TITD ANDREESCD
DORIN ANDRICA
360 Problems for Mathematical Contests
TITU ANDREESCU
DORIN ANDRICA
360 Problems for Mathematical Contests
GIL Publishing House
© GIL Publishing House
ISBN 9739417124
360 Problems for Mathematical Contests Authors: Titu Andreescu, Dorin Andrica Copyright © 2003 by Gil. All rights reserved. National Library of Romania CIP Description
ANDREESCU, TITU 360 Problems for Mathematical Contests/ Titu Andreescu,
Dorin Andrica.  Zalau: Gil, 2003
Contents FOREWORD 3 FROM THE AUTHORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Chapter 1 . ALGEBRA 7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Solutions . . . . . 17 Chapter 2. NUMBER THEORY . . . . . . 47 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Solutions . . . . .. .. .. 57 Chapter 3. GEOMETRY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Chapter 4. TRIGONOMETRY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Problems . . . . . . . . . . . . . . . . . . . . : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Solutions . .. 147 Chapter 5. MATHEMATICAL ANALYSIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 Chapter 6. COMPREHENSIVE PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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B iblio gr. ISBN 9739417124
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1. Andrica, Dorin 51(075.35)(076)
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GIL Publishing House
P.o. Box 44, Post Office 3, 4700, Zalau, Romania, tel. (+40) 260/616314 fax. (+40) 260/616414 email: gi11993@zalau.astral.ro
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IMPRIMERIA
ARTA IV GRAFICA I ��LIBRIS �
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FOREWORD I take great pleasure in recommending to all readers  Romanians or from abroad  the book of professors Titu Andreescu and Dorin Andrica. This book is the fruit of a prodigious activity of the two authors, wellknown creators of mathematics questions for Olympiads and other mathematical contests. They have published innumerable original problems in various mathematical journals. The book is organized in six chapters: algebra, number theory, geometry, trigonometry, analysis and comprehensive problems. In addition, other fields of math ematics found their place in this book, for example, combinatorial problems can be found in the last chapter, and problems involving complex numbers are included in the trigonometry section. Moreover, in all chapters of this book the serious reader can find numerous challenging inequality problems. All featured problems are interesting, with an increased level of difficulty; some of them are real gems that will give great satisfaction to any math lover attempting to solve or even extend them. Through their outstanding work as jury members of the National Mathematical Olympiad, the Balkan Mathematics Contest (BMO) , and the International Math ematical Olympiad (IMO) , the authors also supported the excellent results of the Romanian contestants in these competitions. A great effort was given in preparing lectures for summer and winter training camps and also for creating original problems to be used in selection tests to search for truly gifted mathematics students. To support the claim that the Romanian students selected to represent the country were really the ones to deserve such honor, we note that only two mathematicians of Romanian origin, both former IMO goldmedalists, were invited recently to give conferences at the International Mathematical Congress: Dan Voiculescu (Zurich, 1994) and Daniel Tataru (Beijing, 2002). The Romanian mathematical community unanimously recog nized this outstanding activity of professors Titu Andreescu and Dorin Andrica. As a consequence, Titu Andreescu, at that time professor at Loga Academy in Timi§oara and having students on the team participating in the IMO, was appointed to serve as deputy leader of the national team. Nowadays, Titu's potential, as with other Ro manians in different fields, has been fully realized in the United States, leading the USA team in the IMO, coordinating the training and selection of team contestants and serving as member of several national and regional mathematical contest juries.
One more time, I strongly express my belief that the 360 mathematics problems featured in this book will reveal the beauty of mathematics to all students and it will be a guide to their teachers and professors. Professor loan Tomescu Department of Mathematics and Computer Science University of Bucharest Associate member of the Romanian Academy
FROM THE AUTHORS This book is intended to help students preparing for all rounds of Mathematical Olympiads or any other significant mathematics contest. Teachers will also find this work useful in training young talented students. Our experience as contestants was a great asset in preparing this book. To this we added our vast personal experience from the other side of the " barricade" , as creators of problems and members of numerous contest committees. All the featured problems are supposed to be original. They are the fruit of our collaboration for the last 30 years with several elementary mathematics journals from all over the world. Many of these problems were used in contests throughout these years, from the first round to the international level. It is possible that some problems are already known, but this is not critical. The important thing is that an educated  to a certain extent  reader will find in this book problems that bring something new and will teach new ways of dealing with key mathematics concepts, a variety of methods, tactics, and strategies. The problems are divided in chapters, although this division is not firm, for some of the problems require background in several fields of mathematics. Besides the traditional fields: algebra, geometry, trigonometry and analysis, we devoted an entire chapter to number theory, because many contest problems require knowledge in this field. The comprehensive problems in the last chapter are also intended to help under足 graduate students participating in mathematics contests hone their problem solving skills. Students and teachers can find here ideas and questions that can be interesting topics for mathematics circles. Due to the difficulty level of the problems contained in this book, we deemed it appropriate to give a very clear and complete presentation of all solutions. In many cases, alternative solutions are provided. As a piece of advice to all readers, we suggest that they try to find their own solutions to the problems before reading the given ones. Many problems can be solved in multiple ways and pertain to interesting extensions.
4
This edition is significantly different from the 2002 Romanian edition. It features more recent problems, enhanced solutions, along with references for all published problems. We wish to extend our gratitude to everyone who influenced in one way or another the final version of this book. We will gladly receive any observation from the readers. The authors
Chapter 1 ALGEBRA
6
PROBLEMS
C2 , ,cn }.
be a set of n characters { Cl ' We call word a string of at most m characters, m ::; n, that does not start nor end with Cl . How many words can be formed with the characters of the set C? 1. Let
C
. . •
2. The numbers 1, 2, , 5n are divided into two disjoint sets. Prove that these sets contain at least n pairs such that the number is also an element of the set which contains the pair. . . .
(x, y), x > y,
xy
al, a2 , . . . ,an
3. Let be distinct numbers from the interval permutation of {I, 2, . . . , n } . Define the function f : t as follows:
[a, b] [a, b] x �i' i f(x) { xa i ifotherwIse q(
=
=
)
[a, b] and let
0'
be a
= I, n
Prove that there is a positive integer h such that
fafa... aJ. � htimes
x, where flh]
flh](X)
x, y, z are nonzero real numbers with x y z 0, then xx2 yy2 yy2 Z2 Z2 xx2 xyz3 y3 xyZ3 zx Z Z d Prove that Let a, b, c, d be complex numbers with a b a3 b3 c3 d3 3(abc bcd cda dab). Let a, b, c be nonzero real numbers such that a b c 0 and a3 b3 c3 a5 b5 c5• Prove that 4.
Prove that if
+
+
 + +
+ +

+
+ +

5.
=
+
+
+
+
=
+
+ C + = o. +
=
+ +
+
7. Let a, b, c, d be integers. Prove that a + b + c + d divides
2(a4 b4 c4 +
+
+ at) 
(
a2 b2 c2 d2 )2 8abcd. +
+
=
 +  +.
+
6.
+
+
+
+
+
+
=
1.
10
8. Solve in complex numbers the equation
(x + l)(x + 2)(x + 3) 2 (X + 4)(x + 5) = 360.
9. Solve in real numbers the equation yX + VY
+ 2v'z=2 + v'u + Vv = x + y + z + u + v.
16. Prove that if x, y, z are real numbers such that xS + yS + ZS i= 0, then the
ratio
2xyz  (x + y S+ z) xS + yS + Z equals � if and only if x + y + z = O. 17. Solve in real numbers the equation .vx;:=r
10. Find the real solutions to the equation
(x + y) 2 = (X + 1) (y  1). .
12. Solve the equation
a, b, c
J x + a + Jx + b + J x + c = J x + a + b  c,
are real parameters. Discuss the equation in terms of the values of the where parameters.
13.
a and b be distinct positive real numbers. Find all pairs of positive real (x, y), solutions to the system of equations { xx42  yy42 == axf/a2 byb2 .
Let numbers
14. Solve the equation
[
25x42] = 13x3+ 4'
[a] denotes the integer part of a real number a. 1 +v1s 15. Prove that If. a � 2' then 1+ = n, n = O, 1,2, . . .
where
[ [�]l
1
+ 2J X2  4 + . . . + nJXn  n2 = �(Xl 2 + X2 +
. . .
+
xn).
18. Find the real solutions to the system of equations
11. Solve the equation
x + V4X + V16X + J. . + J4nx + 3  JX = 1.
11
1. 1 . PROBLEMS
ALGEBRA
x1 +y1 = 9 1 + 1 1 + 1 1 + 1  18 (� ?'Y)( �)( ?'Y) 
19. Solve in real numbers the system of equations
y22 + u22 + v22 + w22 = 4x  1 x + u + v + w = 4y  1 x22 + y22 + v22 + w22 = 4u  1 x2 + y2 + u2 + w2 = 4v  1 x + y + u + v = 4w  1 20. Let aI, a2 , as, a4 , a5 be real numbers such that al + a2 + as + a4 + a5 = 0 and max l�i<j� 5 lai  aj l < 1 . Prove that ai + a� + a5 + a� + a� < 10. 21. Let a, b, c be positive real numbers. Prove that 1 1 1 + 2a1 + 2b1 + 2c1 > b+c c+a a + b +22. Let a, b, c be real numbers such that the sum of any two of them is not equal to zero. Prove that _as + bS::5_+ :c35_ .:.,.((aa_ + b +.cc)S).: 5 > 109 (a + b + C)2 

_
a, b, c be real numbers such that abc = 1. Prove that at most two of the 2a  1b ' 2b  c'1 2ca1 are greater than 1. 23.
Let numbers
12
1.
24. Let a, b,
c,
ALGEBRA
1.1.
d be real numbers. Prove that
 d}, d  a2 ) � �. 25. Let aI, a2 , ' . . ,an be numbers in the interval (0, 1) and let k ;::: 2 be an integer. Find the maximum value of the expression n L yt ai (I  ai+ I), i= l where an+1 a . 26. Let m and n be positive integers. Prove that x mn  1 > x n  I  min(a
=
 b2 , b
2  c ,
34. Show that for any positive integer n the number
n ; 1) 22n + n : 1) 22n2 . 3+ . . . + n : 1) 3n C2 C
C
C

x
for any positive real number x.
27. Prove that m! ;::: (n!) [�] for all positive integers m and n. 28. Prove that
n ;::: 2.
1 ++"' 1 +1 > n � I +n +2 1 y'2 v'3 \Iii
29. Prove that
n (1  1/ vn) + 1 > 1 + �2 + �3 + . . . + .!.n > n (\In + 1  1) for any positive integer n. 30. Let aI, a2 , . . . , an (0, 1 ) and let tn al +naa2a+2 . .. .. .a+n an . Prove that n L loga; tn ;::: (n  I ) n . i=l 31. Prove that between n and 3n there is at least a perfect cube for any integer n ;::: 1. 32. Compute the sum n k. Sn L(I) k= l 33. Compute the sums: l a) S n � (k + l) (k + 2) (�); b) T n = � (k + l)(k + 2)(k + 3) (�). E
1
=
=
=
13
is the sum of two consecutive perfect squares.
35. Evaluate the sums:
1
m
for any integer
PROBLEMS
�
Io(k+l) 2
36. Pr.ove that
12 (�) + 32 (�) + 52 (�) + . . . n(n + 1)2n3 for all integers n ;::: 3. 37. Prove that 2Ln [log kJ = (n  2)2n + n + 2 k= l 2 for all positive integers n. 38. Let Xn 22n + 1, n 1,2,3, . . . Prove that 2n1 < 1 Xl1 + X22 + X223 + . . . +Xn 3 for all positive integers n. 39. Let f be a function such that f(z)f(iz) = Z2 for all z Prove that f(z) + f( z) 0 for all z 40. Consider a function f : (0,00) and a real number a > 0 such that f( a) 1. Prove that if f(x)f(y) + f (;) f (�) 2f(xy) for x, y (0, 00), =
=
:
C
=
+
E C.
C
E C
=
+ 1R
=
=
all
E
then f is a constant function.
41. Find with proof if the function f·: 1R + [ 1, 1), f(x) 42. For all i,j IS(i,j)l·
=
=
sin[x) is periodical.
l,n define S(i,j) = kL=n l ki + . Evaluate the determinant � = i
14
43. Let Xij
=
1.
{a
1. 1 .
ALGEBRA
48. Let P(x) be a polynomial of degree n. If
i if i = j 0 if i i j, i + j i 2n + 1 if i + j = 2n + 1
bi where ai , b i are real numbers. Evaluate the determinant � 2 n
evaluate
44. a) Compute the determinant
=
>
=
0, 1 , . . . , n
n.
for all real numbers x. y
y
x
v
z
z
P(m), where m
k for k P(k) k+1
49. Find all polynomials P(x) with integral coefficients such that
= IXij l .
x
15
BROBLEMS
v
z
v
x
y
v
y
50.
Consider the polynomials Pi, i = 1 , 2, . . . , n with degrees at least 1. Prove that if the polynomial
z
x
P(x) PI (xn+l ) + XP2 (xn+l ) + . . . + xn lpn (xn+1 ), i s divisible by xn +x n  l + . . · + x + 1, then all polynomials Pi(x) , i 1, n, are divisible =
abed, bade, cdab, deba are divisible by a prime p, a + b + e + d, a + b   d, a  b + e  d, a  b  e + d, is divisible by p. 45. Consider the quadratic polynomials t1 (x) x2 + PI X + q r and t2 (x) x2 + P2 X + q�, where Pl , P2 , ql , q 2 are real numbers. Prove that if polynomials tl and t2 have zeros of the same nature, then the polynomial b) Prove that if the numbers then at least one of the numbers
e
=
=
51. Let P be a prime number and let
P(x) aoxn + al xn1 . + an be a polynomial with integral coefficients such that an =1= 0 (mod p). Prove that if there are n + 1 integers 0' 1 , 0'2 , , an+ 1 such that P (ar) 0 (mod p) for all 1, 2, . . . , n + 1, then there exist i,j with i i j such that a i aj (mod p). 52. Determine all polynomials P with real coefficients such that pn (x) P(xn ) =
46. Let a, b, e be real numbers with a
>
0 such that the quadratic polynomial
ax2 + bex + b3 + e3  4abe has nonreal zeros. Prove that exactly one of the polynomials Tl (x) ax2 + bx + e and T2 (x) ax 2 + ex + b has only positive values.
+
.
.
• • •
== ==
r =
for all real numbers x, where n
53. Let
has real zeros.
=
by x  1.
>
1 is a given integer.
P(x) aoxn + alxnl + . . . + an , an i 0, =
be a polynomial with complex coefficients such that there is an integer
T(x) =
=
47. Consider the polynomials with complex coefficients
P(x) xn + alxnl + . . . + an and Q(x) xn + bl xn l + . . . + bn having zeros Xl, X2 , . . . ,X n and x ? , x� , . . . , x; respectively. Prove that if al + a3 + a5 + . . . and a2 + a4 + a6 + . . . are real numbers, then bl + b2 + . . . + bn is also a real number. =
=
=
m with
P has at least a zero with the absolute value less than 1. 54. Find all polynomials P of degree n having only real zeros Xl , X2 , . . . , Xn such that n 1 2 tt P(x)  Xi XPnI (X) , Prove that the polynomial
=
for all nonzero real numbers x.
55. Consider the polynomial with real coefficients
P(x) aoxn + alxn l + . . . + an , =
16
1.
an
and
f:. O. Prove that if the equation equation
ALGEBRA
P(x) 0 has all of its roots real and distinct, then the x2 PII (x) 3xP'(x) P(x) 0
has the same property.
=
+
+
=
56. Let R�f and R�) be the sets of polynomials with real coefficients having no multiple zeros and having multiple zeros of order respectively. Prove that if P(x) R�j and P(Q(x)) R�j , then Q'(x) R�]l). 57. Let P(x) be a polynomial with real coefficients of degree at least 2. Prove
SOLUTIONS
n
E
E
that if there is a real number
E
a such that P(a) plI (a) (P'(a)) 2 ,
ai.
aoxn alxn 1
X4  (2 l)x3 m +
+
+ (m
where m is a real parameter.
+ ... +
an
=
0
 l)x2 (2 2 l)x
60. Solve the equation
x2n a1 x2n1 +
if all of its roots are real.
+ ... +
(l) f:. a1 and f(k) f:. a1 For f(l) and f(k) there are  1 possibilities of choosing a character from C2 , . .. ,en and for f(i) , 1 < i < k there are such possibilities. Therefore the number of strings f (l) f ( 2) ... f(k  l) f ( k) is Nk (  1 )2 k 2 It follows that Nl N2 Nm (  1) (  1 ) 2 ( (  1)2 1 1 ) 2 m2 n
with real coefficients Prove that if the equation has all of its roots real, then (n 2:: 2n O . Is the reciprocal true?
59. Solve the equation
+
= n
n
58. Consider the equation
a a2
A
C,
 1. The number that we seek is Nl N2 Nm· {I, 2,...,k}, 1 � k � We need to find out the number of functions
Clearly, Nl
f
P has at least two nonreal zeros.
 l)ar
m.
Ak = m. k + A, k = 2,n with the properties
Let
f:
>
then
1. Let Nk be the number of words having exactly k characters from the set
1�k�
+ ... +
+
m +
a2n_2X2  2 x n
+
= n
+
+m =
0,
+ ... +
= n
+ n
n
O n + n
n + ... + n 
n
=
(Dorin Andrica) 1
=
0,
2.
Suppose, for the sake of contradiction, that there are two sets A and B such that AU B = {I, 2,...,5n}, An B = 0 and the sets contain together less than n pairs > with the desired property. Let k be a given number, k = 1, n. If k and 2k are in the same set A or B the same can be said about the difference 2k  k = k. The same argument is applied for 4k and 2k. Consider the case when k and 4k are elements of A and 2k is an element of B. If 3k is an element of A, then 4k  3k = k E A, so let 3k E B. Now if 5k E A, then 5k  4k = k E A and if 5k E B, then 5k  3k = 2k E B; so among the numbers k, 2k, 3k,4k,5k there is at least a pair with the desired property. Because k = 1 , 2,... , n, it follows that there are at least n pairs with the desired property. Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 75, Problem 3698)
(x, y), x y,

(Dorin Andrica,
3. Note that (1)
17
1.
18
1.2.
ALGEBRA
and furthermore for all integers ml, m2 2:: 1 . Suppose that for all integers k 2:: 1 we have I[k](x) iBecause there are n! permutations, it follows that for k such that positive integers nl >
x.
n2
ai, in=2
Let h = n l since numbers or
(2)
l[n2+h](x)
=
n!
x
E
[a, b],
(3)
= (10 l[n21])(x), X [a, b]. Because I is injective, we obtain l[n2+hl] (x) l[n21] (x), x [a, b] (10 l[n2+hl])(x)
and in the same manner
= I(x), l[h](X) = x, x
l[h+l](X)
or
Alternative solution.
E
E
=
x E
E
[a, b]
[a, b].
Let Sn be the symmetric group of order n and Hn the cyclic subgroup generated by a. It is clear that Hn is a finite group and therefore there is integer h such that a[h) is identical permutation. Notice that if = 1, n l I k ](X) = otherwIse
{ axq[k1(i)
x
�i' i =
= x and the solution is complete. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1978) , pp. 53,
Then I[h](x) Problem 3540)
x + y + z = 0, we obtain x + y z, y + z x, z + x y, or, by squaring and rearranging, x2 + y2 = Z2 2xy, y2 + Z2 = x2 2yz, Z2 + x2 y2  2zx. The given equality is equivalent to x2  2yz y2  2zx x3 y3 Z3 Z2  2xy z= + x + y yz + zx + , xy 4.
Because
=
_

