7 Bruce Shawyer
All communications about this column should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7 In the May 1996 issue of CRUX [1996: 165], we printed the questions of the 1996 Memorial University Undergraduate Mathematics Competition, and invited readers to send in solutions. Here is a complete solution set.
MEMORIAL UNIVERSITY OF NEWFOUNDLAND UNDERGRADUATE MATHEMATICS COMPETITION March, 1996
Solutions
n2 + 3n + 1
1. Prove that if n is a positive integer, then 2 n + 4n + 3 is an irreducible fraction. Solution by Panos E. Tsaoussoglou, Athens, Greece [Shortened by the editors.] Assume that the fraction is reducible. Then there is a positive integer k such that kj(n2 + 3n + 1) and kj(n2 + 4n + 3). Then kj(n2 + 3n + 1) , (n2 + 4n + 3) or kjn + 2, so that there is a positive integer l such that n + 2 = kl.
n2 + 3n + 1 = nl + n + 1 . The latter term is an integer only if k k kjn + 1. Since kjn + 2, this implies that k = 1. Now,
This is a contradiction, proving the result. 2. A jar contains 7 blue balls, 9 red balls and 10 white balls. Balls are drawn at random one by one from the jar until either four balls of the same colour or at least two of each colour have been drawn. What is the largest number of balls that one may have to draw? Solution by Panos E. Tsaoussoglou, Athens, Greece. Assume that three balls of one colour are drawn, then three balls of an other colour are drawn, and then one ball of the third colour. The next ball (of any colour) satis es the conditions of the problem. The answer is 8 balls.
356 3. Find all functions u(x) satisfying u(x) = x +
Z
1 2 0
u(t) dt.
\Ocial Solution" from Maurice Oleson. Z 12 Since u(t) dt is a constant, we have that u(x) = x + c for some 0 constant c. Z 12 Then (t + c) dt = c, which gives c = 41 . 0
Thus u(x) = x + 14 .
1 p
p
1
4. Show that 5 + 2 3 , 5 , 2 3 is a rational number and nd its value. Solution by Panos E. Tsaoussoglou, Athens, Greece. p 1 p 1 Let k = 5 + 2 3 , 5 , 2 3 . Then
k3 =
p
p p p 13 5,2 ,3 5+2 5,2 1 1 p5 + 2 3 , p5 , 2 3 ;
5+2 ,
or
k3 = 4 , 3k or (k3 , 1) + 3(k , 1) = 0: Thus (k , 1)(k2 + 4k + 4) = 0, so that k = 1 (since k > 0). 5. In a quadrilateral ABCD (vertices named in clockwise order), AC and BD intersect in X . You are given that AB k DC , that AB is twice as long as CX and that AC is equal in length to DC . Show that AB and CD are equal in length (and hence ABCD is a parallelogram). Solution by Panos E. Tsaoussoglou, Athens, Greece.
AZ
ZZ Z Z Z Z Z Z
B
X
D C We have AB = 2CX and AC = DC . Triangles 4AXB and 4DXC are similar, so that AX = AB CX DC
357 so that Thus, we have
AX + CX = AB + DC : CX DC
2DC = AB 2DC 2 = 0 = = so that AB = DC .
AB + DC DC AB2 + AB DC DC 2 , AB2 + DC 2 , AB DC (DC , AB )(AB + 2CD);
6. Prove that among any thirteen distinct real numbers it is possible to x , y < 2 , p3. choose two, x and y , such that 0 <
1 + xy
\Ocial Solution" from Maurice Oleson. x,y tan , tan
We recognize the similarity of 1 + xy with tan(, ) = 1 + tan tan . Since twelve fteen degree angles make a straight angle, the pigeonx , y < tan(15). hole principle gives that 0 < 1 + xy We note that
sin2 (15 ) tan2 (15) = cos 2 (15 ) cos(30 ) = 11 , + cos(30 ) p
3 = 1 , p23 1+ 2 p3 2 , = p 2 + p3 = (2 , 3)2 ;
and the result follows. 7. A coastguard boat is hunting a bootlegger in a fog. The fog rises disclosing the bootlegger 4 miles distant and immediately descends. The speed of the boat is 3 times that of the bootlegger, and it is known that the latter will immediately depart at full speed on a straight course of unknown direction. What course should the boat take in order to overtake the bootlegger?