=
=
_
=
=
Z3 (x +y + z) + 2 ( Xyz + yzx + zxY ) = yzx3 + zxy3 + . xy
2(X2 y2 + y2z2 + 2 X2 ) = x4 + y4 + Z4 . On the other side, from x + y + z = 0 we obtain (x + y + Z) 2 = 0 or x2 + y2 + Z2 = 2( xy + yz + zx). Squaring yields X4 + y4 + Z4 + 2(X2y2 + y2z2 + Z2 X2 ) 4(X2 y2 + y2z 2 + Z2X2 ) + 8xyz(x + y + z) Z
(3)
l[n2](x),
and consequently to
The last equality is equivalent to >
0 and observe that for all k the functions I[k] are injective, 1 , n are distinct. From relation we derive that
>
19
SOLUTIONS
=
or as desired.
(Titu Andreescu,
Revista Matematica Timi§oara (RMT) , No. 3(1971), pp. 25, Problem 483; Gazeta Matematica (GMB) , No. 12(1977) , pp. 501, Problem 6090)
a, b,
d are different from zero. Consider the equation x4  (L a) x3 + (L ab) x2  (L abc) x + abed 0 with roots a, b, e, d. Substituting x with a, b, e and d and simplifying by a, b, e, d i 0, 5. We assume that numbers
c,
=
after summing up we obtain
Because
If
L
a=
0, it follows that
L a3 = 3 L abe.
one of the numbers is zero, say
a,
then
b+ e + d = 0, or b e = d. It is left to prove that b3 e3 + d3 = 3bed. Now b3 + e3 + d3 = b3 + e3  (b + e) 3 3be(b + ) = 3bed as desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 12(1979) , pp. 47, +
+
=
Problem 3803)
6. Because
a+b+e
a3 + b3 + e3 3abe =
=
a5 + b5 + e5 5abe( a2 + b2 + e2 + ab + be + ea)
0, we obtain
and
c
=
1.
20 The given relation becomes
3abc = 5abc(a2 + b2 + c2 + ab + bc + ca) or a2 + b2 + e2 + ab + be + = "5'3 since a, b, e are nonzero numbers. follows that "21 [(a + b + e)2 + a2 + b2 + e2 ] ="53 and, using again the relation a + b + e = 0, we obtain a2 + b2 + e2 = �5 ' as desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 59, Problem 3016) It
ca
7. Consider the equation
x4  (L:a) x3 + (L: ab) X2  (L:abc) x + abed = 0, with roots a, b, e, d. Substituting x with a, b, e and d, respectively, we obtain after summation that L:a4 + (L:ab) L:a2 +4abed is divisible by L: a. Taking into account that we deduce that
L: a. Hence 2(a4 + b4 + e4 + d4 )  (a2 + b2 + e2 + d2 )2 + 8abed is divisible by a + b + e + d, as desired. (Dorin Andrica) is divisible by
8. The equation is equivalent to
1.2 . 21 or y3 + 22y2 + 157y = 0, with solutions Yl = 0, Y2 = 11 + 6i, Y3 = 11  6i. Turning back to the substitution, we obtain a first equation, x2 + 6x = 0, with solutions Xl = 0, X2 = 6. The equation x2 + 6x 11 + 6i is equivalent to (x + 3)2 = 2 + 6i. Setting x + 3 = u + iv, u, v we obtain the system { U22uv=v26 = 2 It follows that (u2 + V2)2 = (u2  V2)2 + (2UV)2 = 4 + 36 = 40. Therefore { U2u2 + v2v2 == 2y'lO 2 and u2 = y'lO  1, v2 = vTO + 1, yielding the solutions X3, 4 = 3 ± V..flO  1 ± iV..flO + 1 where the signs + and  correspond. The equation x2 + 6x = 11  6i can be solved in a similar way and it has the solutions X5 = 3 + VvTO  1  iVvTO + 1, X6 = 3  V.J[O  1 + iVvTO + 1. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 3(1972), pp. 26, Problem 1255) SOLUTIONS
ALGEBRA
(x2 + 6x + 5)(x2 + 6x + 8)(x2 + 6x + 9) = 360. Setting x2 + 6x = y yields (y + 5)(y + 8)(y + 9) = 360,
=
E lR,
9. The equation is equivalent to
X  ..;x + y
+  2vz=2 + u  ..jU + v  ..;v = 0, or (VX D 2 + (VV  D 2 + (vz=2  1)' + + (VU  D 2 + (vv  D 2 = 0 Because x, y, U, v are real numbers, it follows that 1 ..;x = VY = ..jU = ..;v = "2 and vz=2 = 1. Hence X = Y = u = v = 4'1 = 3. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1974), pp. 47, Problem 2002; Gazeta Matematica ( GMB ) , No. 10(1974), pp. 560, Problem 14536) 
VY
z
Z,
Z
22
1.2.
1. ALGEBRA
10. Setting X = x + 1 and Y = y  1 yields
(X + y)2 = XY or � [X2 + y2 + (X + y) 2] = o. Hence X = Y = 0, so the solution is x 1 and y = 1. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1977), pp. 40, Problem 2811) =
11. The equation is equivalent to
� x + V4X + V16X + l·· + v'4nx + 3 = v'Z + 1
Summing up relations
SOLUTIONS
(1) and (2) we obtain Jx + b Jx + a + b  c,
23
=
a = c. b = c, x = b. b a c, b c a = c, x = a 2(1978), pp. 26, (Titu Andreescu, 3017) 13. Because a and b are distinct numbers, x and y are distinct as well. The second
and then To conclude, we have found that (i ) If then the equation has the solution ( ii ) If f:. c and f:. there is no solution. ( iii ) If f:. and then is the only solution. Revista Matematica Timi§oara (RMT ) , No. Problem
equation could be written as
Squaring the equation yields
V4x + V16x + l·· + v'4nx + 3 = 2v'Z +
a b. We have a22 b22 = b22y22 + 2b2Y (X24__ y24 )3 + (X4 _ y4)2 a x = b x + X (X y ) .
1
and the system could be solved in terms of and
Squaring again implies
V16X + l·· + v'4nx + 3 = 4v'Z + 1
Subtracting the first equation from the second yields
Continuing this procedure yields
4nx + 3 = 4nx + 2 · 2nVX + 1 1 and 2 . 2n Vx 2 . Hence x = n ' 4 (Titu Andreescu, Revista Matematica Timi§Oara (RMT) , No. 45(1972), pp. 43, Problem 1385) =
12. We distinguish two cases:
1) b = c. The equation becomes Jx + a+ Jx + b = Jx + a, so x:= b. 2) b f:. c. The equation is equivalent to Jx + b + Jx + c = Jx + a + b  c  Jx + a, or bc bc Jx x + a + b  c + Jx + a ' + c Jx + b J so Jx + b  Jx + c = Jx + a + b  c+ Jx + a.
(1)
which reduces to
b b = y3 + 3x2 y a x3 + 3xy2 ) 3 3 2 2 b y  x y a = x  xy ). a = x3 +3xy2 a x(x2 _y22 ) 3b y(y2 _ X2) 3 3 b 3x y + y . a + b = (x + y) a  b = (x  y) . x + y = Va + b x  y = Va  b and its unique solution is x = (Va + b + Va  b)/2, y = (Va + b  Va  b)/2. (Titu Andreescu, Korean Mathematics Competitions, 2001) 13x+4 = y, Y It follows that 14. Let 3 3y  4 x u
Solving the quadratic equation in yields ( and = ( and or = The second alternative is not possible because = and = cannot be both positive. It follows that The system now and = Hence and becomes
E Z.
(2)
=
24
[ �(3Y : 4)  2] = y,
and the equation is equivalent to
or
[75Y 52 126] = Y Using that for any real number a, [a] :s; a < [a] + 1, we obtain y:S; 75y 52 126 < y + 1 , 126 178 or 126 :s; 23y < 178, so 23 :s; y < 23· Note that y Z, therefore y = 6 or y = 7, thus Xl = 1314 and X2 = 1317 ' are the desired solutions. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 3(1972), pp. 25, Problem 1552) 1 15. From a > 1 +V5 2 we obtam. a2  a  I > 0, or a > a + 1. We have 1 +na2 ] = 1 + na  a, O:S;a < 1. [a a E

Hence
That is because and
[ 1 + [�] l = [ l + �:na  a]
= [(1 + �  aH + n] = n, n � O. (1 + a1  a) a1 :s; (a  a)a1 = 1  aa < 1 (1 + a1 ) a1 > a12 > 0. 
 Q
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 45, 3479) 16. First we consider the case when X + y + z = 0. Then x3 + y3 + Z3 = 3xyz and
Problem
the ratio equals
�, as desired.
1 .2 . 2xyz3  (x3+ y +3 z) = ,2 x +y +Z 3
25
SOLUTIONS
1. ALGEBRA
Conversely, if
:��':
then and so
2(x3 + y3 + Z3  3xyz) + 3(x + y + z) = 0. U sing the formula x3 + y3 + Z3  3xyz = (x + y + z)(x2 + y2 + Z2  xy  yz  zx), we obtain by factorization that
(x + y + z)[2(x2 + y2 + Z2  xy  yz  zx) + 3] = and so (x + y + z)[(x  y)2 + (y  Z)2 + (z  X) 2 + 3] = 0. Because the second factor is positive, it follows that x + y + z = 0, as desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1973), pp. 30, Problem 1513) °
17. We write the equation as
Xl  2VXI  1 + X2  2 · 2JX2  22 + . . . + Xn  2nJxn  n2 = 0, or (VXI  1  1)2 + (.JX2  22  2) 2 + . . . + ( Jxn  n2  n) 2 = 0 . Because the numbers Xi , i = 1, n are real, it follows that Xl = 2, X2 = 2 . 22 , . . . ,xn = 2n2 (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1977), pp. 14, Problem 2243) 18. Using the identity
(a + b + C) 3 = a3 + b3 + c3 + 3(a + b)(b + c)(c + a) we obtain (� + � + l) " = � + � + 1 + 3 (� + �) (1 + �) (1 + �) = = 9 + 1 + 54 = 64 Hence
1.2.
1. and so 1 +1 = 3. fIX * The system is now reduced to .!.+.!. x y =9 1 + 1 = 3, fIX *
26
ALGEBRA
and consequently
{
which is a symmetric system, having the solution
x = ,81 y = 1
and
x = 1, Y = .81
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 45(1972), pp. 43, 1386)
Problem
19. By summing up the equations of the system we obtain It follows that
(4x2  4x + 1) + (4y2  4y + 1) + (4u2  4u + 1) + +(4v2  4v + 1) + (4w2  4w + 1) = O.
5 t a�  (t al) " � 50 5 5 Note that I: ai = 0 and so I: ar ::; 10, as claimed. i=l Romanian Mathematica i= l Olympiad  second round 1979; Revista (Titu Andreescu, Matematica Timi§oara (RMT) , No. 12(1980), pp. 61, Problem 4094) 21. The inequality (a + b)2 ;::: 4ab yields 1 1 4 a+b a b  1 1 1 1 4 4 and , sI. mI. larly'  +   '  + b e b+e e a e+a Summing up these inequalities yields 1 1 1 1 1 1 2a + 2b + 2e ;::: a + b + b + e + e + as desired. (Dorin Andrica, Gazeta Matematica (GMB), No. 8(1977), Problem 5966) + >

x,y,u,v,w are real numbers, we obtain x = y = u = v = w = 21 (Dorin Andrica, Gazeta Matematica (GMB), No. 8(1977), pp. 321, Problem 16782)
 >  .
a
a5 + b5 + e5 = (a + b + e) 5  5(a + b)(b + e)(e +
l
ai  aj l ::; lai  ai+l l + . . . + lajl  j l ::;
2 2 i<j�5 (i j) = i1� <j�5 (ai  aj ) ::; 1�I: = (12 + 22 + 32 + 42 ) + (12 + 22 + 32 ) + (12 + 22 ) + 12 = 50 5 a�  2 5 a·a· < 50 4� i= l � ' 3 �
l
� i,i=l i'i'i
) ( a2
+ b2 + e2 + ab + be + 00)
we obtain
a53 + b35 + e5  (a + b + e)35 = 5 (a + b + e + ab + be + 00) a + b +  (a+ b + e) 3 It suffices now to prove that 35 (a2 + b2 + e2 + ab + be + 00) ;::: g10 ( + b + e)2 3(a2 + b2 + e2 + ab + be + 00) ;::: 2(a2 + b2 + e2 + 2ab + 2be + 200) . 2
c3
a
I:
a
'
and
20. From the triangle inequality we deduce
So
>
22. Using the identities
Due to the fact that
Hence
27
SOLUTIONS
2
2
a
or
The last inequality is equivalent to
a2 + b2 + e2 ;::: ab + be + 00,
which is clearly true.
(Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 1(1981), pp. 49, 4295; Gazeta Matematica (GMB) , No. 6(1980), pp. 280, Problem 0148; 11(1982), pp. 422, Problem 19450)
Problem No.
28
1.
1.2.
ALGEBRA
23. Assume by contradiction that all numbers 2a  �b , 2b  �c , 2c  �a are greater
than
1. Then
(2a  D (2b  D (20  D > 1 and (2a  D (2b  D (20  D > 3. From the relation (1) and using abc = 1 we obtain 3 > 2(a b c)  ( � � � ) a b c +
+
1 ( 2)
( )
+
+
+
+
and so
2

+
which is a contradiction. The proof is complete. Revista Matematica Timi§oara (RMT) , No. ( Problem )
2 1986), pp. 72,
24. Assume by contradiction that all numbers are greater than 1/4. Then
d  a2 > 41 41 41 41 hence 0> ar Gbr Gor Gdr G This is a contradiction so the claim holds. a  b2 b  c2 c +
+
�J2
 a +
+++
+
+
+
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1985), pp. 59, Problem 5479) 25. Setting ai = sin2 ai for i = 1, n, where aI, a2 , . . . , an are real numbers, the expression becomes n 2 E =
2:: t/sin
i=l
ai cos2 ai+b an+1 a1· =
Using the AMGM'inequality yields
k k1 2:: bf ;::: b1 b2 . . . bk , bi > 0, i U. i=l For b1 sint ai, b2 = cost ai+1 and b3 = b4 bk = � we obtain ij2 =
=
= ... =
2
,
1e 2

;; and it is reached if and only if a1 a2 = . . . an 1
Hence the maximum value of E is
=
=
=
2'
Problem
+
(Titu Andreescu, 5982
1e

(Dorin Andrica, Revista Matematica Timi§oara (RMT), No. 1(1978), pp. 63, 3266)
2(a b c) ( a1 1b c1 ) > 3 +
n.
,
n(k

On the other hand, relation ( ) gives +
1 ,2 ...,n yields 2) _l � (n k 2 ) > 2;;2; n n · 21 t E< = 2 {14 2
Summing up these relations for k =
(3)
.
+
29
SOLUTIONS
26. Because x and
m
are positive, we have to prove that
x(xmn  1)  m(xn  1) ;::: 0, or (xn  1)[(xn) m1 x (xn) m2 x x  ] � 0. 1 x (xn ) m2x x  and note that if x � 1, then Define E (x) = (x n ) m xn ;::: 1 and E (x) � 0, so the inequality holds. In the other case, when x < 1, we have xn < 1 and E (x) < and again the inequality holds, as claimed. + ... +
+
+ .. . +
+
m
m
°
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 45,
Problem 3480)
27. For
[:].
m ::; n the inequality is clearly true, so consider m > n and define This implies that m = pn + q with q E {O, ...,n  I} and the inequality can be written as (pn + q ! � (n!) P.
p=
1,
)
We have
=
(1 ·
1
)
(pn + q ! � (pn) !
=
2 ... n) (n + ) ...(2n) ... ((P  l)n +
and we are done.
1) . . . (pn)
;::: (n!) P,
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 61,
Problem 3034)
28. We will use the inequality
xmn > n_m11_ (Xl X2 Xn )m , which holds for all positive real numbers X 1 ,X2 , . . . ,X n and all (00 , 0] [1 ,00) . Xm1 x2m +
+ .. .
+
+
+ .. . +
m
E
U
30
1. ALGEBRA
Xl 1, X2 2, . . . , Xn n and _ .!..n We obtain 1 1 . . +1 >1 l [n(n+l)l� n n{;f;' 1 +++ n+l y'2 V'3 vn 2 n �
Now set
=
=
Summing up these inequalities yields
m =
=
.
=

_
n n (a a a ) n 1 Llogai ga tn 2: L o l n i=l l 2 . . . n i= l nl[n + (logal a2 + loga2 al) + . . . + (logal an + logan al) + . . . + n +(logan anl + logan_l an )]. 1 Note that a + > a  2 for all a 0, so nl n + 2(n  1) + 2(n  2) + . . . + 2] (nl)n, � logai t n 2: [ � n i= l =
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1974), pp. 52, 2035)
Problem

29. From the AMGM inequality we deduce
n 1� � i + 1 > TI i +. 1 v n + l, n i=l i= l n1 1 + n1 � i=l � > \In + 1, n + 1  1) ' 1 +21 + 3l + .. . + nI > n (\In 
or
._
�

and so
�
�
=
as claimed.
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 62, 3038)
Problem
; 
31. We begin with the following lemma.
or
i 1 , i 1, n,
nl � 1 1 2 1 ++ 2 3 +n· · · +n > n n _vn_ . =
n (1  _vn1_) + 1 > 1 + !2 + . .. + .!..n
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 62, Problem 3037) 30. Because the numbers aI, a2 ,. " ,an are positive, from the AMGM inequality al + a2 + . . . + an 2: '\I'al a2 . . . an nl n we deduce that tn � (ala2 ' " an ) � Using that numbers ai are less than 1 we obtain nl (al a2 . . . ) logai tn 2: loga. n .
•
an .
Let a > b be two positive integers such that ��> 1. Then between numbers a and b there is at least a perfect cube. Proof. Suppose, for the sake of contradiction, that there is no perfect cube between a and b. Then there is an integer c such that c3 � b < a � (c + 1) 3 . Lemma.
as desired. Observe that the inequality is strict because the numbers � = are � distinct. In order to prove the first inequality we apply the AMGM inequality in the form
Therefore
>
=
,.�
_
=
i
_
as desired.
�
31
1.2. SOLUTIONS
This means so which is false. 0 Now we can easily check that for If 2: then
n 10, 11,12, 13,14, 15 the statement holds. n> (2,5)3 (1,4 1 1) 3 > ({13 1_ 1) 3 ' or � 1 n> {I3r' Hence �  � 1, and using the above lemma the problem is solved. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 12(1990), pp. 59, Problem 4080) n 16,
=
=
>
32
1.
1. 2 .
32. Note that the number k(k 2+ 1)
.
= 4p + 1 or k = 4p + 2 and is
= 4p + 3 = 4p, p = 4m,  1  4p 2 + 4p+ 3 + 4p+4) = 4m. k = �(4p Sn =�(1) k =1 p =O ii) if n = 4m + 1, then Sn = 4m  (4m + 1) = 1. iii) if n = 4m + 2, then Sn = 4m  (4m + 1)  (4m + 2) = (4m + 3). iv) if n = 4m + 3 then Sn = 4m  (4m + 1)  (4m + 2) + (4m + 3) = o. Hence 4m if n = 4m 4m + 1 Sn = �1(4m + 3l ifif nn == 4m +2 if n = 4m + 3 . (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1(1981), pp. 50, Problem 4303) even for k or k where We have the following cases: i) if n then m l n k(k+l) "'"'
"'"'
2
{
33. a) Summing up the identities
(n +k 2) = (n +(n2 + k)2)((nn ++ 1)1  k) (nk)
= 0 to k = n yields (n +k 2) _ (nn ++ 21) _ (nn ++ 22) ) = Sn  (n + 2)(n1 + 1) (� k =O 1 2 n+2  (n + 3) [2 n+2 (n + 2) _ 1] . (n + 2)(n + 1) (n + 2)(n + 1) b) Summing up the identities + 2)(n + 1) (n) (n +k 3)  (n + 3(n +k) 3)(n (n + 2  k) (n + 1  k) k for k = 0 to k = n yields 1 n  (n + 3)(n + 2)(n + 1) for k
_
_
_
T.
_
_
SOLUTIONS
n
odd for k is a positive integer. IS
33
. (�( ; 3) (�:D (�:!) (�:�)) = 1 (2n+3  1 (n2 + 3n + 2) ) (n + 3)( n + 2)(n + 1)
ALGEBRA
_
2
_
n+ 4 3n + 2)  2(n2 + 3)(n(n2++2)(n + 1) .
_
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1975), pp. 43, 2116) 34. Let Sn be the number in the statement. It is not difficult to see that Sn = 41 [(2 + V3) 2n+l + (2  V3) 2n+l ] . The required property says: there exists k 0 such that Sn (k  1) 2 + k2 , or, equi valently, 2k2  2k + 1  Sn = o. The discriminant of this equation is � = 4(2Sn  1), and, after usual computations, we obtain e +l 1 H r = l + .J3)2n ; ( .J3)2n
Problem
=
>
A
Solving the equation, we find 2 n+1 k=
+ (1 + v'.3)2n++l + (1  V3)2n+l
2n 2 Therefore, it is sufficient to prove that k is an integer. Let us denote v'3) v'3) where is a positive integer. Clearly, is an integer for all We will prove that Moreover, the numbers divides satisfy the relation
Em
Em (1 + m + (1  m , [ ] m Em , m = 1,2,3, . . . 2� Emm . Em = 2Em1 + 2Em2 •
The property now follows by induction. Romanian IMO Selection Test,
(Dorin Andrica,
1999)
35. Differentiating the identity sin nx yields
= sinn X ((�) cotn1 X  (;) cotn3 X + (�) cotn5 X 
n cos nx where
= n sinn 1 x cos xP(cot x)  sinn Xsm �X PI (cot X),
• • •
)
1.
34 For
1. 2.
ALGEBRA
x = "4 we obtam v'2) n ( ncosn"47r = 2 (nP(l)  2P'(1)) . 7r
•
we have
nP(l) = n (�)  n (;) + n (;)  . . . 2P'(1) = 2(n  1) (�) + 2(n  3) (;)  2(n  5) (;) + . . . ,
nP(1)  2P'(1) =  [(n  2) (�)  (n  6) (;) + (n  lO) (;)  ... J = = n (G)  (;) + (;)  . . . ) + 28n • To conclude, use that (�) (;) + (;) _ '" = (V2) n sm n; , _
hence
v'2 n )n
Sn  2 (cos ""4n7r + sm. ""4n7r ) (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 89, Problem 3200) 36. Differentiating with respect to x the identities (x + l) n = (�) + (�) x + . . . + (:) xn and (x _l) n = (:) xn  (n � l) xn. + . . . + (1)" (�) _
(
yields
and
and
and
n(x + l)n. + n(n  l)x(x + 1)"2 = (�) + 22 (;) X + . . . + n2 (:) x" ' n(x  1)"' + n(n  l)x(x  1) n2 = n2 (:) x" '  (n  1)2 (n � 1) X"2 + + . . . + (1)" ' (�) Setting x = 1 yields 12 (�) + 22 (;) + . . . + n2 (:) = n(n + 1)2"2 Summing up the last two identities gives
Sn = 12 (n1 ) + 32 (n3 ) + . . . = n(n + 1)2n3 ,
as desired.
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1(1978), pp. 90, 3438)
Problem
37. Note that
n_12i+1_1 [log k] + [log 2n], 2 2 = k= k= 2i and [log2 k] i for 2 i � k < 2i + l . Hence 2" nl 2:l �og2 k] = 2: i . 2i + [log2 2n] = (n  2)2n + n + 2 i=O k= 2>1
2:l [log2 k] = 2: i O 2:
=
as claimed.
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1981), pp. 63, 4585; Gazeta Matematica (GMB ) , No. 23(1982), pp. 83, Problem 19113) 38. Let Yn = 22 >1  1 for all positive integers n . Then 1  2 = 2 1  2 +21 = (22)1)2  2 .)122" + 1 = Yn Yn+l 2 >1  1 2 >1  1 (22)1  1)(22 +1  1) )1  (22)1 _(221)(22_ )1+11) 2 _ 1)  (222)12 )21 1 = 22>1 1+ 1 = x1n and therefore 1 1 2 Problem
n(x  1} "' = n (:) x n.  (n  1) (n : 1) xn2 + . . . + ( _ 1) " ' (�). Multiplying by x gives nx(x+ 1) "' = (�) x + 2 (;) x2 + . . . + n (:) xn
35
Differentiating again we obtain
Because and
SOLUTIONS
_
_
>1
1.
36
1.2.
ALGEBRA
Consequently,