358 \Ocial Solution" from Maurice Oleson. We look at two diagrams:
V
A q
dr &
ds
d & 
B q

A q
B q
The distance from A to B is 1 mile. dr From ds2 = dr2 + (r d)2 , ds dt = 3v and dt = v , we get
2 9v 2 = v 2 + r2 d dt :
Thus
p8v = r d = r d dr = r d v; dt dr dt dr dr 1 yielding = p d. Hence ln(r) = p + c. r 8 8 p Since = 0 when r = 1, we have c = 0, giving r = e= 8.
This is the spiral path that the coastguard must follow if she wishes to intercept the bootlegger.
PHOTO PROBLEM
Can you identify this regular CRUX contributor?
359
BOOK REVIEWS Edited by ANDY LIU The Universe in a Handkerchief by Martin Gardner, published by Copernicus, an imprint of SpringerVerlag New York Inc., 1994, ISBN 038794673X, x+158 pages, US$19.00. Reviewed by Richard Guy, University of Calgary. A book by Martin Gardner is always welcome, and most mathematicians and many other people are interested in Lewis Carroll. Although it is almost a century since he died, there is still much to learn about him. Morton Cohen's biography [2] only appeared last year and Edward Wakeling's edition of Lewis Carroll's diaries [4] is still in process of appearing. The present book contains a bibliography of more than thirty items: and there is little overlap with the nearly fty items which relate to Carroll in John Fisher's book [3]. Many mathematicians are interested in word play of all sorts, and this makes up the major part of the book: Carroll's `Doublets' or `Word Links' are still a popular pastime: can you get from `Bread' to `Toast' with less than 21 links? But if you're looking for speci cally mathematical items, then they'll need picking out for you. The gravityoperated train from Sylvie and Bruno concluded. Carroll's attachment to, if not obsession with, the number 42. The Butcher's piece of algebra in Fit 5 of The Hunting of the Snark. The quadratic equation in the poem in Rhyme? and Reason? The puzzle of the monkey and weight over a pulley. John Conway's \Doomsday Rule" for nding what day of the week any date falls on has its original inspiration from Lewis Carroll. Rhymes for remembering logarithms to seven decimal places. The Telegraph Cipher. The game of Arithmetical Croquet. Conjuring tricks depending on `casting out the nines'. Probability paradoxes which anticipate the `motor car and goats' debate of recent years. And mathematical recreations of all the traditional kinds, such as are listed in Rouse Ball [1], the rst three editions of which appeared in Carroll's lifetime. It was probably in the fourth edition that Rouse Ball added a footnote to his geometrical fallacies, saying that `they particularly interested Mr. C. L. Dodgson; see the Lewis Carroll Picture Book, London, 1899, pp. 264, 266, where they appear in the form in which I originally gave them.' The following problem, from a letter by Carroll to Enid Stevens, was thought by Morton Cohen to be too ambiguously stated to have a precise answer: Three men, A, B and C , are to run a race of a quarterofamile. Whenever A runs against B , he loses 10 yards in every hundred; whenever B runs against C , he gains 10 yards in every hundred. How should they be handicapped?
360 Isn't the following a reasonable interpretation? For distances run in equal times, A : B = 90 : 100 and B : C = 110 : 100, so that
A : B : C = 99 : 110 : 100 = 396 : 440 : 400; and B should start from scratch, with C at the 40 yard mark and A at 44 yards.