=

f(iz)f(z) =
_ Z2 .
C,
=
(_
)
C
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1976), pp. 56, 2583) 40. Setting X = Y = 1 yields f2 (1) +f2 (a) 2f (1 ) and {f(l)  1 )2 = 0 so f(l) = 1. Substituting Y = 1 gives f(x)f(l)+f (�) f(a) = 2f(x) Problem
x.
or
Y Xa and observe that =

f(x)f (�) +f (�) f(x) = 2f(a).
f(x)
f
1 12(1977), 45, 18455) 10(1980), 439,
f
(Titu Andreescu, 2849;
T 0 f(x
+
It follows that
T) =
f(x) or sin[x +T] = sin [x] , for all x E
[x +T]  [x] 2k(x)7r, X =
k
where : 1R + Z is a function. Because all E 1R and therefore
x
7r
R
E lR,
is irrational, we deduce that
[x] [x +T] for x all
=
k(x) 0 for =
E 1R
which is false, since the greatest integer function is not periodical. Revista Matematica Timi§oara ( RMT ) , No. Problem
(Dorin Andrica, 3430)
1(1978), pp. 89,
42. Considering the determinant
=
Take now
12 (0) +12 (�) 2/(t)
= and because the lefthand side is positive, it follows that is positive and for all Then is a constant function, as claimed. pp. Revista Matematica Timi§oara ( RMT ) , No. Problem Problem pp. Gazeta Matematica ( GMB ) , No.
41.
f(iz) = 0 or fez)+f(z) = o . From the relation f(z)f(iz) = Z2 we deduce that fez) = 0 if and only if z = o. Hence if z f:. 0, then f(iz) f:. 0 and so fez) + f( z) = 0 and, if z = 0, then fez) + f(z) = 2f(0) O. Clearly, fez) + f(z) = 0 for all numbers z E as desired. Remark. A function f : + satisfying the relation f(z)f(iz) = Z2 is fez) = V22 +iV22 z. C
> o. = 0, that gives
The function is not periodical. Suppose, by way of contradiction, that there is a number > such that
f(iz)(f(z)+f( z)) = 0,
so
=
=
39. Substituting z with iz in the relation f(z)f(iz) = Z2
Summing up gives
x Y

(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 12(1980), pp. 67, Problem 4135)
yields
f(x)f (�) = 1,
f2 (X) = 1, x
therefore Now set
xn Yn Yn+l Summing up these relations yields 1 + 2 + 22 +. . . + 2nl 1  2n < 1 xn Yl Yn+l Yl Xl X2 X3 for all positive integers n, as desired.
37
SOLUTIONS
we have
12 22 32 n 2 2 1 n 3 8= I n 2 n 3 n nn 11 122 133 In 2 n = 82 , � = IS(i,j)1 = 8· 2 2 2 n
l n2 n3
n
n
because the second determinant is obtained from 8 by interchanging rows and columns.
1.
38
1.2.
ALGEBRA
On the other hand,
1 1 8 = n!
1 n 1
1 2
1 3
Because the coefficient of X4 is equal to
1 n
2n1 3n1
43. The determinant is al
0 0 �2n = 0 0
b2n
0 a2 0 0 b2nl 0
0 0
a3 b2n2
0 0
0 0
0 b2 b3 0 0 a2n2 0 a2nl 0 0
b1
0 0 0 0
a2n
y x v z
z v z x y y x
=
a b e d
b a d e
w
=
(
1, we have = 1 and so ..\
x + y + z + v) (x + y  z  v) (x  y + z  v) (x  y  z + v)
b) As shown above, we have
(here we used the known result on Vandermonde determinants). Therefore
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1(1982), pp. 52, Problem 3862)
x y z v
39
SOLUTIONS
�
e d a b
d e b a
= (a b + e + d) (a + b  e  d) (a  b + e  d) (a  b  e + d) +
1000,
100,
On the other hand, multiplying the first column by the second by and adding all these to the fourth, we obtain on the last column the third by the numbers abed, bade, edab, debao Because all those numbers are divisible by the prime number p, it follows that p divides � and therefore p divides at least one of the numbers a b e d, a b  e  d, a  b e  d, a  b  e d.
10
+++ + (Titu Andreescu)
+
+
x
x
Because the quadratic polynomials h ( ) and t2 ( ) have zeros of the same nature, it follows that their discriminants have the same sign, hence
45.
Expanding along the first and then the last row we obtain
� 2n which gives
�2n
= (al a2n  bl b2n)�2n2 '
= k=lITn (ak a2nk+l  bk b2nk+l )
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 90, Problem 3201; Gazeta Matematica (GMB) , No. 8(1977), pp. 325, Problem 16808) 44. a) Adding the last three columns to the first one yields that x + y + z + v
divides the determinant. Adding the first and second columns and subtracting the last two columns implies that divides the determinant. Analogously we can check that and divide the determinant, and taking into account that it has degree 4 in each of the variables, the determinant equals
x +y  z v
where
..\
xy+zv xyz+v
..\(x + y + z + v) (x + y  z  v) (x  y + z  v)(x  y  z + v),
is a constant.
Consequently,
(pIP2
+ 4ql q2 )2  4(plq2 + P2Ql )2 2:: O.
Note now that the lefthand side of the inequality is the discriminant of the quadratic polynomial and the conclusion follows. Revista Matematica Timi§oara (RMT), No. pp. Problem Gazeta Matematica (GMB), No. pp. Problem
(Titu Andreescu, 3267;
t
5(1979), 191,
1(1978), 63, 17740)
46. Because the quadratic polynomial T has nonreal zeros, the discriminant � is negative. a bserve that
= b2 e2  4a (b3 + e3  4abe) =
0,
� ( b2  4ae) (e2  4ab) < where � l b2  4ae and �2 e2  4ab are the discriminants of the quadratic polynomials Tl and T2 • Hence exactly one of the numbers � l and � 2 is negative and since a > the conclusion follows .
= = 0, (Titu Andreeseu, Revista Matematica Timi§oara (RMT), No. 1(1977), pp. 40, Problem 2810)
1. 47. Observe that al +a2 + . . · +an and al a2 . . . + ( _ l) n l an are real numbers, that is P(l) and P(  1) are real numbers. Hence (1) P(l) = P(l) and P( 1) = P( 1) Because P(x) = (x  xd . . . (x  x n ), the relations (1) become (1  xd . . . (1  xn ) = (1  xd . . . (1  xn) and (1 + xd . . . (1 xn ) = (1 xd . . . (1 + xn )
1.2.
ALGEBRA
40
+
+
+
Multiplying these relations yields
(1  xi) . . . (1  x� ) = (1  xi) . . . (1  x�), or Q(l) = Q(l). Therefore b1 b2 + . . . + bn is a real number. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1977), pp. 47, Pro blem 2864) 48. Because P(O) 0, there is a polynomial Q with P(x) xQ(x) . Then Q(k) = +1 l ' = 1,n. Define H(x) = (x + l)Q(x)  1. It is clear that degH(x) = n and H(k) = 0 for all = !,n, hence H (x) = (x + l)Q(x)  1 = ao(x  l)(x  2) . . . (x  n) (1) Setting x = m n in relation (1) yields Q (m) = ao (m  1) (m m 2)+ .1. . (m  n) + 1 . On the other hand, setting x = 1 in the same relation implies ao = ((n_ l)+n+I)!l Therefore  2) . . . (  n) + 1 Q(m) = ( _ l)n+l (m(n l)(m l)!(m + 1) m 1 and then  1) . . . (m  n) m ' P(m)  ( _ l)n+l(nm(m + l)!(m + 1) + m+l (Dorin Andrica, Gazeta Matematica (GMB), No. 8(1977), pp. 329, Problem 16833; Revista Matematica Timi§oara (RMT) , No. 12(1980), p. 67, Problem 4133) +
=
=
k
k
k
m,
>
m
+
49. We are looking for a polynomial with integral coefficients
P(x) = aoxn + alxn 1 + . . . + an , ao i= O.
+
4
SOLUTIONS
1
We have
P'(x) = naoxn 1 + (n  1)alxn2 + . . . + an l and by identifying the coefficient of x(n l) n in the relation P(P'(x)) = P'(P(x)), we obtain or
aonn 1 = 1. Hence
ao = nn1 l and since ao is an integer, we deduce that n = 1 and ao = 1. Then P(x) = x aI, P'(x) = 1 and P(P'(x)) = P'(P(x)) yields 1 + al = 1 or al = O. Therefore P(x) = x is the only polynomial with the desired property. (Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 12(1979), Problem 3902) 50. Let (}l , (}2 , . . . , (}n be the roots of the equation xn + xn 1 . . . + x + 1 = O. They are all distinct and Or+1 = 1, i = !,n. Because P(x) is divisible by xn + xn  1 + . . . + x + 1, it follows that P( Oi ) = 0, i = !,n, hence PI (1) (}1P2 (1) + . . . + O� lPn (l) = 0 PI (1) + 02 P2(1) + . . . + O� lPn (l) = 0 +
+
+
The above system of equations has the determinant
1 01 on1 l1 V = 1 02 0'2 1 On (}nn l Because all of the numbers 01 , 02 , . . . , On are distinct, it follows that V i= 0 and so the system has only the trivial solution PI (1) = P2 (1) = . . . = Pn (l) = O. This is just another way of saying that x  I divides pi(X) for all i = 1, n. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 75, Problem 3120; Gazeta Matematica (GMB), No. 8(1977), pp. 329, Problem 16834)
1.
42
1. 2.
ALGEBRA
51. Consider the determinant =
an nII+l (ak  a,) 10 , 1 = 1 10 > 1
anI, the third by an2 , ... , the last by ao and P(ad P(a2 ) P(an+d a +l al a2 v= an;+l = an nII+l (ak  at) a� a� a�+l af a� On the other hand, P (ar ) 0 (mod p), for all r = 1, n 1 and an 0 (mod p) II< +l (ak  a,) 0 (mod p). Therefore there are at least two numbers implies l<l <k ai , aj, i f=. j sunch that ai  aj 0 (mod p) and so ai aj (mod p), as desired.
Multiplying the second row by adding all to the first yields
10 , 1 = 1 10 > 1
==
;j.
+
==
==
==
(Dorin Andrica, Gazeta Matematica (GMB), No. 8(1977) , pp. 329, Problem
16835)
52. Let m = degP(x) and let
P(x) = aoxm Q(x), ao f:. 0 If follows that degQ(x) = r m  1. From pn (x) = P(xn ) we obtain a�xmn (�) ag l xm(nl ) Q(x) . . . Qn (x) = aoxmn Q(xn ), or a�xmn R(x) = aoxmn S (x), where degR(x) = m(n  1) r and degS(x) = nr. ' Because ao f:. 0, it follows that ao = 1 if n is even and ao = 1 or ao = 1 if n is odd. Moreover, degR(x) = degS (x) or m(n  1) r = nr and so (n  l)(m  r) = O. +
>
=
+ . .. +
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 12(1979) , pp. 59,
Problem 3884)
53. From the relations between the zeros and the coefficients we obtain
aamo = ( _ l) m L XIX2 .. ' Xm
+
+
+
=
+
:::;
and
+
+
43
Q(x) f:. 0, because n 1 and m > r, therefore Q(x) = 0 and P(x) = xm , for n even and P(x) = ±xm , for n odd. Alternative solution. Let degP(x) = m and let P(x) aoxm alxm l am . If P(x) = x k Q(x) with k a positive integer, then xkn Qn (x) = xkn Q(xn ) or Qn (x) = Q(xn ) Note that Q satisfies the same condition as P. Assume that P(O) f:. O. Setting x = 0 in the initial condition yields a� = am ' Then am = 1 if n is even and a m = ± 1 if n is odd. Differentiating the relations implies npnl (x)P'(x) = nP'(xn)x n l . (1) Setting now x = 0 gives P'(O) 0 and so am  l = O. Differentiating again in relation ( 1 ) yields analogously a m 2 = 0 and then am3 = am4 = . . . = ao = O. The polynomials are P(x) = xm , if n is even and P(x) = ±xm , if n is odd.
This is impossible if the polynomials are
V=
+
SOLUTIONS
It follows that
+
and by applying the triangle ' s inequality for complex numbers, we deduce that 1
L I xd l x21 .. · I xnm l > (mn ) . Consider Xo = min l x I , I X 2 1 , . . . , I xn  m l } . Then {
+
so
Xo
<
1, as claimed.
I
1. ALGEBRA
44
1.2. SOLUTIONS
(Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 2(1978), pp. 52, 3531)
Problem
54. We have
PI(X) = ;n2 . P(x) i= l  X·
�
L...J
p,
Li=l ln I P (x)  xii = n2 In C lx l ,
or
C>
0
IIi=nl IP(x)  xii = In Cn2 Ixl n2 Ill. (F(x)  Xi) I = klxl n' , k > 0 IP (P(x))1 = k l xl n2 . In
Hence
or
Eliminating the modules gives
P(P(x))
2 = AX n ,
A E JR.
P (x) = axn with a E (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1977), pp. 47, Problem 2863; Gazeta Matematica (GMB), No. 1(1977), pp. 22, Problem 17034) 55. Define Q(x) = xP(x). Because an f:. 0, the polynomial Q has distinct real zeros, so the polynomial QI has distinct real zeros as well. Consider H(x) = XQI(X). Again, we deduce that HI has distinct real zeros, and since HI (x) = x2 PI (x) 3xPI (x) P(x) the conclusion follows. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 52, Problem 3530) R
Therefore
+
56. Let
m
a of order k. Since P(Q(a)) = 0, we have m ao II(Q(a)  Xi) = 0, i=l and so there is an integer 1 � � such that Q(a)  xp = O. Observe that = Q(a)  xp f:. 0, for all j f:. otherwise Xj xp , which is false. Hence Q(x)  Xj has the multiple zero a of order k and so QI(X) = (Q(x)  xp )' = QI(X) has a multiple zero of order k  1. This concludes the proof. (Dorin Andrica, Romanian Mathematical Olympiad  final round, 1978; Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 67, Problem 3614) 57. Assume by way of contradiction that P(x) has less than two nonreal zeros. As a polynomial with real coefficients P(x) cannot have only one nonreal zero, hence all of its are real. Let Xl , X2 , . . . ,X n be the zeros of P(x). Then PI(X)  n _1_ P(x) � X  Xi has a multiple zero
p,
�
Integrating the equation yields n
+
= degP (x) and let
P(x) = ao (x  XI)(X  X2 ) . . . (x  xm ). Because P(x) E R[x], X l , X 2 , . . . , x m are distinct zeros. Now P(Q(x)) = (Q(x)  xt )(Q(x)  X2 ) . . . (Q(x)  xm )
45
p
m,
_
and differentiating we obtain
PII (x)P(x)  [PI(X)] 2 =  n
1 � (x  Xi)2 Setting X = a we reach a contradiction, therefore P(x) has at least two nonreal zeros, as claimed. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 12(1979), pp. 59, Problem 3883) 58. Let P(x) = aoxn al xn l . . . an be a polynomial with real coefficients. If all of its zeros are real, then the same is true for the polynomials pI , P", . . . , pCn 2) . Because (n  2)! P n2) (x) = 2 [n (n  l) ao x2 2(n  l)alX + 2a2] P2 (X)
+
+
+
C
+
is a quadratic polynomial with real zeros, we have or
(n  l)a� ;::: 2naO a2 . The reciprocal is not always true, as we can see from the following example: 2 + (a + l)x + a, = x3 + (a + with
a
P(x) E ( 00 1] [2, (0). ,
U
1)x
1.
46
2
( (Dorin Andrica)
ALGEBRA
(
Observe that 2 a + 1) ;::: 2 . 3 a + 1), or (a + l) (a  2) ;::: 0, so the inequality holds. On the other hand, P (x ) = ( x + a ) ( x2 + x + 1) does not have all zeros real.
59. For m = 0 the equation becomes
X4  x3 x2 X 0 and has roots X l 0, X2  1, X3 X4 1. If m :f. 0, we will solve the equation in terms of m. We have 2xm2 ( x2  2x3 l ) m X4  x3  x2 X 0 
=
=
=
=
=
+
+
+
+
+
=
and
Chapter 2
It follows that
NUMBER THEORY
The initial equation becomes
[
2
x2  1 [m  ( x  x )] m  �
Hence and
]
=
0.
x 2  x  m = 0, with solutions
=
1±� 2
2
=
m±
x  2mx  1 = 0,
Xl,2 with solutions X3 , 4
VI m2 . +
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977) ,
Problem 3121)
pp.
75,
60. From the relations between the zeros and the coefficients we obtain L
XIX2 · · · X2n1
=
2n and
XIX2 · · · X2n 1. =
Hence
2n 1 2n, Xk l k= X2n ' so we have the equality case in the AMGM inequality. Therefore Xl X2 Since Xl X2 ... X 2n and 2n  > 0 "' ' �X k l = k X2 1. we have Xl X 2 . (Titu Andreescu, Revistan Matematica Timi§oara (RMT) , No. 2(1977) , 52, L
=
=
Problem 2299)
=
=
=
1
. .

= .. . =
1
=
=
pp.
PROBLEMS
1. How many 7digit numbers that do not start nor end with 1 are there? 2. How many integers are among the numbers l路m 2路m n ' n
p路m
where p, m, n are given positive integers?
3. Letn
divides 1 p
p +
2
n
> be a prime number and let n be a positive integer. Prove that n n 2p + . . l) p .
. + (p _
p
4. Prove that for any integer n the number 55
n +1
+ 55 n + 1
is not prime.
5. Let
n
be an odd integer greater than or equal to 5. Prove that
is not a prime number.
6. Prove that
345 + 456
102002 . 7. Find all positive integers n such that [\1111] divides 111. 8. Prove that for any distinct positive integers a and b the number 2a(a2 + 3b2 )
is a product of two integers, each of which is larger than
is not a perfect cube.
9.
Let p be a prime greater than 5. Prove that p of an integer.
10.
(x,y)

4 cannot be the fourth power
Find all pairs of nonnegative integers such that simultaneously perfect squares.
49
x2 +3y and y2 +3x are
50
2.
2.1.
NUMBER THEORY
11. Prove that for any positive integer n the number (17 + 120) n _ (17 _ 120) n
is an integer but not a perfect square.
22. Prove that the equation
4v'2
has infinitely many solutions in positive integers such that U and
24. Solve in nonnegative integers the equation
un+
Un
x + y + z + xyz = xy + yz + zx + 2
+5
25. Solve in integers the equation
+ 3n + 3
xy(x2 + y2 ) 2z4 .
15. Let p be a prime. Prove that a product of 2p + 1 positive consecutive numbers
2
cannot be the p + Ipower of an integer.
16. Let p be a prime and let a be a positive real number such that pa2 Prove that [n0i  �] = [n0i+ �]
for all integers
<
=
26. Prove that for all positive integers n the equation
1
4'
n [ y'1 a2avp] + 1.
x2 + y2 + Z2 59n =
has integral solutions.
27. Let n be a positive integer. Prove that the equations
2::
17. Let n be an odd positive integer. Prove that the set 18. Find all positive integers and n such that (:) = 1984. 19. Solve in nonnegative integers the equation m
x2 + 8y2 + 6xy  3x  6y
and
xn + y n + zn + u n = vn+ 1
have infinitely many solutions in distinct positive integers.
contains an odd number of odd numbers.
20. Solve in integers the equation
are both primes.
x2 (y  z) + y2 (z  x) + Z2 (x  y) = 2.
n 6 between and 1 there is a perfect square. 13. Prove that for all positive integers n the number n ! is not a perfect square. 14. Prove that if n is a perfect cube then n2 cannot be a perfect cube. 2::
v
23. Find all triples (x, y, z) of integers such that
12. Let (Un) n�l be the Fibonacci sequence: Prove that for all integers
51
PROBLEMS
=
3
(x2 + 1)(y2 + 1) + 2(x  y)(1  xy) = 4(1 + xy) 21. Let p and q be prime numbers. Find all positive integers x and y such that 1 1 1  +  = . x y pq
28. Let n be a positive integer. Solve in rational numbers the equation
x n  1 + yn  1 . 29. Find all nonnegative integers x and y such that x(x + 2)(x + 8) = 3Y• xn + y n =
30. Solve in nonnegative integers the equation
(1 + x!)(1 + y!) = (x + y)!.
31. Solve the equation
x! + y! + z! = 2v! .
52
2.
2. 1 .
NUMBER THEORY
32. Find all distinct positive integers X l , X2 , . . . ,Xn such that
Xl + 2XIX2 + . . . + (n  1)XIX2 . . . Xnl XIX2 . . . Xn . 33. Prove that for all positive integers n and all integers aI, a2 , . . . , an , bl , b2 , . . . , bn the number n II k=l (a�  b�) 1+
34. Find all integers x, y,
v, t such that X + Y + + v + t xyvt + (x + y)(v + t) xy + + vt xy(v + t) + vt(x + y). 35. Prove that for all nonnegative integers a, b, d such that a and b are relatively prime, the system ax  yz 0 bx  yt + d 0 Z,
=
Z
Z
=
c,

=
c
=
has at least a solution in nonnegative integers.
36. Let p be a prime and let Xl , X2 , . . . ,xp be nonnegative integers.
Prove that if
Xl + X + . . . + x 0 (mod p) x� + x�2 + . . . + x�p 0 (mod p) Xpl l + Xp2 l + . . . + xp  I 0 (mod p) then there are k, l E {I, 2, . . . , p}, k =I l, such that Xk  Xl 0 (mod p) . 37. Prove that for any odd integers n, aI, a2 , . . . ,an , the greatest com mon divisor of numbers aI, a2 , . . . ,an is equal to the greatest common divisor of al +a2 ' a + a an + al 2 2 23 2== ==
P
==
==
3.8 . Let <p(n) be the number of numbers less than n and relatively prime with n.
Prove that there are infinitely many positive integers n such that
n <p(n) = 3'
39. Let 7r (x) the number of primes less than o r equal t o x. Prove that 7r (n) for all positive integers n.
40. Let Pk denote the kth prime number. Prove that pf + pr + . . . + pr;:
=
can be written as a difference of two squares.
<
n 3+
2
53
PROBLEMS
>
nm
+
l
for all positive integers m and n.
41.
Let n be a positive integer. Find the sum of all positive integers less than 2n and relatively prime with n.
1
42.
Prove that any number between and n! can be written as a sum of at most distinct divisors of n!. n
43.
Find the largest value of n such that the complementary set of any subset contains at least two elements that are relatively with n elements of {I, prime.
aln.
2, . . . , 1984}
44. Find all positive integers n such that for all odd integers a, if a2 � n then
45. Consider the sequences (Un ) n� l' (Vn ) n� l defined by UI 3, VI 2 and Un+l 3un + 4vn , Vn+l 2un + 3vn , n � 1. Define Xn Un + Vn , Yn Un + 2vn , n � 1. Prove that Yn [x nV2J for all n � 1. 46. Define Xn 22,, 1 + 1 for all positive integers. Prove that (i) Xn XIX 2 . . . Xn  l + 2, n E N (ii) (X k' Xl) 1, k, l E N, k =I l (iii) X n ends in 7 for all n � 3. 47. Define the sequence (an ) n � l by al 1 and an+l 2an + J3a�  2 for all integers n � 1. Prove that an is an integer for all n. 48. Define the sequences (an ) n�O and (bn ) n �o, by ao 1, an 1 +2a2na2n1 l and bn 1 21a2n l for all positive integers n. Prove that all terms of the sequence (a n ) n �O are irreducible fractions and all terms of the sequence (bn ) n �o are squares. 49. Define the sequences (Xn ) n �O and (Yn) n �O by Xo 3, Yo 2, Xn 3Xn l + 4Yn 1 and Yn 2Xn 1 + 3Yn 1 for all positive integers n. Prove that the sequence (zn ) n �O' where n 1 + 4x;y;, =
=
=
=
=
=
=
=
=
=
=
=
=
=
=
_
=
=
I
=
=
Z
contains no prime numbers.
=
2.
54
p be a positive integer and let Xl be a positive real number. Define the (Xn ) n�l by Xn+l Vp2 1xn pvx; 1 for all positive integers n. Prove that among the first terms of the sequence there 50.
Let sequence
=
are at least
+
+
+
m
[�] irrational numbers.
51. Define the sequence (Xn )n�o by 1) Xn = 0 if and only if n = 0 and 2) Xn+l X[�] (l) nx[�] for all n 2:: O. =
Find
2 3
52. Define the sequence (an ) n �O by ao = 0, al = 1, a2 = 2, a3
=
6 and
n divides an for all n > O. 53. Let X l X2 X3 1 and Xn+3 Xn + Xn+I Xn+2 for all positive integers n . Prove that for any positive integer there is an integer k > 0 such that divides Xk · 54. Let (an ) n�O be the sequence defined by ao 0, al 1 and an+l  3an + an l ( _ l) n 2 for all integers n O. Prove that an is a perfect square for all n 2:: O. 55. Let al a2 97 and an+l ananl V(a;  l)(a;_ 1  1), n > 1. Prove that a) 2 + 2an is a perfect square. b) 2 J2 2an is a perfect square. 56. Let k � 2 be an integer. Find in closed form for the general term an of the sequence defined by ao 0 and an  a [ �] 1 for all n O. 57. Let ao al 3 and an+l tan  an l for n 2:: 1. Prove that an  2 is a perfect square for all n 2:: 1. 58. Let a and {3 be nonnegative integers such that 0'2 + 4{3 is not a perfect square. Define the sequence (x n ) n �O by =
m
m
=
=
=
>
=
=
=
+
+
+
=
=
=
=
=
>
(4n
+
yields all odd positive integers less than or equal to
Prove that
=
=
±1 ± ± ± . . . ±
+
=
55
60. Prove that for different choices of signs + and  the expression
Xn in closed form.
=
2. 1 . for all integers n 2:: 0, where X l and X2 are positive integers. Prove that there is no positive integer no such that X� XnoIXno+I ' 59. Let n > 1 be an integer. Prove that there is no irrational number a such that \/'a + Ja2  1 + \fa  Ja2  1 is rational. PROBLEMS
NUMBER THEORY
1),
(2n 1) (4n +
+
1).
SOLUTIONS
1. The problem is equivalent to finding the number of functions
{1,2,3,4,5,6, 7} {0, 1,2, . . . ,9} such that 1(1) f. 0, 1(1) f. 1 and 1(7) f. l. Because 1(1) E {2, 3, . . . , 9}, there are 8 possibilities to define 1(1) . For 1(7) there are 9 possibilities and for 1(2),1(3),1(4),1(5),1(6) there are 10 . To conclude, there are 8 . 9 . 105 72 . 105 numbers with the desired property. (Dorin Andrica, Gazeta Matematica (GMB) , No. 11(1979), pp. 421, Problem 17999) I:
t
=
m and n. Hence m ml d and n nl d 1 · ml , 2 · ml p · ml nl nl nl and, since m" are relatively prime, there are [ :, ] integers among them. Because gCd(m, n)p] .mtegers. n n It lo11ows that there are [ nl d gcd(m, n) n d
2.
Let be the greatest common divisor of and n l . for some integers The numbers are
=
ml

=

n,
=

'
=
.r
(Dorin Andrica, Gazeta Matematica (GMB ) , No. 11(1979), pp. 429, Problem 0:89)
pn and note that k is odd. Then dk (p d) k p[dkl dk2 (p d) . . . (p d) kl ] Summing up the equalities from d 1 to d [�] implies that p divides 1 k 2k (p  l ) k , claimed. 3. Define k
=
+
=
_
_
_
+
+
=
=
as
+ ...+
_
+
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 12(1979), pp. 49, Problem 3813) 4. Define m 55n • Then 55n +1 55n 1 m5 + m 1 (m2 + m 1)(m3  m2 + 1) 57 =
+
+
=
+
=
+
2.
58
1, (Titu Andreescu, 2001) 5. Let N = (�)  5 (�) + 52 (�)  + 5n ' (�). Then 5N = 1  1 + 5 (�) _ 52 (�) + 53 (�) _ . . . + 5n (�) = 1 + (1 + W. Hence N = � W + l) = � [C2n + 1)2  (2� n = = � [2n  2 � + 1] [2n + 2� + 1] = = � [ (2 ";'  1r + 2n ' ] [ (2 ";' + 1r + 2n' ] . Because n is greater than or equal to 5, both factors of the numerator are greater than 5. One of them is divisible by 5, call it 5N1 , N1 > 1, the other being N2 . Then N = N1 N2 , where N1 and N2 are both integers greater than 1, and we are done. (Titu Andreescu, Korean Mathematics Competition, 2001) 6. The given number is of the form m4 + � n4 , where m = 344 and n = 4� = 2� .
(Titu Andreescu)
and, since both factors are greater than the conclusion follows. Korean Mathematics Competition, ' "
The conclusion follows from the identity
2m  mn + in1 2 > n ( in1  m) = 2 �2 (2 562_1  344 ) > 61 . �( 54 � 1 ( 56 1 5 1
and the inequalities
> 2 2 2 2  2 2 4 ) > 2 2 . 25 2 2 2 2  1 ) > > 210. � 251 2 > 21 0.54 . 210.50 > 103.54 . 103'50 > 102002 (Titu Andreescu, Korean Mathematics Competition, 2002) 7. The positive divisors of 111nare 1, 3, 37, 111 . So we have the following cases: 1) [yl111] = 1 or 1 $ 111 2 , hence n � 7. 2) [yl111] = 3, or 3n $n 111 4n , son n = 4. 3) [yl111] = 37, or 37 $ 111 38 , impossible. 4) [yl111] = 111, or 111n $ 111 112n , and so n = 1. Therefore n = 1, n = 4 or n � 7 . .
<
<
2. 2.
NUMBER THEORY
<
<
59
SOLUTIONS
8. Note that
2a(a2 + 3b2 ) = (a + b)3 + (a  b)3 The Fermat equation for n = 3 x3 +y3 = z3 has no solution in positive integers (see T. Andreescu, D. Andrica, "An Introduction to Diophantine Equations" , GIL Publishing House, 2002 , pp. 8793) . Hence there is no integer c such that
if
a > b.
On the other hand, if
b > a then there is no integer
c
such that
This concludes the proof. Revista Matematidi Timi§oara (RMT ) , No. Problem
(Titu Andreescu, 1911)
1(1974) , pp. 24,
4 = q4 for some positive integer q. Then p = q4 + 4 and q > 1 . We obtain p = (q2 2q + 2)(q2 + 2q + 2), a product of two integers greater than 1, contradicting the fact that p is a prime. (Titu Andreescu, Math Path Qualifying Quiz, 2003) 9. Assume that p

_
10. The inequalities
0 � x + y + 8, which is x2 +23y (x + 2 y2 + 3x (y + 2)2 is true. Without x + 3y (x + 2)2 . x2 x2 + 3y (x + x + 3y = (x + 1)22 , hence 3y2 = 2x + 1 . x = 3k 1 y = 2k + 1 k 0 y + 3x = 4k + 13k + 4. k > 5, (2k + 3)2 4k2 + 13k + 4 (2k + 4)2 so y 2 + 3x cannot be a square. It is easy to check that for k E {O, 1,2,3,4}, y2 + 3x is not a square but for k = 0, y 2 + 3x = 4 = 2 2 . Therefore the only solution is (x, y) = (1, 1) . (Titu Andreescu) cannot hold simultaneously because summing them up yields false. Hence at least one of 2 ) 2 or < < loss of generality assume that < From < < 2 ) 2 we derive Then + and for some integer � and so If then <
<
2. 11. Note that 17 + 12V2 (V2 + 1) 4 and 17  12V2 (V2 _ 1) 4 , so (17 + 12V2) n _ (17  12V2) n  (V2 + 1) 4n (V2 _ 1) 4n 4V2 4V2 2n 2 2 2 n n n 1) 1) 1) _ + (V2 (V2 1) (V2 + (V2 + 2 2V2 Define 2n  (V2 _ 1) 2n 2 2 A (V2 + 1) n +2 (V2 _ 1) n and B (V2 + 1) 2V2
60
2. 2.
NUMBER THEORY
=
If
=
n > 2v'UUn1+1 >y""n+1  yUn  v'UUnn+1+1 + U..;u;;, n 1 1 U � . n 2y ....Un1 2v3. fi)v'Un1 > 1 and so between Un and U n+1 there is a perfect square. (Dorin Andrica)
_
�=!....�
that
( V2 + 1 ) 2n = X + yV2, (V2  1) 2n = X  yV2
Then X=
AB
(V2 + 1) 2n + (V2  1) 2n 2
and
and so is as integer, as claimed. Observe that
A B
so and are relatively prime. not a perfect square. We have
It
y such
2n  (V2 _ 1) 2n y  (V2 + 1) 2V2 _
is sufficient to prove that at least one of them is
2n 2 A (V2 + 1) n +2 (V2  1) =
[(V2 + 1) nV2 
=
Since only one of the numbers
(y'2
 1) n ] 2 + 1
(2)
l) n + (v'2 _ l) n (v'2 + l) n  (v'2 _ 1) n V2 V2 is an integer  depending on the parity of n  from the relations (1) and (2) we derive that A is not a square. This completes the proof. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 1(1981), pp. 48, Problem 4285) 12. The claim is true for n 6 and n 7, because U6 8 9 U7 13 16 Us 21. =

=
=
<
<
=
<
<
If
n � 5,
n! + 5 5(5k + 1) for some integer k and therefore is not a perfect square, as (Dorin Andrica, Gazeta Matematidi (GMB) , No. 8(1977), pp. 321, Problem 16781; Revista Matematica Timi§oara (RMT) , No. 1(1978), pp. 61, Problem 3254) 14. Suppose by way of contradiction that n2 +3n+3 is a cube. Hence n(n2 +3n+3) is a cube. Note that n (n2 + 3n + 3) = n3 + 3n2 + 3n = (n + 1)3  1 and since (n + 1) 3  1 is not a cube, we obtain a contradiction. (Dorin Andrica, Gazeta Matematica (GMB ) , No. 8(1977), pp. 312, Problem E5965; Revista Matematica Timi§oara (RMT) , No. 12(1979), pp. 28, Problem 3253) 15. Consider the product of 2p + 1 consecutive numbers P(n) (n + l)(n + 2) . . . (n + 2p + 1) Observe that P(n) > (n + 1) 2p+1 . On the other hand, P(n) [(n + l) + (n + 2)2p++ . 1. . + (n + 2P + l) ] 2P+1 (n +p+ l) 2p+1 from the AMGM inequality. If P(n) m 2 p+1 , then m {n + 2, ... ,n + pl. Assume by way of contradiction that there is k {2,3, . . . ,p} such that P(n) (n + k) 2 p+1 . Then (n + 1)(n + 2) ... (n + k  1)(n + k + 1) ... (n + 2p + 1) (n + k)2P. (1) We have two cases: I. k p 1) If n 0 (mod p), then (n + k)2p is divisible by p2p . The lefthand side of the equality (1) is clearly not divisible by p2p , hence we reach a contradiction. 2) n r (mod p), r ::j:. 0, then the lefthand side of the equality (1) is divisible by p2 , because of the factors n + p  r and n + 2p while the righthand side is not, since (n + k) 2p r2 P. This is a contradiction. II. k {2,3, . . . ,p l} then desired.
=
<
(v'2 +
=
�
=
(1) and
.
13. If n = 1,2,3 or 4 then n! + 5 = 6,7, 11 or 29, so it is not a square.
. �= �.:..=!...
U sing the binomial expansion formula we obtain positive integers x and
u .�
� =
_
=
n � 8, then
61
SOLUTIONS
=
E
=
E
=
=
=
==
If
==

==
E
r,
62
2.
NUMBER THEORY
SOLUTIONS
1) If n k (mod p), then the righthand side of (1) is divisible by p2p , but the lefthand side is not. 2) n q (mod p), q f. k and q {O, 1, . . ,p  I}, then the lefthand side of (1) is divisible by p. On the other hand (n + k)2p (q  k)2 (mod p) � 0 (mod p), because 0 < Iq  kl < p. Both cases end up in contradictions, so the problem is solved. (Dorin Andrica) ==
If
E
==
.
==
16.
[ JI Q'2Q'vp] + 1.
It suffices to prove that there are no integers in the interval
(nvp  �n ,ny'P + �n ] for n �
Assume by way of contradiction that there is integer
ny'P Q'n < k < ny'P+ Q'.n
Hence
2.2. (  1) 1984, wIth. no mtegra . · I soI utIOns. If n 2, then 2 ( 1)(  2) Hence 1 1904 >If n � 3, then ( ) � ( ) ' so 1984 � n 6 3 m3 3m 2 + 2m or (  30)(m2 + 27m + 812) � 12456 < 0, and so < 30. This (  1) (  n + 1) does not contam· the implies that ( ) f. 1984, because n! n factor 31 of 1984. To conclude, the solutions are 1984, n 1 and 1984, n 1983. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1985), pp. 80, Problem 5)
k such that
m m
=
n2p  1 < k2 < n2p + 1 It follows that k2 pn 2 or Vp � , which is false, since p is prime. n (GMB) , No. 8(1977), pp. 324, Problem (Dorin Andrica, Gazeta Matematica 16804) 17. For n 1 the claim is clear, so let n � 3. Define Sn (�) + (; ) + + (n�l) . Then =
=
=
=
...
S!, 2n 1  1. Sn Sn ) 2(1984), 71, (Titu Andreescu, 5346) 18. Because (:) ( ': n) ' we can assume that m :5 [iJ . If n 0, then 1 1984, false. If 1, then 1984.
or = Because is odd it follows that the sum contains an odd number of odd terms, as desired. Revista Matematica Timi§oara (RMT , No. pp. Problem =
= n=
= m=
m
m m
m
m
mm
m

m
.
m
... m =
m=
m=
=
19. The equation is equivalent to
(x + 2y)(x + 4y)  3(x + 2y) 3, (x + 2y)(x + 4y  3) 3. =
or
=
(.) { xx ++ 4y2y  33 1 , with solution (x, y) (2, 21 )
We have
=
1
so
=
m

63
=
=
. (x,y) (4, 2'5 ) . (ii) { xX ++ 4y2y  31 3 , with solutIOn Note that there are no solutions in integers, as claimed. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1971), pp. 20, Problem 312)
or
=
=
=
20. The equation is equivalent to
x2y2 2xy + 1 + x2 + y2  2xy + 2(x  y)(I  xy) 4, or (xy  1)2 + (x  y) 2 + 2(x  y)(1  xy) 4. Hence (1  xy + x  y) 2 4 and, consequently, 1(1 + x)(1  y)1 2. We have two cases: I. (1 + x)(1  y) 2. Then a) 1 + x 2, 1  y 1, so x 1, y O . b) 1 + x 2, 1  y 1, so x 3, y 2. c ) 1 + x 1, 1 2, so x = 0, y = 1. d ) 1 + x 1, 1  y 2, so x 2, y 3. II. (1 + x)(1  y) 2 . Then _
=
=
=
=
=
= =
=
=
=
y
=
=
=
=
=
=
=
=
=
=
2. a) 1 + x = 2, 1  y = 1, so x = 1, y = 2. b) 1 + x = 2, 1  y = 1, so x = 3, y = O. c) 1 + x = 1, 1  y = 2, so x = 0, y = 3. d) 1 + x = 1, 1  y = 2, so x = 2, y = l. (Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 45(1972), pp. 43, Problem 1383)
64
NUMBER THEORY
21. The equation is equivalent to
65 2. 2 . Observe that ( x  y ) + (y  z ) = x z . On the other hand, 2 can be written as a product of three distinct integers in the following ways i) 2 = (1) . (1) . 2, ii) 2 = 1 . 1 . 2, iii) 2 = (1) · 1 · (2). SOLUTIONS
a) We have the cases:
1) x pq = 1, y pq = p2 q2 2 , so X = 1 + y = pq(l + pq). 2) x pq = p, y pq = pq , so X = p(l + q), y = pq(l + q). 3) x pq = q,2 y pq = p2 2q, so x = q(l + p), y = pq(l + p) . 4) x pq = p , Y pq = q , so X = p(p + q), Y = q(p + q). 5) x  pq = pq, y pq = pq, so x = 2pq, y = 2pq. The equation is symmetric, so we have also: 6) x = pq(l + pq), Y = 1 + pq. 7) x = pq(l + q), Y = p(l + q). 8) x = pq(l + p), Y = q(l + p). 9) x = q(l + q), y = p(p + q). (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 45, Problem 3486) 
pq,


b)


l.
Xn+l = 2xn + Yn
and
Yn+l = Xn + 2Yn
for all n � By induction we obtain that
x = xp", y = YP,,' = 3, = y2 = . Alternative solution. p q p 3. pq = 2k + 1, 2k+1 = (pq + 1 pq  1 ) ( x, y, ) =  ' ' p, q satIsfy the equatIOn. (Dorin Andrica)2 2
Denote by Pk the kth prime number. Then U v Pk is a V solution of equation x2 U for any integer k > O Let and be two arbitrary primes, � Then for some positive integer k. Because (k+ 1) 2  k 2 , it follows that all quadruples
{ x� � �z =: 21 { y: = z: =: l=� { xxy  yzz === l1l
Since in the first case any two factors do not add up to the third, we only have three POSS b i e so (
x,y, z) = (k + 1, k, k  1) for some integer k;
so (
x, y,z) = (k  1,k + 1, k) for some integer k;
so (x, y, z ) = (k, k  1, k + 1) for some integer k.  z = 2. Revista Matematica Timi§oara (RMT) , No. 12(1989), pp. 97, yAndreescu, (Titu Problem 2) c)
24. We have
xyz  (xy + yz + zx) + x + y + z 1 = 1, and, consequently, ( x  1) ( y  1) (z  1) = 1. Because x, y, z are integers, we obtain x  I = y 1 = z  1 = 1, so x = y = z = 2. (Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 3(1971), pp. 26, Problem 487) 25. Multiplying by 8 yields 