Something I learned while reviewing this book, though not while reading it, and which Canadians might be particularly interested in, is that Carroll, with such de nitions as `OBTUSE ANGER is that which is greater than Right Anger', anticipated the ideas in Stephen Leacock's Boarding House Geometry. And where does the title come from? In Sylvie and Bruno concluded Mein Herr gives instructions, not in fact practicable in three dimensions, to Lady Muriel for sewing three handkerchiefs into the Purse of Fortunatus, which is a projective plane. \Whatever is inside that Purse, is outside it; and whatever is outside it, is inside it. So you have all the wealth of the world in that leetle Purse!"
References [1] W. W. Rouse Ball & H. S. M. Coxeter, Mathematical Recreations & Essays, 12th edition, Univ. of Toronto Press, 1974. [2] Morton Cohen, Lewis Carroll: A Biography, Knopf, New York, 1995. [3] John Fisher (ed.), The Magic of Lewis Carroll, Penguin Books, Harmondsworth, 1975. [4] Edward Wakeling (ed.), Lewis Carroll's Diaries, Vols. 1, 2 & 3., Lewis Carroll Society, Luton, England, 1993, 1994, 1995.
361
PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be typewritten or neatly handwritten, and should be mailed to the EditorinChief, to arrive no later that 1 June 1997. They may also be sent by email to cruxeditor@cms.math.ca. (It would be appreciated if email proposals and solutions were written in LATEX). Graphics les should be in epic format, or encapsulated postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication.
2189. Proposed by Toshio Seimiya, Kawasaki, Japan.
The incircle of a triangle ABC touches BC at D. Let P and Q be variable points on sides AB and AC respectively such that PQ is tangent to the incircle. Prove that the area of triangle DPQ is a constant multiple of BP CQ. 2190. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Determine the range of
sin2 A + sin2 B + sin2 C
A
B
C
where A; B; C are the angles of a triangle. 2191. Proposed by Sefket Arslanagic, Berlin, Germany. Find all positive integers n, that satisfy the inequality
1 < sin < p1 : 3 n 2 2
362
2192. Proposed by Theodore Chronis, student, Aristotle University of Thessaloniki, Greece. Let fan g be a sequence de ned as follows: an+1 + an,1 = aa2 an; 1
an
a2
Show that if
2, then
n. a1 a1
n 1:
2193. Proposed by Luis V. Dieulefait, IMPA, Rio de Janeiro, Brazil.
(a) Prove that every positive integer is the dierence of two relatively prime composite positive integers. (b) Prove that there exists a positive integer n0 such that every positive integer greater than n0 is the sum of two relatively prime composite positive integers. 2194. Proposed by Christopher J. Bradley, Clifton College, Bristol, UK. Prove or disprove that it is possible to nd a triangle ABC and a transversal NML with N lying between A and B, M lying betweenA and C , and L lying on BC produced, such that BC , CA, AB , NB , MC , NM , ML, and CL are all of integer length, and NMCB is a cyclic inscriptable quadrilateral. 2195. Proposed by Bill Sands, University of Calgary, Calgary, Alberta. A barrel contains 2n balls, numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw. What is the probability that the threeelement sequence obtained has the properties that the smallest element is odd and that only the smallest element, if any, is repeated? For example, the sequences 453 and 383 are acceptable, while the sequences 327 and 388 are not. (NOTE: this problem was suggested by a nal exam that I marked recently.) 2196. Proposed by JuanBosco Romero Marquez, Universidad de Valladolid, Valladolid, Spain. Find all solutions of the diophantine equation with x > 0, y > 0.