•

.
u, v
23. The equation is equivalent to (
.
x  y) (x  z) (y  z) = 2.
x + y)4 _ (x _ y)4 = (2z)4 , and so ( x  y ) 4 + (2z) 4 = (x + y ) 4 . This is Fermat's equation for the case = 4 and it is known that this equation has solutions only if x y = 0 or 2z = 0 (see T. Andreescu, D. Andrica, "An Introduction to Diophantine Equations" , GIL Publishing House, 2002, pp. 8587). or
(
n
2.
66
Case I.
If
x
2 . 2.
NUMBER THEORY
Y = 0, then x = Y and Z = ±x. The solutions are x = y = m,
for any integer m. Case II. If z = are
is a solution to the equation
z = ±m
x = O,
y = m,
z=o
x = m,
y = O,
z=o
for any integer m .
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1978), Problem 2813; Gazeta Matematica (GMB) , No. 11(1981), pp. 424, Problem 0:264) 26. Consider the sequences (Xn ) n �l' (Yn ) n�l' (Zn ) n�l, defined by Xn+2 = 592xn ' Xl = 1, X2 = 14 Yn+2 = 592Yn ' YI = 3, Y2 = 39 Zn+2 = 592zn , Zl = 7, Z2 = 42 for all � 1. It is easy to check that x� + y� + z; = 59 n , for all integers � 1. (Dorin Andrica, Romanian, Mathematical Olympiad  second round, 1979, Revista Matematica Timi§oara (RMT) , No. 12(1980), pp. 58, Problem 4075) n
27. Observe that the equation has infinitely many solutions in distinct nonnegative integers, for example
(1)
Xk = (1 + kn)n 2 , Yk = k(l + kn )n2 , Zk = (1 + kn)n l , for any integer k � O . Let (Xk l , Ykl , Zkl ) and (Xk2 ' Yk2 ' Zk2 ) be two solutions of equation (1) with kl =1= k2 . Thtm and mUltiplying yields
(2)
has infinitely many solutions in distinct nonnegative integers, for example
and
(Xkl Xk2 )n + (Xkl Yk2 )n + (Ykl Xk2 )n + (Ykl Yk2 )n = (Zkl Zk2 )n1
This means that
kl =1= k2
Since are arbitrary positive integers, the conclusion follows. For the second equation, the proof is similar, based on the fact that the equation
0, then (x  y)4 = (x + y)4 and so x = 0 or y = O . The solutions
n
67
SOLUTIONS
Xk = 1 + kn , Yk = k(l + kn), Zk = 1 + kn , for any integer k � O . (Dorin Andrica) 28. It is clear that x = 0, Y = 0 is a solution to the equation xn + yn = xn l + yn l Let a =1= 1 be a rational number such that Y = ax. Hence
1 + annl ' Y = a 1 + annl 1+a 1+a U sing the symmetry of the equation, we also have the solution 1 + ann l , Y = 1 + annl x=a 1+a 1 +a with a =1= 1 rational. a =  1 and > 1, then again x = Y = O . This concludes the solution. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1981), pp. 62, Problem 4578) 29. Let x = 3u, x + 2 = 3v , x + 8 = 3t so + + t = y . Then 3v  3u = 2 and 3t  3u = 8. It follows that t 3U (3V U  1) = 2 and 3U (3 u 1) = 8. Hence = 0 and 3v  1 = 2, 3t  1 = 8, therefore = 1, t = 2. The solution is x = 1, Y = 3. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 47, Problem 2812; Gazeta Matematica (GMB), No. 12(1980), pp. 496, Problem 18541) 30. If x, Y � 2 then 1 + xl and 1 + yl are both odd and (x + y) l is even. Hence
so
x=
If
n
U
v

u
the equation has no solutions.
v
2. NUMBER THEORY
2.2. SOLUTIONS
1. The equation becomes 2(1 + yl) = (1 + y)l and it is not difficult to notice .the solution y = 2. If y � 3, then 3 divides (1 + y) l but not 2(1 + yl) and y = 1 does not satisfy the equality. Hence x = 1, y = 2 or x = 2, = 1 due to the symmetry of the equation. (Titu Andreescu, Revista Matematidl. Timi§oara (RMT) , No. 2(1977), pp. 60, Problem 3028; Gazeta Matematica (GMB), No. 2(1980), pp. 75, Problem 0:118)
1, x = 2 or x = 3 and all those cases lead to a contradiction. 3, then xl = 2vl  8. Then v � 3 and x! = 2 (2VI   1) � 2 7. It follows that x � 5 and because x = 5 does not yield a solution and x � 6 implies 23 (2VI 3  1) 0 (mod 16), which is false, we do not obtain a solution here. In case II we have found only x = 1, y = 1, z = 2, v = 2,
68
Consider the case x =
Y
31. Without loss of generality we may assume that x � y � z. The equation is
Hence x = If
3
y =
 1) . . . (z + 1) + y (y  1) . . . (z + 1) + 1] = 2vl . If z � 3, then the righthand side is divisible by 3 but the lefthand side is not, so z � 2. We have two cases. I. z = 1. Then we have xl + yl = 2vl  1, or yl [x(x  1) . . . ( y + 1) + 1] = 2 vl  1. y � 2, then the righthand side is an even number but the lefthand side is odd, so y = 1. Then xl = 2(2VI  1  1). If x � 4, then 2(2VI  1  1) 0 (mod 8), false. It remains to examine the cases x = 1, x = 2, and x = 3. If x = 1, then 1 = 2(2VI  1  1), impossible. If x = 2, then 2 = 2(2VI  1  1) or vI = 2, so v = 2. x = 3, then 2vl  1  1 = 3 or vI = 3, false. Hence the only solution in this case is x = 2, = 1, z = 1, v = 2 II. z = 2. Now we have xl + y! = 2 vl  2, or y! [x(x  1) . . . (y + 1) + 1] = 2(2VI  1  1). If y � 4, the 2(2VI  1  1) 0 (mod 8), false. It follows that y = 1, y = 2, or y = 3. y = 1, then xl = 2vl  3. Since x � 2 implies 2vl  3 0 (mod 2) false, then x = 1, v = 2. If y = 2, then xl = 2vl  4. We must have v � 3 so xl = 4(2VI 2  1). If x � 3, then 4(2VI 2  1) 0 (mod 8), false. If
x=
If
==
==
•
2,
Y
=
1,
z=
1, v = 2
and, due to the symmetry of the equation, we also have
1, x = 1, x=
and
2, y = 1, y=
v=2 z = 2, v = 2. z = 1,
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1981), pp. 62, 4576)
Problem
32. The equation is equivalent to or XI [X2 . . . Xn
Hence X l = 1 and X 2 [X3 . " Xn
 (n  l) x  (  1) x n
Since X2 f:. Xl , it follows that X3 [X4 . " Xn
Y
==
3
which does not satisfy the condition x � y � z . To conclude, we have the solution  from case I 
If
==
3
==
equivalent to
zl [x (x
69
2 . . . Xn l
3 . . . Xn  l
=
X2
 (n  1) x
2 and
4 . . . Xn l
 ...  2X2  1] = 1.  . . .  3X3  2] = 2.  . . .  4X4  3] = 3. = 3.
Because X3 f:. X2 and X3 f:. Xl , we obtain X3 Continuing with the same procedure we deduce that Xk = k for all k. Turning back to the equation we find the identity
Remark.
1 + I! . 1 + 2 ! . 2 + . . . + (  1)!(n  1) = nIno (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 3(1973), pp. 23, Problem 1509) 33. We proceed by induction on For n = 1 the claim is true. Using the identity (x ) = (xu + )  ( xv + ) )( n
2
y
2
n.
2 2 U v
YV
2
YU
2
2.
70
and the fact that the claim holds for as desired. We have
n we deduce that the property is valid for + 1, n
Alternative solution. n n n Pn = kII=l (a�  b�) = kII= l (ak  bk) kII= l (ak + bk) = where An , B n are integers. Hence
as claimed.
(
Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1975), pp. 45, 2239; Gazeta Matematidi (GMB ) , No. 7(1975), pp. 268, Problem 15212)
Problem
34. Subtracting the equalities yields
(x + y  xy) + (v + y  vt) = (x + y  xy)(v + t  vt) , or [(x + y  xy) 1][(v + t  vt)  1] = 1, so (1) (1 x)(l  y)(l  v) (l  t) = 1. It follows that 11  xl = 11  yl = 1 1  v i = 1 1  tl = 1, and using (1) we obtain (x, y, z, t) = (0,0,0,0), (0,0, 2,2), (0,2,0,2), (0,2, 2,0), (2,0,0,2), (2,0,2,0), (2, 2,0,0) and (2,2, 2,2). Turning back t o the system we obtain (x, y, z, v, t) = (0,0,0,0,0), (0,0, 4, 2, 2), (0,2,0,0,2), (0,2,0,2,0), (2,0,0,0, 2), (2,0,0,2,0), (2, 2, 4,0,0) and (2,2, 24,2,2) ( Titu Andreescu, Revista Matematica Timi§oara (RMT ) , No. 2(1978), pp. 46, Problem 3431; Gazeta Matematica ( GMB ) , No. 5(1981), pp. 216, Problem 18740) 

35. We start with a useful lemma.
a and b are relatively prime positive integers, then there are positive integers u and v such that au  bv = 1. Lemma. If
2. 2 .
NUMBER THEORY
71
SOLUTIONS
Proof. Consider the numbers (1) 1 . 2, 2 · a, . . . , (b  1) . a When divided by b the remainders of these numbers are distinct. Indeed, otherwise we have kl f. k2 E { I, 2, . . . , b  I} such that kla =Plb+ r, k2 a = P2 b+r for some integers PI , P2 ' Hence a b Ikl  k2 1 = (mod b), which is 1 I kl  k2 1 b. (1) is divisible by b. Indeed, if k 2, . k . a = P . b for some integer p. Let d be the greatest common divisor of k and p. Hence k = k l d, P = PI d, for some integers PI ,kl with ged(pI ' kl ) = 1. Then kl a = P I b and since ged( a, b) = 1, we have kl = b, PI = a. This is false, because kl b . It follows that one of the numbers from (1) has the remainder 1 when divided by b so there is u E { I , 2, . . . , b I} such that au = bv + 1 and the lemma is proved. We prove now that the system { axbx yt  yz  e = O +d= with a, b, e, d nonnegative integers and ged(a, b) = 1 has at least a solution in nonneg ative integers. Because ged(a, b) = 1 using the lemma, there are positive integers u and v such that au  bv = 1. Hence x = cu + dv, y = ad + be, z = v, t = u, is a solution to the system. ( Titu Andreescu, Revista Matematica Timi§oara ( RMT ) , No. 2(1977), pp. 60, Problem 3029)
Since and are relatively prime it follows that false because � < On the other hand, none of the numbers listed in so, then there is E {I, . , n I} such that .
°

<

°
36. Consider the determinant
1 Xl �= Xpl  l
1 X2 Xp2 l
1 xp X�  l
P
II (Xi  Xj )
i , i=l i>i
72
2.
NUMBER THEORY
SOLUTIONS
Summing up all columns to the first one and applying the hypothesis yields A == 0 p
p), hence II ( ) 0 (mod p) . Because P is a prime number, it follows that there are distinct positive integers k, I E {I, 2, . . . ,p} such that Xk 0 (mod p) . (Dorin Andrica) (mod
',j= 1 i >j
Xi  Xj
==
 Xl ==
37. Let
a = d(al, a2 , . . . , an ) and b = d ( al +2 a2 ' a2 +2 a3 ' . . . , an +2 al ) Then ak = aka, for some integers ak, k = 1,2, . . . , n. It follows that ak + ak+l ak + ak+l a, (1) 2 2 where an+l = al and an+l = al. Since ak are odd numbers, ak are also odd, so ak + ak+ l are mtegers. . 2 ak + ak+1 for all so a divides b. From relation (1) it follows that a divides ak + ak+1 = f3kb, for some 2mtegers . f3k. Then On the other hand, 2 ak + ak+ l 0 (mod 2b) for all k E {I, 2, . . . , n}. Summing up from k = 1 to k n yields 2(al + a2 + . . . + an) 0 (mod 2b) Since n, al, a2 , . . . , an are all odd al + a2 + . . . + an ¢ 0 (mod 2), hence al + a2 + . . . + an 0 (mod b) . (3) Summing up for k = 1,3, . . . , n  2 implies al + a2 + . . . + an l 0 (mod 2b) and furthermore al + a2 + . . . + an I 0 (mod b) . (4) 0 (mod b), then using relation (2) we obtain Subtracting (4) from (3) implies a ak 0 (mod b) for all k. Hence bla andn the proof is complete. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1978), pp. 47, Problem 2814) gc
gc
=
==
=
==
==
==
==
==
==
38. It is known that
cp(kl) = cp(k)cp(l) for any relatively prime positive integers k and l .
73 2 . 2. On the other hand, it is easy to see that if p is a prime number, then cp(pl ) = pl pl  I for all positive integers Let n 2 · 3m , where is a positive integer. Then cp(n) = cp(2 . 3m ) cp(2)cp(3m ) = 3m  3m 1 = 2 · 3m 1 �3 for infinitely many values of n, as desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1(1978), pp. 61, Problem 3255) 39. For n 1 to n = 6 it is easy to check the claim. For n � 7 note that the number of even positive integers less than n is [�]. Moreover, the number of positive multiples of 3 less than n which are odd is [iJ  [i]. Then 7r (n) < n  [�J  ([iJ  [i]) , n � 7. Since x  I < [x] � x, it follows that ( n) < n  (�  1)  (�  1) + � = � + 2 ' 2 3 6 3 as desired. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 1(1978), pp. 61, Problem 3256) 40. Summing the inequalities Pk+1  Pk � 2 from k = 1 to k = n  1 yields Pn  2 � 2 (n  1) and so Pn > 2n  1, n � 1 . Then PI + P2 + ' ' ' +Pn > 2 n(n2+ 1)  n = n2 . The inequality a\" + ar : . . . + a::' ;::: ( a, + a2 : . . . + an ) m holds for any positive real numbers al, a2 , . . . , an ' Hence PmI + Pm2 + . . . + Pmn � n (Pl + P2 +n . . . +Pn ) m > n ( n;2 ) m = nm+ , as desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 45, Problem 3483) 41. Let S = L d and let cp(n) be the number of numbers less than n and relatively prime with n. _
I.
m
=
=
=
=
7r
1
d <2n gcd(d, n ) = l
2.
74
2. 2.
NUMBER THEORY
{:}
{:}
cp( )
cp( )
cp( ) 
hence d<n gcd(d, n ) = l
On the other hand,
d = ncp(n) 2 .
d = L (n + d) = ncp(n) + L d = = 3ncp(n) = ncp(n) + ncp(n) . 2 2 Therefore 3ncp(n) = ncp(n) 2 + 2 = 2ncp(n). (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1981), pp. 61, Problem 4574) 42. We proceed by induction. For n = 3 the claim is true. Assume that the hypothesis holds for n  1. Let 1 < k < n! and let kl , q be the quotient and the remainder when k is divided by n. Hence k = k n + q, 0 � q < n and 0 � kl < � < n n! = (n  I)!. n From the inductive hypothesis, there are integers di < d� < . . . < d�, s � i = 1,2, . . . , s and kl = di + d� + . . . + d�. Hence n  1, such that. .dil(n I)!, k = ndi + nd� + . + nd� + q. If q = 0, then k = d1 + d2 + . . . + ds, where di = ndi, i = 1,2, . . . , s, are distinct divisors of n!. If q f. 0, then k = d1 + rh + . . . + ds+1 , where di = ndL iI, 2, . . . , s, and ds+1 = q. It is clear that dil n !, i = 1, 2, . . , s and ds+1 I n !, since q < n. On the other hand, dS+1 < d1 < d2 < . . < ds, because dS+1 = q < n � ndi = d1 . Therefore k can be written as a sum of at most n distinct divisors of n!, as claimed. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1983), pp. 88, Problem C4:10) L
d<n gcd( d , n ) = l
n < d <2n gcd( d , n ) =l
d<n gcd(d , n ) = l
s
i
.
.
75
43. If n � 992, take the set with all 992 odd numbers from {I, 2, . . . , 1984}.
Note that
gcd(n, d) = 1 gcd(n, n  d) = 1 gcd(n, n + d) = 1 (1) Let d1 , d , . . . ,d be the numbers less than n and relatively prime with n. From (1) we deduce2 that n d1 + d n = n d2 + d n l = n
SOLUTIONS
Its complementary set has only even numbers, any two of them not being relatively Let be the complementary set of a subset with elements prime. Hence � of the set Define D E If n D 0, then C U D has elements but u D C which is false. Hence n D f. 0, so there is an element E n D. It follows that E and + E and since and + are relatively prime we are done. Revista Matematica Timi§oara (RMT) , No. pp. Problem
n 991. c 991 {I, 2, . . . , 1984}. = {c + 11 c C}. C = C {I, 2, . . . , 1985}, 2 · 993 = 1986 C a C a C a a 1 a 1 C (Titu Andreescu, 1(1984), 102, C4:7) 44. Let a be the greatest odd integer such that a2 < n, hence n < (a + 2) 2 . If a � 7, then a  4, a  2, a are odd integers which divide n. Note that any two of these numbers are relatively prime, so (a  4)(a  2)a divides n. It follows that (a 4)(a  2)a < (a + 2) 2 so a3 6a2 + 8a < a2 + 4a + 4. Then a3  7a2 + 4a  4 � 0 or a2 (a  7) + 4(a  1) � O. This is false, because a � 7, hence a = 1,3 or 5. If a = 1, then 1 2 � n < 32 , so n E {I, 2, . . . , 8}. If a = 3, then 32 � n < 52 and 1 . 31 n , so n E {9, 12, 15, 18,21, 24}. If a = 5, then 52 � n < 72 and 1 . 3 . 51 n so n E {30,45}. Therefore n E {2,3, 4,5,6, 7,8,9, 12,15, 18,21,24,30,35}. (Dorin Andrica and Adrian P. Ghioca, Romanian Winter Camp 1984; Revista Matematica Timi§oara (RMT) , No. 1(1985), pp. 78, Problem T. 2 1) 45. We prove by induction that
u�  2v� = 1, n � 1.
(1)
n = 1 the claim is true. Assuming that the equality is true for some n, we U�+l  2V�+1 = (3un + 4Vn)2  2(2un + 3Vn )2 = u�  2v� = 1 hence (1) is true for all n � 1. We prove now that (2) 2x�  Y� = 1, n � 1 Indeed, 2x�  Y� = 2(un + Vn )2  (un + 2vn)2 = u�  2v� = 1,
For have
as
claimed. It follows that
(Xn V2  Yn) (Xn V2 + Yn) = 1, n � 1. Notice that xn ..,fi + Yn > 1 so 0 < xn V2 Yn < 1, n � 1. Hence Yn = [Xn V2] , as claimed.
2.
76
2.2. Taking into account that al = 1, a 2 = 3 and inducting on SOLUTIONS
NUMBER THEORY
(Dorin Andrica, Gazeta Matematica (GMB), No. 11(1979), pp. 430, Problem 0: 9 7) 46. (i) We have
(1) Xk  2 = Xk l (Xk l  2). MUltiplying the relations (1) from k = 2 to k = yields Xn  2 = Xn l ... X2 Xl (Xl  2), and since Xl 3, we obtain Xn = Xl X2 ... Xn l + 2 (2) for all 2:: 2. For a different proof use the identity X2x I 1 = nIIl (X21o  1 + 1).  k= l (ii) From relation (2) we obtain gcd(x n ' xd gcd(xn ' X2 ) = . . . = gcd(xn , Xn  l ) 1 for all hence gcd(Xk' xz) = 1 for all distinct positive integers k and l . (iii) Since X2 5 and X lX2 ... X n  l is odd, using relation (2) it follows that X n end with digit 7 for all integers 2:: 3. n
=
n
..
=
1
n,
=
(Dorin Andrica)
n
n we reach the conclu
sion.
(Titu Andreescu, Gazeta Matematica (GMB), No. 11(1977), pp. 453, Problem
16947; Revista Matematica Timi§oara (RMT) , No. 1 (1978) , pp. 51, Problem 2840) 48. For 2 we have a2 1 +2 2 = 2 Suppose that an = Pn , where Pn , qn are integers and gcd(pm qn ) = 1. qn Then an+l = qn22p+nq2nPn2 ' and it suffices to prove that gcd(2Pn qn ' q; + 2p�) = 1 . Assume by way of contradiction that there are integers d and kl ' k2 such that 2Pnqn = kld and q� + 2p� = k2 d. If d = 2, then q; = 2k2  2p�, so qn is even. This is a contradiction since ql , q2 , · .. , qn are all odd. If d 2 then q� = (k2qn  kl Pn )d so dl qn ' Since q� + 2p� k2 d it follows that dlPn , which is false, because gcd(Pn ' qn ) = 1. This proves our claim . n
hence
77
=
=
3'
>
=
We prove that
q�  2p� = 1, 2:: 2. For 2 we have q�  2p� = 32  2 . 22 = 1. Suppose that q;  2p� = 1. Because Pqn+l+l q�2p+nq2pn � n n
n
(1)
=
=
47. We have
is irreducible, we obtain
Then Hence
and
a�+2 + a�+l  4an+2an+l =  2, � 1. a�+2  a�  4an+l (an+2  an ) = 0 and (an+2  an)(an+2 + an 4an+d = 0, 2:: 1 . Since the sequence (a n ) n 2: l is increasing, it follows that n
S ubtracting these relations yields
n
as claimed. Hence
1 2 q2 q�2l 2 = q�  l bn = 1  21a2nl 1 2Pq�nll n l Pn l and so bn is a perfect square for all 2:: 2. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 63, Problem 3083; Gazeta Matematica (GMB) , No. 1(1981), pp. 44, Problem C:88) _
n
=
2.
78
2. 2.
NUMBER THEORY
49. Inducting on n we obtain
Subtracting these relations implies
x;  2y; = 1, n 2:: 1.
Hence
79
SOLUTIONS
= x� + ; = hence (1) = + X2nl , n 2:: 1. We induct on n to prove that (2) n 2:: 1. = + Indeed, X2 = + and assume that (2) is true up to n. Then = 2 2 2 + 2 = ( + Xn } 2  X2n  Xn2 +l + = � + x; = X2n+l , ( as claimed (the equality ( ) holds because of (1) and the induction hypothesis ) . From relations (1) and (2) it follows that = Xn+l +xn for all n 2:: o. Because 's sequence, hence is the Fibonacci = 1 the sequence (x ) = 0 and Xl n n2:0  vs + vs = )g (e 2 f  e 2 f) . (Titu Andreescu, Romanian Winter Camp 1984; Revista Matematica Timi§oara (RMT) , No. 1(1985), pp. 73, Problem T. 3) 52. From the hypothesis it follows that = 12, = 25, = 48. We have l ' = 5 ' 6 = 8 so n = Fn for all n = 1,2,3,4,5,6, where ' 4 = 3 ' 's5sequence. 2  1 ' is3 the= 2Fibonacci for all n. Indeed assuming that k = kFk We prove by induction that = for k ::; n + 3, we have = 2(n + ) + (n + )  2(n +  nFn = = 2(n + 3)Fn+3 + (n + 2)Fn+2  2(n +  n (Fn+2  Fn+l ) = = 2(n + 3)Fn+3 + 2F 2  (n + 2)F = = 2(n + 3)F 3 + 2Fn+2  (n + 2)(F +3  Fn+2 ) = = (n + 4)(Fn+3 + Fn+2 ) = (n + 4)Fn+4 , as desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1(1986), pp. 106, Problem C8: 2 ) 53. Observe that setting o = 0 the condition is safisfied for n = O. We prove that there is integer k ::; m 3 such that Xk divides m. Let rt be the remainder of when divided by m for t = 0,1, . . , m 3 + 2. Consider the triples (ro, rl , r2 ) , (rl , r2 , r3 ) " .. , (rm3, rm3+2 ) . Since rt can take m values, it follows X2n+ 1  X 2n
Zn = 1 + 4x;y; = (x;  2y�)2 + 4x�y; = (x� + 2y�)2  4x;y� = = (x; + 2y� + 2xnYn)(x; + 2y�  2xnYn), for all n 2:: 1. Since both factors are greater than 1, it follows that all numbers Zn are not prime. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1976), pp. 54, Problem 2571}
X2 n
X2n+ 1
X2n1
X2n
Xl
X2n+2  X2n
X2n2 ,
Xo
X n+2  X n  X n+ l + Xn+l  Xn)
50. We have
2
X nl
(*)
Xn+ l
X +l
*
or
X2n l ,
X _l
Xn+2
Xo
Xn
Hence and Xn
= Jp2 +
From hypothesis we have Xn+ l >
= Jp2 +
lXn+ 1 ± P Xn ,
hence
J;
+ 1.
J;
+ 1.
X +1
(1) But (2) and so (3) by summing up the relations and (2). Because p is a positive integer, it follows that Jp2 + 1 is irrational. From (3) we deduce that among any three consecutive terms of the sequence there is at least an irr�tional term. Hence there are at least [;] irrational terms among the first m terms of the sequence. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 12(1980), pp. 68, Problem 4139; Gazeta Matematica ( GMB ) , No. 6(1980), pp. 281, Problem C: 48) 51. Setting n = 0 and n = 1 yields = x� and X2 = x�, hence = X2 = 1. Xn
lXn+ 1  P
X +1
( I)
Xl
From the given condition we obtain
as
a4
Xl
a2
_
a4
a3
as
al
a6
an
a6
(Fn )n > l
an
an+4
a
nFn
3 Fn+3
2 Fn+2
I)Fn+1
I)Fn+ 1
n+ 1
n+
n
n+
X
Xt
.
rm3 + 1 ,
2.
80
2. 2 .
NUMBER THEORY
by the Pigeonhole Principle that at least two triples are equal. Let p be the smallest number such that triple is equal to another triple p < q :::; m3 • We claim that p = Assume by way of contradiction that p 1. Using the hypothesis we have
(rp ,rp+1 ,rp+2 ) O.
(rq,rq+1 ,rq+2 ),
2::
rp+2 rp 1 + rprp+1 (mod m) and rq+2 rq  1 + rqrq+1 (mod m). Since r rq, r +1 rq+1 and r + rq+2 , it follows that rp 1 rq  1 so (rp 1 , rp , rpp+d (rq p 1 , rq, rq+ d, whichpis2 a contradiction with the minimality of Hence 0, so rq ro 0, and therefore Xq 0 (mod m). (Titu Andreescu and Dorel Mihet, Revista Matematica Timi§oara (RMT) , No. 1(1986), pp. 106, Problem C8: 1) 54. Note that a2 1, a3 4, a4 9, a5 25, so ao F(f, a1 F12 , a2 Fi, a3 Fi, a4 Fi, a5 Fl, where (Fn ) n2:o is the Fibonacci sequence. We induct on n to prove that an F� for all n 2:: O. Assume that a k F� for all k :::; n. Hence (1) ==
b
=
c2 yields (c + � ) 2
c + � 4. Let 2 + .;3. We will prove by an � (C4Fn + C4�n ) ' n � 1,
=
induction that
Fn
p=
=
=
=
=
=
=
=
=
=
nth
where is the Fibonacci number. Indeed, this is true for = 1, =
n n 2 assuming that ak HC4F• + � ) k $; n =
and,
4 . c
,
p.
==
=
=
=
=
=
=
c
=
16, thus =
==
=
81
SOLUTIONS
=
=
=
=
and
1 2 2 + J2 + 2an 2 + e2Fn + e21Fn (eFn + eFn ) Let us mention that em + � is an integer for all positive integers m. em Olympiad Shortlist, 1997) (Titu Andreescu, USA Mathematical 
=
From the given relation we obtain
=
56. We induct on m to prove that
and
=

(2) we obtain an+1 2F� 2F�_ 1  F�_2 (Fn + Fn_d2 + (Fn  Fn_ 1 ) 2  F�_2
(2)
Using the relations (1) and
=
+
=
55. The expressions a2  1 and 2 + 2a for the substitution a � (b2 + b� ) 2 196, hence b + � 14. Setting The equality � ( b2 + � ) 97 implies ( b + � ) b =
.
=
+
=
<
<
=
=
Problem C8:8)
=
=
=
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1986), pp. 108, =
=
<
as desired.
ask
=
<
=
Summing up these equalities yields
=
(1) km :::; n km+1 , then an 1 + m. For m 0 the claim is true, since if 1 :::; n < k, then [ � ] 0 hence an l ao 1 + 0. Assume that (1) holds for m. Then if km+1 :::; n k m+ 2 , we have km :::; � k m+1 , so km :::; [�] km+1 . Using the inductive hypothesis, we deduce that a[]!] 1 + m so an 1 + a[ �) 1 + (m + 1) as claimed . Therefore for any positive integer n with m :::; logk n m + 1, we have an 1 + m thus an 1 + [log kn). (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1982) , pp. 66, if
an 3an  1 + an2 2(I) n r, n 2:: 2.
<
=
=
Problem 4997) all
57. We prove that an 2 n 2:: 1. This is clear for n
o
b� , where b 0 and = 1.
=
=
n
=
1,
b1
=
1 and
bn+1 3bn  bn 1 for =
82 2. Assume that for any k � n we have ak  2 = b�. Subtracting the relation an = 7an l  an2 from an+l = tan  an  l yields an+l = 8an  8an l + an 2 or an+l  2 = 8(an  2)  8(an l  2) + (an 2  2) Hence an+l  2 = 8b�  8b�_ 1 + b�_2 = 8b�  8b�_1 + (3bn1  bn ) 2 = = 9b�  6bn bn1 + b�_l = (3bn  bn_ 1 ) 2 = b�+l Alternative solution. The general term of the sequence is given by ( v'5) 4n2 + ( 1 _ v'5) 4n2 n 2:: 0 an = 1 +2 ' 2
2. 2.
NUMBER THEORY
Hence
( 1 + 15) 2nl + ( 1  15) 2nl
and an inductive argument proves that is an integer for all n.
2
2
From the relations (1) and (2) we deduce that
that
59. Assume by way of contradiction that there is an irrational number a such
is rational. Define a = Va + Ja2  1 and observe that Va  Ja2  1 = .!a . Hence A = a + .!a is rational. We prove that a n + � an is rational. Indeed, 2
U
a2 + a_12 = (a + a1 )  2
3
+ :3 = ('" + � r  3 ('" + �) is rational.
sing the identity
1 1 ak + �ak = (ak  1 + _ ak 1 ) (a + .!a ) _ (ak2 + a k2 ) it follows by induction that ak + \a is rational for all positive integers k, hence na + a1n . ratlOna Thus a + Ja2  1 + a  Ja2  1 = 2a is rational, which is false. The solution is complete. (Titu Andreescu, Romanian IMO Selection Test 1977; Revista Matematidi Timi§oara RMT) , No. 1(1978), pp. 78, Problem 3344) 60. We induct on n. For n = 1, from ±1 ± 2 ± 3 ± 4 ± 5 we obtain all odd positive integers less than or equal to (2 + 1)(4 + 1) 15: +1  2 + 3 + 4  5 = 1 1 + 2 + 3 + 4  5 = 3 1 + 2 +3  4 + 5 = 5 _
IS
.
I.
(
The last equation has no rational roots since the discriminant = a2 + 4/3 is not a perfect square. This is a contradiction and hence the problem is solved. Alternative solution. From the condition of the problem we obtain aXn = Xn+l  /3Xn l , n 2:: 1. (1) d
(2)
(a? + 4(3)x�o = (Xno+l + /3xno  d2 which is false, since a2 + 4/3 is not a square. This completes the proof. (Dorin Andrica)
",
58. Note that all terms of the sequence are positive integers. Assume by way of contradiction that there is a positive integer no such that
X�o = Xno  lXno +l . Then Xno+l  �  t, xno Xno  l where t is rational. Because Xno+l = aXno + /3Xno  l , it follows that Xno+l = a + /3 � xno Xn o  l or t2  at  /3 = O.
Assume by way of contradiction that there is a positive integer no such that
is rational and
(Titu Andreescu)
83
SOLUTIONS
=
84
2.
NUMBER THEORY
1 + 2  3 + 4 + 5 = 7 1  2 + 3 + 4 + 5 = 9 +1  2 + 3 + 4 + 5 = 11 1 + 2 + 3 + 4 + 5 = 13 +1 + 2 + 3 + 4 + 5 = 15 Assume that from ± 1 ± 2 ± . . . ± (4n + 1) with suitable choices of signs + and we obtain all odd positive integers less than or equal to (2n + 1)(4n + 1). Observe that (4n+2) + (4n+3) + (4n+4)  (4n+5) = O . Hence from ±1±2± . . . ± (4n + 5) for suitable choices of signs + and  we obtain all odd positive numbers less than or equal to (2n + 1)(4n + 1) . It suffices to obtain all odd integers m such that (1) (2n + 1)(4n + 1) < m � (2n + 3)(4n + 5) There are (2n + 3)(4n + 5)  (2n + 1)(4n + 1) = 8n + 7 2
such odd integers m. We have
(2n + 3)(4n + 5) = +1 + 2 + . . . + (4n + 5) (2n + 3)(4n + 5)  2k = +1 + 2 + . . . + (k  1 )  k+ +(k + 1) + . . . + (4n + 4) + (4n + 5), k = 1, 2, . . . , 4n + 5 and (2n + 1)(4n + 5)  2l = + 1 + 2 + . . . + (l  1)  l + + (l + 1) + . . . + (4n + 4) + (4n + 5), l = 1, 2, . . . , 4n + 1. Hence all numbers m from (1) are obtained, desired. (Dorin Andrica, Gazeta Matematica (GMB) , No. 2(1986), pp. 63, Problem C570) as
Chapter 3 GE OMETRY
PROBLEMS
1. Triangle ABC has side lengths equal to c. Find a necessary and sufficient condition for the angles of the triangle such that c2 can be the side lengths of a triangle. 2. Prove that the triangle whose side lengths are equal to the length of the medians of a triangle ABC has area equal to 3/4 of the area of triangle [ABC]. 3. In triangle ABC the median AM meets the internal bisector BN at P. Let Q be the point of intersection of lines CP and AB. Prove that triangle B NQ is isosceles. 4. In triangle ABC the midline parallel to AB meets the altitudes from A and B at points D and E . The midline parallel to AC intersects the altitudes from A and C at points F and G. Prove that DCI I BFI I GE. 5. Let M be a point in the interior of triangle ABC. Lines AM, BM, CM intersect sides BC,CA,AB at points A',B', C', respectively. Denote by the areas of triangles MA' B, MA' C, MB' C, MB' A, MC' A and MC' B, respectively. Prove that if a, b, 2 2 a ,b ,
81 , 82 , 83 , 84 , 85 , 86
81
82
+
83 84
+
85
3
86 
,
then M is the centroid of the triangle ABC. 6. Let M be a point in the interior of triangle ABC and let M, P, Q be three colinear points on sides AB, BC and line CA. Prove that if SMAN
8MBN
+
5MBP 8MCP
=
2J
SMAQ
8McQ
,
then MP is skewparallel with AC. 7. Let c and be the sides lengths and the area of a triangle AB C. Prove that if P is a point interior to triangle ABC such that a , b,
8
aPA + bPB + cPC 4 =
87
8
3.
88
then
3. 1.
GEOMETRY
8. Let 11 , 12 be the incenters of triangles AlBl Cl and A2 B2 C2 . Prove that if II
Al A�, A2 A� in the same ratios they divide the AlBl Cl and A2 B2 C2 are similar. BlBL B2 B�, 9. On side BC of a triangle ABC consider points M and N such that ifAjJ CAN. Prove that MB + NB 2 AB MC NC 2: AC 10. Let M be a point on the hypotenuse BC of a right triangle ABC and let points N, P be the feet of the perpendiculars from M to AB and AC. Find the position of M such that the length NP is minimal. 11. Let ABC be an equilateral triangle and let P be a point in its interior. Let the lines AP, BP, CP meet the sides BC, CA, AB at the points AI, Bl, Cl respectively. and 12 divide the internal bisectors then triangles internal bisectors
==
Prove that
S
be the set of all triangles
ABC for which
3 =6 5 ( AP1 + BQ1 + CR1 )  min{P, BQ, CR}
r'
P,
where T i s the imradius and Q , R are the points of tangency of the incircle with sides respectively. Prove that all triangles in S are isosceles and similar to one another.
AB, BC, CA,
13.
ABC ABC.
Let be a triangle inscribed in a circle of radius R, and let P be a point Prove that in the interior of
PA PB PC 1 BC2 + CA2 + AB2 2: R ' 14. Let I be the incenter of an acute triangle ABC. Prove that AI· BC + BI· CA + CI · AB if and only if triangle ABC is equilateral. 15. Let ABC be a triangle such that (8AB  7BC  3CA) 2 6(AB2  BC2  CA2 ). Prove that A 60° . = SABe
PB and PC are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to PA. Prove that BAlJ is acute. 17. Triangle ABC has the following property: there is an interior point P such that PAJj = 10°, PEA = 20°, PcA 30°, and PAC 40°. Prove that the triangle ABC is isosceles. 18. Let ABC be a triangle such that max{A,B} = C + 30°. Prove that ABC is right angled if and only if !!. J3 + 1. T 19. Prove that in the interior of any triangle ABC there is a point P such that circum radii of triangle PAB,PBC,PCA are equal. the 20. The incircle of the triangle ABC touches the sides AB, BC, CA at the points C', A', B', respectively. Prove that the perpendiculars from the midpoints of A'B', B'C', C'A' to AB, BC, CA, respectively, are concurrent. 21. Let AlA2 A3 be a nonisosceles triangle with the incentre I. Let Ci , i = 1,2,3, be the smaller circle through I tangent to AiAi+ l and Ai Ai+ 2 (the addition of indices being mod 3). Let Bi , i = 1,2,3, be the second point of intersection of Ci+ l and Ci+ 2 . Prove that the circumcenters of the triangles AlBlI, A2 B2 I, A3 B3 I are collinear. 22. On sides AB and AC of a triangle ABC consider points B' and C' such that AB' + AC' k is constant. B ' B C' C Find the locus of the intersection point of lines BC' and C B'. 23. Let ABC be an equilateral triangle of side length 1 . Find the locus of points P such that · PB · PC max{PA,PB,PC} = PA · PB +2PA PB · PC + PC · PA  1 . =
=
=
=
24. Prove that in any acute triangle
Ja2 b2  4S2 + Ja2 c2  4S2 = a2 .
25. Prove that a triangle in which r,;.
. a yT
=
=
89
16. Let P be a point in the plane of triangle ABC such ' that the segments PA,
P is the orthocenter of the triangle.
12. Let
PROBLEMS
is equilateral.
+ y. Tb + y. Tc r,;::
�
JTaTb Tc T
= 
3 . 1 . PROBLEMS
3. GEOMETRY
90
26. Prove that a triangle is equilateral if and only if
1 + 1 + 1  1 + 1 + 1 . m� m� m� r� r� r� 27. A triangle with side lengths a, b, c has circumradius equal to 1. Prove that a + b + c 2:: abc.
28. Prove that in any triangle
2
a � 6R L (p  b)(p  c) ;:.
29. Prove that in any triangle
1 + 1 + 1 2:: 1
h�
30. Prove that in any triangle Jrarb
h�
h�
3r2
+ Jrbrc + Jrcra 2:: gr.
31. Prove that in any triangle 32. Prove that
2p2 2:: 27Rr
for any triangle. 33. Let ABC be a triangle. Prove that A � i if and only if
(p  b)(p  c) � 4bc '
34. Three equal circles of radii r are given such that each one passes through the centers of the other two. Find the area of the common region. 35. Let ABCD be a nonisosceles trapezoid with bases AB and CD. Prove that
AC2  BD2 AB + CD AD2  BC2  AB  CD ' 36. Let ABC D be a trapezoid with bases AB and CD. Prove that if (AB + CD) 2 BD2 + AC2 =
91
then the diagonals of the trapezoid are perpendicular. 37. Prove that if in a trapezoid with perpendicular diagonals the altitude is equal to the midline, then the trapezoid is isosceles. 38. Let ABCD be a trapezoid with bases AB and CD. Prove that
AB22  BC22 + AC22 = AB22 _ AD22 + BD22 = AB CD  AD + AC CD BC +BD CD'
39. Prove that a trapezoid whose difference of the diagonal lengths is equal to the difference of nonparallel side lengths is isosceles. 40. Let ABCD be a cyclic quadrilateral. Prove that
lAB  CDI + l AD  BCI 2:: 21AC  BDl ·
41. Prove that a right trapezoid whose altitude length is equal to the geometrical mean of the lengths of its bases has perpendicular diagonals. 42. Let E and F be the projections of the vertices A and B of a trapezoid ABCD on line CD. Let M and N be the projections of E and F onto BD and AC, respectively and let P and Q be projections of E and F onto BC and AD, respectively. Prove that the quadrilateral MNPQ is cyclic. 43. Let ABCD be a convex quadrilateral that is not a parallelogram and let M and N be the midpoints of the diagonals AC and BD. Prove that numbers
AB + CD, BC +AD, AC +BD, 2MN
can be the side lengths of a cyclic quadrilateral. 44. Let ABCD be a cyclic quadrilateral and let lines AB and CD meet at point E. Point F is the reflection of C across E. Prove that lines AF and BD are perpendicular if and only if lines AB and CD are perpendicular. 45. Find all cyclic quadrilaterals having the side lengths odd integers and the area a prime number. 46. A cyclic quadrilateral has the side lengths a, b, c, d, the diagonal lengths f and the semiperimeter p. Prove that e,
3.
92
GEOMETRY
47. A cyclic quadrilateral has the side lengths and e I. Prove that if max{l l Ib  dl} ::; 1, then
a,
b, d and the diagonal lengths c,
ac ,
Ie