2(x + y ) + xy = x2 + y 2;
363
2197. Proposed by Joaqun Gomez Rey, IES Luis Bu~nuel, Alcorcon, Madrid, Spain. Let n be a positive integer. Evaluate the sum: 1 X
,2k k
: 2k+1 k=n (k + 1)2
2198. Proposed by Vedula N. Murty, Andhra University, Visakhapatnam, India. Prove that, if a, b, c are the lengths of the sides of a triangle, 2 1 2 1 2 1 2 2 (b , c) bc , a2 + (c , a) ca , b2 + (a , b) ab , c2 0; with equality if and only if a = b = c. 2
2199. Proposed by David Doster, Choate Rosemary Hall, Wallingford, Connecticut, USA. Find the maximum value of c for which (x + y + z )2 > cxz for all 0 x < y < z. 2200. Proposed by Jeremy T. Bradley, Bristol, UK and Christopher J. Bradley, Clifton College, Bristol, UK. Find distinct positive integers a, b, c, d, w, x, y , z , such that z2 , y2 = x2 , c2 = w2 , b2 = d2 , a2
and
c2 , a2 = y2 , w2:
Bonus Problem for 1996 220A?. Proposed by Ji Chen, Ningbo University, China. Let P be a point in the interior of the triangle ABC , and let 1 = \PAB, 1 = \PBC , 1 = \PCA. p Prove or disprove that 3 =6. 1 1 1
364
SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems.
2030. [1995: 91, 1996: 132] Proposed by Jan Ciach, Ostrowiec etokrzyski, Poland. Swi For which complex numbers s does the polynomial z 3 , sz 2 + sz , 1 possess exactly three distinct zeros having modulus 1? Editor's comment. Readers who found #2030 interesting might want to look at \What Is the Shape of a Triangle" by Dana N. Mackenzie [NOTE DI MATEMATICA 13:2 (1993) 237250; MATH REVIEWS 96c: 51029]. The author notes that an arbitrary triangle ABC is similar to a triangle whose vertices are represented by three complex numbers zi with jzi j = 1 and z1 z2 z3 = 1. He de nes the SHAPE of the triangle to be = z1 + z2 + z3 . It follows that the shape appears as the coecient of z 2 in the cubic 3 z , ()z2 + ()z , 1, whose zeros all lie on the unit circle. His solution to the question, \what are the possible values of ?" is essentially the solution to CRUX 2030 (appearing more than two years before it appeared in CRUX). The author goes on to discuss properties of his \shape invariant"; quite surprisingly, during that discussion the Morley triangle of triangle ABC plays a key role. 2076. [1995: 278] Proposed by John Magill, Brighton, England. 4 AC C 24 This is a multiplicative magic square, where the product of each row, column and diagonal has the same value, ABCD. Each letter represents a digit, the same digit wherever it appears, and each cell contains an integer. Complete the square by entering the correct numbers in each of the nine cells. Solution by Toby Gee, student, the John of Gaunt School, Trowbridge, England. Letting AC = 10A + C and letting x be the value in the bottom left hand corner we have 24Cx = 4(10A + C )x, implying C = 2A. If we
365 further let z be the value of the upper left hand corner and y be that of the middle top row box, we also have 4yz = 24(10A + C )z = 24 12Az , which yields y = 72A. The product of the entries in the middle column is then 72A 12A 2A = 1728A3, which must be a four digit integer, whence A = 1 and we can complete the square as below:
6 72 4 8 12 18 36 2 24 Berlin, Germany; CARL BOSLEY, Also solved by SEFKET ARSLANAGIC, student, Washburn Rural High School, Topeka, Kansas, USA; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; ROBERT GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; POLLY KONTOPOULOU, student, Aristotle University of Thessaloniki, Greece; KEEWAI LAU, Hong Kong; J.A. MCCALLUM, Medicine Hat, Alberta; WOLFGANG GMEINER, Millstatt, Austria; P. PENNING, Delft, the Netherlands; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; HRISTOS SARAGHIOTES, student, Aristotle University, Thessaloniki, Greece; HEINZJURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia; DAVID TASCIONE, student, St. Bonaventure University, New York; PANOS E. TSAOUSSOGLOU, Athens, Greece; CHRIS WILDHAGEN, Rotterdam, the Netherlands; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer.