93
3 . 1 . PROBLEMS
II ::; vi
48. A cyclic quadrilateral has area S and semiperimeter p. Prove that if S = (�) 2 , then the quadrilateral is a square. 49. Let ABCD be a convex quadrilateral and let P be the intersection point of its diagonals. Prove that SPAB + SPCD = SPBC + SPDA if and only if P is the midpoint of AC or BD . 50. Let M be a point on the circumcircle of the cyclic quadrilateral ABCD and let points A', B', C', D' be the projections of M onto AB, BC, CD, DA, respectively. Prove that (i) lines A' B' , C'D' and AC are concurrent (ii) lines B' C', D' A' and BD are concurrent. 51. A convex quadrilateral ABCD with area 2002 contains a point P in its interior such that PA = 24, PB = 32, PC = 28, and PD = 45. Find the perimeter of ABCD . 52. Find the locus of points P in the plane of a square ABC such that
max(PA,PC) = v'21 (PB + PD) .
53. Let P be the set of all quadrialterals with the same diagonal lengths and let )'1 (P) and A (P) be the lengths of the segments determined by the midpoints of two opposite sides2 of a quadrilateral p E P . Prove that for all p E P the sum A� (P) + A� (P) is constant and find the value of this constant. 54. Let ABCDEFGHIJKL be a regular dodecagon and let R be the circumra dius. Prove that AB + AF = 4 and AB2 +AC2 +AD2 +AE2 +AF2 = 10R2 . AF AB
55. Prove that if inside a convex poligon there is a point such that the sum of the squares of its distances to the vertices of the poligon is twice the area of the poligon, then the poligon is a square. 56. Let Al A2 . . . An be a cyclic polygon and let P be a point on its circumcircle. Let Pl, P2 , . .. ,Pn be the projections of P onto the sides of the polygon. Prove that the product n PA� IIl P�� i= is constant. 57. Let AlA2 . . . A2n be a cyclic polygon and let M be a point on its circumcircle. Points Kl, K2 , . . . , K2n are the projections of M onto sides AlA2 ,A2 A3 , . . . ,A2n Al and points Hl,H2 , . . . ,Hn are the projections of M onto diagonals AlAn+ l,A2 An+2 , ' " ,An A2n . Prove that l
MKl · MK3 · . . MK2n l= MK2 · MK4 . . . MK2n = MHl · MH2 o o . MHn o 58. Find the circum radius of a cyclic polygon with 2n sides if n sides have the length and n sides have the length b. 59. Let P be a point in the interior of a tetrahedron ABCD such that its projec tions Al,Bl,Cl,Dl onto the planes (BCD), (CDA), (DAB), (ABC), respectively, a
are all situated in the interior of the faces. S is the total area and the inradius of the tetrahedron, prove that SBCD + SCDA + SDAB + SABC � PAl PBl PCl PDl When does the equality hold? 60. Let AlA2 A3 A4 be a tetrahedron, G its centroid, and A�, A�, A�, A� the points where the circumsphere of AlA2 A3 A4 intersects GAl, GA2 , GA3 , GA4 respectively. Prove that If
r
>

and
r
1 1 1 1 1 1 1 1 GA� + GA� + GA� + GA4 ::; GAl + GA2 + GA3 + GA4 °
SOLUTIONS
1. The positive real numbers a2 , b2 , c2 can be the side lengths of a triangle if and only if
(1)
Because
b2  c2 cosC a2 +2ab relation (1) is equivalent to cosA 0, cosB 0, cos C O. Hence the necessary and sufficient condition is that triangle ABC is acute. (Dorin Andrica, Revista Matematica TimiÂ§oara (RMT) , No. 2(1978), pp. 48, Problem 3507) =
>
>
'
>
2. Let A', B', C' be the midpoints of sides BC, CA, AB respectively. Construct point M such that BC MC' is a parallelogram. Note that AC' C M and BB' MA' are also parallelograms hence AM CC' and A'M BB' . Hence the triangle determined by the medians is AA'M. =
=
M
A
.cďż˝
A'
C Let N be the intersection point of lines B' C' and AA'. Because AC' A' B' is a parallelogram, we have C'N B'N !B'M 2 ' B
=
=
95
3.
96
3.2.
G EOMETRY
97
SOLUTIONS
SO B' is the centroid of triangle AA' M. Hence
A
Alternative solution: Consider the vectors = BC, b = AC, c = AB, ma = AA', mb = BB' and observe that a
We obtain
ma mb = 41 (b +c) (b  2c) = 43 (c b), . follows. hence I ma mbl = :t3 Ic bl and the concluslOn (Titu Andreescu) 3. If BP meets AC at P, then by Ceva's theorem we obtain QA MB NC QB . MC . NA = 1 . Because MB = MC, we have QA NA QB NC ' Therefore QNI I BC and then NBC = QiiB. On the other hand, because BN bisects angle ifiiC, it follows that NBC = QiiN. Hence QiiB = QiiN and triangle B NQ is isosceles, desired. (Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1(1978), pp. 66, Problem 3286) 4. Let H be the orthocenter of triangle ABC and let A' be the midpoint of side BC: Because FA'I I AC, it follows that FA' ...L BH. Moreover BA' ...L HF, so A' is the orthocenter of triangle BHF. Hence (1) HA' ...L BF. On the other hand, A'DII A B, so A'D ...L HC. Since HD A'D it follows that D is the orthocenter of triangle HA' C thus DC ...L HA' . (2) x
x
B
x
x
x
as
...L
���;
F
C
Note that GH ...L AB and EH ...L AC. Since ABI I A'E and ACI I GA', we have GH ...L A' E and EH ...L GA', so H is the orthocenter of triangle A' EG. Therefore (3) A'H ...L GE. From relations (1), (2) and (3) we obtain DCII BFI I GE, as desired. (Titu Andreescu) 5. Using Ceva's theorem, we obtain
C'A A'B · B'C C' B · A' C B ' A = 1 ' S1 . S3 . S5 = 1. S2 S4 S6
hence
The given condition now reads S1 S2
so
+ S4S3 + S5S6 = 3 V.s/S1S2 . SS43 . S6S5
This implies It
C'A = A'B = B'C = 1. C 'B A'C B'A follows that A', B', C' are the midpoints of the triangle's sides and M is the
centroid, as claimed. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1974), pp. 23, Problem 1904; Gazeta Matematica (GMB ) , No. 2(1979), pp. 63, Problem 0:16)
are strict.
6. Using Menelaus' theorem yields
NA AB QO NB . BC . QA = 1. SMAN 5MBA SMCQ 5MBN . SMCA . SMAQ  1 , SMAN 5MBP SMQA 5MBN . SMCP = SMCQ ' A
Hence or
N k
P
�.L .....
B
���� P C Q
From the condition in the hypothesis it follows that
SMAN 5MBP SMAN 5MBP SBMN + SMCP = 2 5MBN . SMCP , and so SMAN  5MBP ' SMCP 5MBN Thus NA PB PC NB hence NP and AC are skewparallel. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1978), pp. 66, Problem 3287; Gazeta Matematica (GMB), No. 2(1979), pp. 56, Problem 17607) . 7. Let AI, Bl , Cl be the feet of triangle's altitudes and let A', B', C' be the pro jections of P onto its sides. If P is the orthocenter of the triangle, then the equality holds. Assume by way of contradiction that P is not the orthocenter of the triangle and aPA + bPN + cPC = 4S. PA+ PA � AAl, PB + PB' � BBl, PC + PC' � COl
C
A'
It follows that
L________
Then at least two of the inequalities
A B'
_ _
B
99
3.2. SOLUTIONS
3. G EOMETRY
98
aPA + bPB + cPC > a(AAl  PA' ) + b(BBl  PB' ) + c(CCl  PC') or 4S > aAAl + bBBl + cCCl  (aPA' + bPB' + cPC' ). Hence 4 S > 6 S  2S = 4S , which is false. (Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 2(1978), pp. 74, Problem 3689; Gazeta Matematica (GMB), No. 2(1980), pp. 64, Problem 18122) 8. Using the angle bisector theorem and Van Aubel's theorem, it follows that
IA B'A C'A b + c lA' = B' C + C'B = a = kl and IB = A'B + C'B = c+a = k2 IB ' A' C C' A b Let IC  a+ b  k IC ' c and note that kl + 1 = �2p , k2 + 1 = b2p ' k + 1 = 2cp , where p is the semiperimeter. Hence 1 + 1 + 1 =1 kl 1 k + 1 k2 + 3 + 1 so k  klklk+ 2k2+1 2 . Furthermore, since + 1) (k3 + 1) 1 ' cos A = �2 [ kk23 ++ 11 + kk32 ++ 11 (k2 (kl + 1) 2 _
_
3
3
3
_
_
3.
100
3. 2.
G EOMETRY
kl k3 •
Hence cos A = cos Al and, analo and we observe that cos A depends only on gously, cos B = cos Bl , thus triangles ABC and AlBl Cl are similar. Romanian Regional Mathematical Contest " Grigore Moisil" , 1999; Revista Matematidi Timi§oara (RMT), 1999, pp.
(Dorin Andrica,
87)
101
SOLUTIONS
Similarly, we obtain Bl Cr ;:: BI A · CIA and Cl A� ;:: CI B · Al B. MUltiplying the three inequalities together, we obtain Now the lines AAl , BBl , CCI concur, so
9. Using Steiner's theorem, we obtain MB · NB MC · NC Applying the AMGM inequality yields MB MC
. NB JMB NB ;:: 2 + NC MC . NC
and after substituting and taking square roots, we have AI BI . BI CI . CI AI ;:: AlB · BI C · CIA, =
2
AB AC '
as desired. Note that the equality case occurs only if AM and AN coincide with the internal bisector of angle A.
(Titu Andreescu)
10.
The quadrilateral APMN is a rectangle, therefore NP = AM. Hence NP is minimal if AM l BC, so M is the foot of the altitude from A.
(Titu Andreescu)
11. Applying the law of cosines to triangle Al BI C we obtain and using the inequality get
x2 + y2  xy ;:: xy, which holds for all real numbers x, y, we Al B� ;:: Al C , Bl C. A
the desired inequality. Equality holds if and only if CAl = CBI , ABI = ACI and BCI = BAI , which in turn holds if and only if P is the center of triangle ABC. IMO 1996 Shortlist)
(Titu Andreescu,
12. We start with the following lemma.
Lemma.
Let A, B, C be the angle of a triangle ABC. Then A C C B B A tan + tan  + tan  tan  + tan  tan 2 2 2 2 2 2
=

1.
Proof. We present two arguments. First approach. Since tan(a + ,8) [1  tan a tan ,8] tan a + tan ,8, tan(90°  a) cot a l/ tan a, and A/2 + B/2 + C/2 90°, the desired identity follows from C A B C B B A tan  tan  + tan  tan  tan  (tan  + tan  ) 2 2 2 2 2 2 2 tan !!" tan (� + � ) [ 1  tan � tan � l 2 2 2 2 2 =
=
=
=
=
=
= tan
Second approach.
=
=
� tan (900  � ) [1  tan � tan �l = =
b, r, s
A C 1  tan '2 tan '2 '
denote the side lengths, inradius and semiperime Let a, c, ter of triangle ABC, respectively. Then AP =  a, and tan(A/2) = =  a ) . Hence A tan  =  .  a) 2 Likewise, tan !!" = and tan � = 2  c) 2 
r/(s
SABe
s(s b)
s
SABe rs, S s(sABe
S
s(sABe .
Hence
A
B tan "2 tan "2
B C tan "2 tan "2
We can also use halfangle formulas to calculate
C A tan "2 tan "2
+ + = = S�SBC2 ( (S (sC) +a)(s(S  a)b)(s+ (S ) b) ) = S� s(s  a)(s BCb)(s  ) = 1 ,
sin B
c
by Heron 's formula.
c
0
103
3.2. SOLUTIONS
3. G EO METRY
102
C 2 tan 2 C 1 tan2 "2
= sin C + =
=
3
"5 '
= q yz
From this it follows that AQ : QB : BA = 3 : 4 : 5 and BC : AC : BC 5 : 5 : 8. By introducing the variables and =  1, relations (1) and (2) become 2 p = and = respectively. Eliminating yields
Alternative solution.
A
p = y+z x + 5 6 xp + q 0, p(6 5p) + 2q = O.
x
(3)
y and z are the roots of the equation (4) t2  pt + (q + 1) O. Expressing q in terms of p in (3) , and substituting in (4), we obtain the following quadratic equation in t: t2 pt+ 5p2  26p + 2  0. This equation has discriminant (3p  2) 2 $ O. Hence the equation has real solutions only if p = 2/3, and y = z = 1/3. Note. We can also let x = AP, y = BQ , z CR and use the fact that r(x + y + z) SABC = Jxyz(x + y + z) Note that
Q
B
C
=
x
Without loss of generality assume that AP min{AP, BQ, CR} . Let = tan(A/2) , tan(B/2) , and = tan(C/2) . Then AP = BQ and CR The condition given in the problem statement becomes
= r/z.
y=
z
r/x,
2x + 5y + 5z = 6,
= r/y,
(1)
_
and the equation in the lemma is
xy + yz + zx = 1.
Eliminating
(2)
x from (1) and (2) yields 5y2 + 5z2 + 8yz  6y  6z + 2 = O.
=
Completing squares, we obtain
(3y _ 1)2 + (3z _ 1) 2 = 4(y  Z) 2 . Setting 3y  1 u, 3z  1 = v ( Le., y = (u + 1)/3, z = (v + 1)/3) gives 5u 2 + 8uv + 5v 2 = O. Because the discriminant of this quadratic equation is 82  4 25 < 0, the only real solution to the equation is u = v = O. Hence there is only one possible set of values for the tangents of halfangles of ABC (namely x = 4/3, y = z = 1/3) . Thus all triangles in S are isosceles and similar to one another. Indeed, we have x = r/AP = 4/3 and y = z = r/BQ r/CQ = 1/3 = 4/12, so we can set r = 4, AP AR = 3, and BP = BQ = CQ = CR 12. This leads to AB AC = 15 and BC = 24. By scaling, all triangles in S are similar to the triangle =
x
=
=
with side lengths 5, 5, 8.
=
=
=
=
to obtain a quadratic equation in three variables. Without loss of generality, we may set 1. Then the solution proceeds as above. USA Mathematical Olympiad,
x= (Titu Andreescu, 13. Let a, b,
2000)
A, B, C be the side lengths and angles of triangle ABC. Let X, Y, Z be the feet of the perpendiculars from P to lines BC, CA, AB, respectively. Recall the inequality (the key ingredient in the proof of the ErdosMordell inequality) : c,
PA sin A � P Y sin C
+ PZ sin B.
(1)
3.
104
3. 2.
G EOMETRY
105
SOLUTIONS
A
A
B' Z
.
•M
.
C'
'N� " " ' : : :
I
. p" . B
This says that the length of Y Z is greater than equal to its projection onto BC, the latter being equal to the sum of the lengths of the projections of PY and P Z onto BC. In fact, since AYP Azp 90°, AZPY is cyclic with AP as a diameter of its circumcircle. By the Extended Law of Sines, Y Z P A sin A. Let M and N be the feet of perpendiculars from Z and Y to the line P X. Since 1fZj5 BxP 90° , . PZBX is cyclic. Hence x:iPZ 13 and ZM PZ sin B. Similarly, YN PY sin C. Thus (1) is equivalent to Y Z � Y N +M Z. Multiplying by 2R and using the Extended Law of Sines, (1) becomes
=
=
=
=
=
=
=
=
aPA � cPY + bPZ. Likewise, we have bPB � aPZ + cPX and cPC � bPX + aPY. Using these inequalities, we obtain
(
b c PA PB PC ;;:;: + b2 + 7 � P X c3 + b 3
> 
) + PY ( c3 + a ) + PZ ( ba3 + ab3 ) �

2PX 2PY 2PZ +  bc + ab ca
Al A'
B
x
a
We prove that I == H. Assume by way of contradiction that points I and H are distinct. Let AI , Bl , Cl be the feet of the altitudes from A, B , C and let A' , B' , C' be the projection of I onto the sides BC, CA, AB, respectively. Hence at least two of the inequalities IA + lA' � AAl , a · IA + b · IB + c · IC
AH · BC + BH · CA + CH · AB
= 4SABC .
> a(AAl  IA') + b(BBl  IB') + c(CCl  IC' ) ,
which is false. Therefore
1 = H and ABC is an equilateral triangle, as desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1981) , pp. 67,
Problem 4616)
15. Using standard notations, we have 64c2 + 49a2 + 9b2  112ac  48bc + 42ab = 6c2  6a2  6b2 . This is equivalent to
4SABC = 1
14. It is known that if H is the orthocenter of a triangle ABC then
IC + IC' � CCI
or
(AMGM inequality)
= = (Titu Andreescu,
IB + IB' � BBl ,
are strict. It follows that
c3
abc R' Equality in the first step requires that Y Z be parallel to B C and so on. This occurs if and only if P is the circumcenter of ABC. Equality in the second step requires that a b c. Thus equality holds if and only if ABC is equilateral and P is its center. USA IMO Team Selection Test, 2000)
C
= O. Viewing this as a quadratic equation in b, the condition � � 0 is satisfied. That 441a2  1008ac + 576c2  825a2 + 1680ac  870c2 � O. 15b2 + 2b(21a  24c) + 55a2  112ac + 58c2
is
The last relation is equivalent to
6(64a2  112ac + 49c2 ) � 0,
:$ O.
or (8a  7C) 2 It follows that 8a yields 3a 7b. We obtain
=
= 7c. Substituting back into the given condition ,
106
hence triangle ABC is similar to triangle A' B' C' having sides 7 , 3, 8. In this triangle 3 2 + 82 72 1 cos AI = 2' .3.8

2
It follows that A = A7 = 60°. Korean Mathematics Competition, 2002)
(Titu Andreescu, 16. First solution. By the CauchySchwarz Inequality, ..jPB2 + PC2 ..jAC2 + AB2
Applying the (Generalized) Ptolemy's Inequality to quadrilateral ABPC yields
2:: PA . BC.
Because PA is the longest side of an obtuse triangle with side lengths PA, PB, PC, we have PA VPB2 + PC2 , and hence
>
P A . BC
2:: ..jPB2 + PC2 . BC.
AQ 2
2:: AQ2 _ QC2 =
>
=
(Ap2 _ PQ 2 ) _ (Cp 2 _ PQ 2 )
AP2  PC2
> BP2 2:: DQ2 .
=
l,
l,
o
P o
O��O��OrO
A
D
Q
c
+
+
It
=
+
=
z · z
C
=
z 
z
+
=
z ·
+
z
=
0
Applying the Lemma to points A, B , C, P gives
o � AB 2 + Bp2 + PC 2 + CA2 _ AP 2  BC2 =
>
Let I be the ray AC minus the point A. Note that, since PA PC, Q lies on ray D did not lie on then AQ would be less than or equal to DQ, a contradiction. Thus, D lies on and angle BAC is acute. B
l. If
=
c ,
=
=
2:: BC,
implying that angle BAC is acute. With some careful argument, it can be proved that quadrilateral ABPC is indeed convex. We leave it as an exercise for the reader. Let D and Q be the feet of the perpendiculars from B and P to line AC, respectively. Then DQ � BP. Furthermore, the given conditions imply that Ap2 Bp2 + PC2 , which can be written as Ap2  PC 2 BP 2 . Hence,
Note. Second solution. >
=
=
z
Combining these three inequalities yields
..jAB2 + AC2
Third solution. Set up a coordinate system on the plane with A (0, 0) , B (a,O), C (b, ) and P (x, y). Without loss of generality, we may assume that a > 0 and that > O. Proving that angle BAC is acute is equivalent to proving that b > O . Since PA2 > PB2 PC2 , x2 + y2 > (x _ a)2 + y2 (x _ b)2 + (y _ C)2 . Hence o > (x  a) 2  2bx + b2 (y  C) 2 2:: 2bx . Since P A > P B, we have x > � > O. follows that b > 0, as desired. Fourth solution. We first prove the following Lemma. Lemma. For any four points W, X, Y, Z in the plane, WY2 + X Z2 � WX2 + Xy2 + YZ2 + ZW2 . Proof. Pick an arbitrary origin 0 and let w, x, y, denote the vectors from 0 to W,X,Y, Z, respectively. Then W X 2 + X y 2 + Y Z2 + Z W 2  W y 2  X Z2 I w  x l 2 + I x  Yl 2 + I y  Zl 2 I w l 2  Iw  Yl 2  I x  Zl 2 w · w + x· x + y . y +  2(w , x + x · y y . + w  w · y  x · ) I w + y  x  Z1 2 , which is always nonnegative. Equality holds if and only if w y x + which is true if and only if W XY Z is a (possibly degenerate) parallelogram. +
2:: PB · AC + PC · AB.
P B . AC + PC . AB
107
3.2. SOLUTIONS
3. G EOMETRY
(PB2 + PC 2
_
PA2 ) + (AB 2
< 0 + (AB2 + AC2  BC2 )
=
+
AB 2
=
z,
=
AC2  BC2 ) < +
AC 2  BC 2 .
Therefore angle B AC is acute. In this solution, sin  1 takes on values between 0° and 90° . Note that p;fjj < 90° , since PB < P A. Applying the Law of Sines to triangle PAB yields  PB  PB sin PAB = sin ABP � PA PA ' It follows that .  1 PB PAii < sm PA ' Since P A2 PB 2 + PC 2 , we have similarly
Fifth solution.
>
. _ 1 VPA2 _ PB2 < sm '  1 PC < sm P AC _ PA PA
108
the circumcircle of ABPC, arcs AB and AC are bigger than arcs PC and PB, respectively. Thus, IiPC > IfAiJ Because these two angles are supplementary, angle BAC is acute.
Thus 
If () =