2087. [1995: 307] Proposed by Toby Gee, student, the John of Gaunt School, Trowbridge, England. Find all polynomials f such that f (p) is a power of two for every prime p. Solution by Luis V. Dieulefait, IMPA, Rio de Janeiro, Brazil. The only solutions are constant polynomials. Suppose instead that a polynomial f with deg(f ) > 0 were a solution. Since f (x) takes positive values for every prime value of x, it is clear that limx!1 f (x) = 1. Also there is a real number x0 such that f (x) is increasing for x > x0 . If we enumerate the sequence of primes: p1 , p2, p3 , : : : in increasing order, then f (pi+1) > f (pi) for pi > x0: () Lemma. Given a real number r > 1, every suciently large pi satis es
r pi pi+1.
366 Proof. Let us suppose, on the contrary, that there exist in nitely many primes pj satisfying pj +1 > r pj . Then (r pj ) , (pj ) = 0 where (x) is the number of primes less than or equal to x. This means that ((rpp)) = 1 for in nitely many primes. Applying the prime number theorem, we see that j
j
(r pi) = lim r ln(pi) = r > 1: lim i!1 ln(r p ) i!1 (p ) i
i
This contradiction proves the lemma. Let d = deg(f ) > 0. Choose r > 1 such that rd < 2. From (*) above we see that f (pi+1 2 f (pi) for every i large enough. Applying the lemma we have f (r pi) > 2 f (pi) for every i large enough (because f is increasing for such i). This implies limi!1 ff(r(pp)) 2. But f is a polynomial of degree d, so we know that the limit above equals rd and rd < 2. This contradiction proves that no polynomial f with deg(f ) > 0 can solve our problem. Then the only solutions are constant, and obviously we must have f (x) = 2n, for n a nonnegative integer. ~ and EMILIO FERNANDEZ Also solved by MANUAL BENITO MUNOZ MORAL, I.B. Sagasta, Logro~no, Spain; RICHARD I. HESS, Rancho Palos Verdes, California, USA; KEEWAI LAU, Hong Kong; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer. i
i
2089. [1995: 307] Proposed by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain. Let ABCD be a trapezoid with AB k CD, and let X be a point on segment AB . Put P = CB \ AD, Y = CD \ PX , R = AY \ BD and T = PR \ AB. Prove that 1 = 1 + 1 : AT AX AB
Solution by Waldemar Pompe, student, University of Warsaw, Poland. Using Menelaus's Theorem on the triangle ABD, we get
AT BR DP TB RD PA = 1: Since BR : RD = AB : DY and DP : PA = DY : AX , we obtain TB = AB DP = AB DY = AB ; AT DY PA DY AX AX whence
AB = TB + 1 = AB + 1 or 1 = 1 + 1 ; AT AT AX AT AX AB
367 as we wished. Berlin, Germany; MANUEL Also solved by SEFKET ARSLANAGIC, ~ BENITO MUNOZ and EMILIO FERNANDEZ MORAL, I.B. Sagasta, Logro~no, Spain; CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; TIM CROSS, King Edward's School, Birmingham, England; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut,USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEEWAI LAU, Hong Kong; P. PENNING, Delft, the Netherlands; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.
2091. [1995: 343] Proposed by Toshio Seimiya, Kawasaki, Japan.
Four points A, B , C , D are on a line in this order. We put AB = a, BC = b, CD = c. Equilateral triangles ABP , BCQ and CDR are constructed on the same side of the line. Suppose that \PQR = 120 . Find the relation between a, b and c.
I. Solution by Jordi Dou, Barcelona, Spain. If the middle triangle is largest then b is twice the arithmetic mean of a and c; if smallest, b is half the harmonic mean of a and c. Proof: We suppose that BCQ is xed. Let S be the image of Q under re ection in BC . P and R will lie on the prolongations of SB and SC , which we will call p and r respectively. As in the gure, K , L are the points where the line through Q parallel to BC intersects p, r. P0 is the foot of the perpendicular to p from Q, and R00 is the point where that line meets