JPA2  PB2 PB + sin  1 BAC � BAP + PAC < sin PA PA 1 sin then
��,
sin(90°  ()) Hence
=
1
.
'\.
JPA2  PB2 . cos () = V'1  sin2 () = PA
.  1 JPA2 _ PB2 .  1 PB + sm BAC < sm PA
PA
and angle B AC is acute.
=
C '\.
1
B
B
As we mentioned at the end of the first solution, the conditions in the problem imply that quadrilateral ABPC is indeed convex. Thus, the diagram on the righthand side is not possible, but this solution does not depend on this fact.
( \
...I
I
\
\
"
I
\
'...
""
,/
'\.
I
/
17.
A
/
A
All angles will be in degrees. Let x = F0B. Then PiiC Law of Sines (or the trigonometric form of Ceva's Theorem) ,
P
I
,/
'
\
(Titu Andreescu, USA Mathematical Olympiad, 2001)
P
/'
\
P
90° ,
C
Sixth solution.
109
3.2. SOLUTIONS
3. G EOMETRY
_
/
_
;'
B \
P C
'"
\
P A PB P C PB PC PA
./
I
=
� !!.!!. �
sin sin sin sin PAB sin PBC sin PCA
80  x. By the
=
4 sin x sin 40 cos 10 sin(80  x)
sin 20 sin x sin 40 sin 10 sin(80  x) sin 30 B
\ A
,/
...
\
=
=
x
/
Note that PA2 > PB 2 + PC2 . Regard P as fixed and A, B, C as free to rotate on circles of radii PA, PB, PC about P, respectively. As A, B, C vary, IfAiJ will be maximized when B and C are on opposite sides of line PA and Jfijp and ;[(j'p are right angles, i.e. , lines AB and AC are tangent to the circles passing through B and C. Without loss of generality, we assume that PA > PB � PC. In this case, ABPC is cyclic and AB 2 = P A2  PB 2 > PC2 , and similarly AC 2 > PB 2 . Hence on
A
C
The identity 2 sin a cos b = sin( a  b) + sin( a + b) (a consequence of the addition formula) now yields 1=
2 sin x(sin 30 + sin 50) sin(80  x)
=
sin x(1 + 2 cos 40) , sin(80  x)
so 2 sin x cos 40
=
sin(80  x)  sin x = 2 sin(40  x) cos 40.
This gives x = 40  x and thus x triangle ABC is isosceles.
=
20. It follows that AcE
=
50
=
IfAiJ, so
110
3 . G EOMETRY
3.2. SOLUTIONS
Alternative solution. Let D be the reflection of A across the line BP. Then triangle D is isosceles with vertex angle APi5 = 2(180  EPA) = 2(PAii + AiiP) = 2(10 + 20) = 60, and so is equilateral. Also, YSiiA = 2PBA = 40. Since 1f.AC = 50, we have D B AC. AP
.L
B
it follows that
111
r = 4( v. 3 + 1)r sm. "2A sm. 2"B sm. "2C In
or 1 2 ( va + 1) Then 0 1 4
= 2 (sm. 2"A sm. "2C ) sm. "2B '
AC A + C) . B = (cos   cos 2 sm 2" 2
= 30° , we obtain J3  1 = ( V6 +4 V2 sin �2 ) sin �2 . 4 . B = x Y1e . lds ' sm Lettmg "2 V6 +4 V2x + 04 1 _ 0 , x  V2 and x = 2 ..J2 ' It follows that "2B = 15 o or 2"B = 45° . whose solutions are x = V6 4 The second solution is not acceptable, because A 2: B. Hence B = 30°, A = 90° and C 60° . Thus triangle ABC is right angled. ( Titu Andreescu, Korean Mathematics Competition, 2002)
and, since A  C x
_
A
C
D
E be the intersection of DB with CPo Then PEi5 = 180  CEi5 = 180  (90  AcE) = 90 + 30 = 120 and so PEi5 + J5Ai5 = 180. We deduce that the quadrilateral APED is cyclic and ' therefore D EA = D PA = 60. Finally, we note that I5EA = 60 = DEC . Since AC .L DE, we deduce that A and C are symmetric across the line DE, which implies that BA = BC, as desired. ( Titu Andreescu, USA Mathematical Olympiad, 1996) 18. Let max{A, B} = A. If triangle ABC is rightangled, then A = 90° , B = 30° R and C = 60° . In order to find  , we may assume that ABC is the triangle with sides = 2, b 1 , = J3. We haverR 1 and va S ABC 2 r = s = 2 + 1 + J3 _ 3 +J3J3 ' Let
a
=
so !!:.
3 + J3 J3
r=
=
c
2
= va + 1.
Conversely, assume that
r R
=
r=
J3
+ 1. From the identity
. B . C · A sm 4R sm "2 "2 sm "2
2
_
=
19.
Construct in the exterior of triangle ABC three circles equal to the circum circle ABC that pass through two vertices of the triangle. By the fivecoin theorem the circles will have a common point P, as desired (see Dorin Andrica, Csaba Varga, Daniel Vacare�u, " Selected Topics and Problems in Geometry" , PLUS, Bucharest, 2002, pp. 5156) . Let H be the orthocenter of triangle ABC. The reflections of H across the sides of the triangle are points of the circumcircle of triangle ABC. Therefore the circum circles of HAB, HBC, HCA are equal to the circumcircle of ABC and for P H the claim holds. ( Revista Matematica Timi§oara (RMT), No. 2(1978) , pp. 74)
Alternative solution. = Dorin Andrica,
20.
Denote the midpoints of A'B', B'C' , C'A' by Co , Ao , Bo, respectively, and the three perpendiculars in question by , Consider the centroid of triangle A'B' C'.
lc, lA lB .
11 2
3.
C
B A C' Since AoG : GA' BoG : GB' CoG : GC' 1 : 2, the dilatation h with center G and coefficient 2 takes Ao, Bo, Co to A', B', C', respectively. Since dilatations carry straight lines into parallel lines, h transforms le into the line through C' perpendicular to AB. But C' is the point of tangency of the incircle and AB, so this line passes through the incenter of triangle ABC. The same applies to the images of lA and lB under h. Since the images of lA , lB , le under h are concurrent, so are lA , IB , le themselves. (Titu Andreescu, Romanian IMO Selection Test, 1986) 21. Because triangle A1 A2 A3 is not isosceles, it is not difficult to see that the circumcenter of the triangles A 1 B 1 I, A2 B2 I, A3 B 3 1 are defined. We start with a sim pIe lemma. Lemma. Let ABC be a triangle with the incenter I. Let T be the circumcenter of the triangle BIC. Then T lies on the internal bisector of the angle A. Proof. Let us draw the external bisectors of the angles B and C as shown in the figure below. B =
=
A
C
The lemma is proved.
..1
113
SOLUTIONS
=
They intersect at the excenter E, which lies on the internal bisector of the angle A. Since BE BI and GE GI, the quadrilateral BEGI is cyclic with the center of the circumscribed circle on IE . This center will be also the circumcenter of BIC. ..1
3.2.
GEOMETRY
i 1, 2, 3 Qi Ai+1 IAi+2 . Oi Ai ' Ti T1 T2T3 I. 0i+1 0i+2 Ti+1 Ti+2 ' Ti+l Ti+2 Bili AlB1 I, A2B2I, A3 B3 I, 101 02 , 102 03 , 103 01 (T1 ' T2 , Q3 ), (T2 ' T3 , Q1 ), (T3 ' TI, Q2 ), 01 T1 . IT2 . 02 Q3  1 , IT1 02T2 01 Q3 IT 0 T Q Q 1 03 T33 . IT2 22 . 023 Q1  1 , ITI 0 T 0 1 Q OlTl . IT3 33 . 03 Q22  1 . Multiplying them all one gets 02 Q3 . 03 Q1 . 01 Q2 1, 0 1 Q 3 0 2 Q 1 03 Q 2 which means that the points Q 1 , Q 2 , Q 3 are collinear. Alternative solution. This proof will be based on inversion. We take the incenter I to be the center of the inversion and the power of the inversion is arbitrary. Using
Let us prove the main statement. For = we denote by the center of the circle Ci and by the circumcenter of the triangle Clearly, lies on the internal bisector of the angle By the lemma, also lies on the same bisector. Thus the triangles and are perspective from the point By Desargues' theorem these triangles are perspective from a line. This is to say that if we denote then and = to be the point of intersection of the lines are collinear. But since is the perpendicular bisector the points of and is the perpendicular bisector of these points are exactly respectively. the circumcenter of the triangles A student not familiar with Desargues' theorem may proceed from the point as follows. Applying Menelaus' theorem to the triangles and to the triples of points respectively, one can, observing usual agreement about the signs, write:
Ti 0 1 02 03 Qi , i 1,2,3, Q 1 , Q , Q3 Ail 0i+120i+2 Remark.
_
=
primes to denote images of points under the inversion we have the following " dual" figure shown below.
3.
11 4
3. 2.
G EOMETRY
SOLUTIONS
11 5
23. Without loss of generality assume that
B'I
PC = max{PA, PB, PC}. The condition in the hypothesis is PB · PC + PA · PC = PA · PB + 1 or
B'2 Indeed, the image of the circle Ci is a straight line Bi+ 1 Bi+2 ' with these lines forming the triangle Bi B�B� . The line AiAi + 1 will be transformed into the circle ri+ 2 , with the side AiAi+ 1 becoming the arc AiAi+ 1 which does not contain I. Note that all these circles have equal radii since the distances from I to the sides of AI A2 A3 were equal. Let us note that if �I ' � 2 ' � 3 are three circles passing through the common point I and no two of them touch, then their centres are collinear if and only if there is another common point f:. I through which all these three circles pass. We will use this observation for �i being the circumcircle of AiBiI. Since the inversion takes :Ei to the line AiB�, the desired result is to show that the lines A� Bi , A� B� , A�B� are concurrent. For this, it suffices to show that the triangles A� A� A� and Bi B�B� are homothetic, which is the same to say that their corresponding sides are parallel. Since the radii of the circles r I , r 2 , r3 are equal, the triangle PI P2 P3 formed by their centre has its sides parallel to the corresponding sides of the triangle Bi B�B�. The homothety of ratio centred at I takes the triangle A� A� A� into the triangle whose vertices are the midpoints of the triangle PI P2 P3 . Therefore the corresponding sides of the triangles A� A�A� and PI P2 P3 are also parallel and the result follows. IMO 1997 Shortlist)
J
PC PA · PB + l · 1 1 PA . 1 PC . 1 . From the converse of the second theorem of Ptolemy it follows that P ACB is a cyclic quadrilateral. Note that P cannot be A, B or C otherwise the denominator of the righthand side equals Hence the locus of point P is the circumcircle of triangle ABC without the vertices A, B, C. Revista Matematidl Timi§oara (RMT), No. 1(1985) , Problem C7:3)
(Titu Andreescu,
24. We have
Ja2 b2  4S2 + Ja2 c2  4S2 = Ja2 b2  a2 b2 sin2 C+ + Ja2 c2 2 a2c22 sin22 B = ab2 cos C + ac cos B = = ab a +2abb  c +ac a +2acc2  b2 = a2 , as desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 8(1971), pp. 25,
Problem 1006)
25. The relation is equivalent to
Fa + y'rb + Fc = 1 , 
from Van Aubel's theorem, therefore line segment parallel to BC.
(Titu Andreescu)
::, is constant. Hence the locus of point I is a
..jrarbre
or
(Titu Andreescu,
intersection point of lines AI and BC. We have AI AB' AC' = + B' B C' C IA"
+
O.
1/2
22. Let I be the intersection point of lines CB' and C' B and let A' be the
=
On the other hand,
so
(
1 1 1 1  +  +  = ,
ra rb re r 1 1 1 1 1 1 +  +  =  + +  . rb re Fay'rb y'rbFc FcFa ra 
so Then
r 1 1 1 1 = . + + .;r;.fo vr;.;r;. .;r;.y'rb r
2 2 1 )  1 ) 2 + ( 1 1 ) = 0, + ( vr;.  .;rb fo Fc vr;  .;r;. 1
1
ra = rb = re ' It follows that the triangle is equilateral, as desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1974) , pp. 21,
Problem 1903)
116
3. 26.
If the triangle is equilateral the conclusion is true. To prove the converse, we assume by way of contradiction that the triangle is not Then equilateral and say that
b f:. c. b2 +2 c2 a2 (b + C)42  a2 =p(p  a) m2a   4 > and likewise m� 2: p(p  b), m� 2: p(p  c). It follows that 1 ) 1 1 1 = + + + 1 + 1 < 1 ( m� m� m� p p  a p  b p  c p2 + ab +2 bc + ca _

Problem 3513; Gazeta Matematica ( GMB ) , No . 11(1981), Problem 0258; No. 2(1988) , pp. 78, Problem 21353)
�____ _
2"
28. For any positive real numbers x, y, Z we have
(Titu Andreescu,
Setting gives so
A . B . C 1 . < ' sm  sm  sm 8 2 2 2 C B A ' C. ' B sm ' A sm cos "2 cos "2 cos "2 2: sm
xyz ::; (x + y)(y + z)(z + x) x = a + b + c, y = a  b + c, Z = a + b  c, ( a + b + c) (a  b + c) (a + b  c) ::; abc, 8
2'
which is a contradiction. Revista Matematidi Timi§oara (RMT ) , No. 2(1977) , pp. 66, Problem 3063)
so
=:=
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp. 49,
8 r� + r� + r� = 82 [(p  a) 2 + (p _ b)2 + (p _ C)2] = p2 + a282+ b2 + c2 Since b f:. c then ab + bc + ca < a2 + b2 + c2 hence 1 + m12b + m1e2 < r21a + r1b + r1e m2a 
On the other hand 1 1 1 1
27. We know that
117 3.2. Alternative solution. Let Zl, Z2 , Z3 be the afixes of points A, B, C such that I zd = IZ2 1 = I Z3 1 = 1 . We have a BC = I Z2  z3 1 , b = AC = IZ3  zd, c = AB = IZI  z2 1 · U sing the identity Z; (Z2  Z3 ) + Z� (Z3  zd + Z� (ZI Z2 ) = (Zl  Z2 )(Z2  Z3 )(Z3  zd and triangle inequality it follows abc = IZI  z2 1 z2  z311 z3  zd ::; IZll2 1 z2  z3 1 + I Z2 12 1 z3  zll + IZ3 1 2 1 z1  z2 1 = = I Z2  z3 1 + IZ3  zd + I ZI z2 1 = a + b + c. SOLUTIONS
GEOMETRY
It follows that (1)
On the other hand,
' A + sm' B + sm' C = 4 cos "2A cos "2B cos "2C '
sm
then Hence
3
sin A
+ sin B + sin C 2: 4 sin A sin B sin C.
Since the circumradius is 1, we have
a = 2sinA, b = 2 sin B, c = 2 sin C, and relation (2) yields a + b + c 2: abc, as
claimed.
pabc
2 ' 32
so inequality (1) gives (2) as
desired.
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1974) , pp. 51,
Problem 2028)
29. Using the inequality 3(
a2 + b2 + c2 ) � (a + b + C)2 ,
11 8
3.
we obtain
a2 + b2 + C2 � p2 ' 4S2
Hence as
3.2.
GEOMETRY
1
h�
desired.
+
1
h�
3S2
+
1
h�
�
1 3r2 '
as
desired.
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1973), pp. 43,
Problem 1585)
33. We have
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 66,
It follows that
Problem 3062)
30. From the inequality we obtain
1 1 1 1  >  +  +  . ..jrbrc ..jrcra ..jrarb
r
Then
hence as


desired.
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1984), pp. 67,
Problem 5221)
hence ..jrarb + ..jrbrc + ..jrcra � 9r, as
) 1 V (P  b)'bc(P C' < 2 ' (p  b) (P  c) :s; 4bc ' '
1 1 1 1 1 1  + +  >  +  + ra rb rc  ..jrbrc ..jrcra ..jrarb '
119
SOLUTIONS
34. Let 0 1 , O2 , 03 be the centers ofthe three circles and S the area of the common
region .
claimed.
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 64,
Pro blem 3277)
31. We have ma =
and likewise
b2 +  a42 � J(b + C) 2  a2 = Vp(p  a) J4 2 ,?
It follows that ma m b m c � as
desired.
pVp(p  a)(p  b) (P  c) = pS = rarbrc,
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 (1978), pp. 64,
The three sectors with centers 0 1 , O2 , 03 which subtend the arcs 02 03 , 0 1 03 ,02 0 1 , respectively, cover the 2 surface of area S and twice more r the surface of triangle 0 1 02 03 ( which is f3 ). On the other hand, the area of these three circular sectors equals the area of a semicircle, which is
Problem 3276)
p3 � 27abc. Hence
r2 J3
1 7fT = S + 2 .  ' 4 2

32. By the AMGM inequality, so 8
2
�7fT2. Hence
(a + b + c)3 � 27abc, abc · S = 27Rr, 2p2 > 274S p 
therefore
S
= 21 r2 (1/'  V3).
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp. 50,
Problem 3522)
1 20
3. G EOMETRY
35. The parallel BD through C meets AB at point E. By Stewart's formula, we
D F N
obtain
AC2 · BE + CE2 . AB  CB 2 · AE = AB · BE · AE Because CE = BD and BE = CD, we deduce AC 2 . CD + BD 2 . AB  BC 2 . (AB + CD) = AB . CD . (AB + CD) D
121
3.2. SOLUTIONS
C
I (1)
C
B
E
M
A Then
AB + CD = EF, 2 which is the length of the midline and hence the length of the altitude. It follows that IM and IN are also altitudes in triangles lAB and ICD therefore lAB and ICD are isosceles. Thus ABCD is isosceles, as claimed. Revista Matematidi Timi§oara (RMT), No. 1 (1978) , pp. 48, Pro blem 28 17) IM + IN =
A
B
E
(Titu Andreescu,
Drawing the parallel to AC through D and using similar computations yields BD 2 . CD + AC2 . AB  AD 2 . (AB + CD) = AB · CD · (AB + CD)
(2)
Subtracting the relation (2) from (1) gives
38. From the Law of Cosines we deduce that
2AB . AC cos IiAC = AB 2  BC2 + AC2
(AC 2  BD 2 ) (AB  CD) = (AD 2  BC2 ) (AB + CD) ,
2DC · AC cos DcA = CD 2  AD 2 + AC2 2AB . DB cos ISiiA = AB 2  AD 2 + DB 2
as desired.
(Dorin Andrica, Gazeta Matematidi (GMB) , No. 9(1977) , Problem 6852; Revista
Matematidi Timi§oara (RMT), No. 12(1980) , pp. 64, Problem 4119)
36. Let I be the intersection point of the diagonals AC and BD. Since AB · AC AB · BD IA = and IB = AB + CD AB + CD the condition in the statement becomes Hence AiB = 90° , as desired . Revista Matematidi Timi§oara (RMT) , No. 2(1978) , pp. 59, Problem 3524)
(Dorin Andrica,
37. Let ABCD be the trapezoid. Point I is the intersection of diagonals and
M, N are the midpoints of AB and DC. In a right triangle the length of the median corresponding to the hypothenuse is half of lenght of the the hypothenuse. Hence AB CD IM = and IN = . 2 2
Note that
2DC · DB cos CiJij = CD 2  BC2 + DB 2
(1) (2) (3) (4)
IiAC = DcA, ISiiA = CiJij,
so dividing relations (1) and (2), (3) and (4) yields AB AB 2  AD 2 + DB 2 AB 2  BC2 + AC2 = = DC ' CD2  AD2 + AC2 CD2  BC2 + DB2 as desired.
(Dorin Andrica) 39. Let a, b the lengths of the bases, c, d be the lengths of the nonparallel sides
and d 1 , d2 be the lengths of the diagonals. From Euler's theorem for quadrilaterals, it follows that Hence
c
(d1  d2 ) 2 + 2d1 d2 = (  d) 2 + 2(ab +
cd) ,
3.
1 22
d1  d2
d
3. 2.
GEOMETRY
d1 d2 ab cd.
= + = C  implies From Ptolemy's Theorem, we deduce that the trapezoid is cyclic and so isosceles, as claimed. Revista Matematica Timi§oara (RMT) , No. 1(19 78), pp. 48, Problem 281 7) and
(Titu Andreescu,
40. First solution.
Assume the opposite. Then lAC  BD I > lAB  CD I or lAC  BDI > lAD  BCI. Without loss of generality, l AC  BD I > lAB  CDI , otherwise switch B and D. We have AC2  2AC . BD + BD 2 > AB 2  2AB . CD + CD 2
(1)
and, from Euler's relation, AB 2 + BC2 + CD 2 + AD 2 = AC2 + BD2 + 4MN2 ,
AD 2 + BC2  2AC · BD > 4MN2  2AB · CD .
lAC  BDI = 2 1 sin(a
+,6)  sinCa + ,) 1 = I Sin a � , cos ( a � , +,6) I .
Since 0 (a + ,)/2 ::; (a + ,)/2 + ,8 ::; 7r/2 (by the assumption ,8 8) and the cosine function is nonnegative and decreasing on [0, 7r /2] , we conclude that l AB _ CD I � l AC  BDI , and similarly lAD  BCI � lAC  BD l . USA Mathematical Olympiad, 1999)
�
�
(Titu Andreescu,'
41.
Let E be the intersection point of the diagonals. Consider AD basis of the trapezoid and AB the altitude. Since
<
BC the
AB 2 = AD . BC, then
( 3)
Let P be the midpoint of AB. Then NP = AD/2, MP = BC/ 2 and since MN � INP  MPI , it follows that 4MN 2 � (AD  BC) 2 .
and
(2)
where M and N are the midpoints of AC and BD, respectively. From (1) and (2) ,
1 23
SOLUTIONS
AB BC AD AB ' so the right triangles ABC and ABD are similar. On the other hand we have
(4)
(3)
and (4) , 2AC · BD > 2AB · CD  2AD · BC, in contradiction with From Ptolemy's Theorem. We are done. The cyclicity is essential. The inequality fails if ABCD is a parallelogram. Let E be the intersection of AC and BD. Then the triangles ABE and DCE are similar, so if we let x = AE, y = BE, z = AB, then there exists k such that kx = l)E, ky = CE, kz = CD. Now
Note. Second solution.
hence It follows that ill = 90° so the diagonals are perpendicular, as claimed. Revista Matematica Timi§oara (RMT) , No. 2 (19 72) , pp . 28, Problem 1 164)
(Titu Andreescu,
lAB  CDI = Ik  l i z and
lAC  BDI = I (kx + y)  (ky + z ) 1 = I k  1 1 · Ix  y l · Since Ix  yl ::; z by the triangle inequality, we conclude that l AB  CDI � lAC � BD I , and similarly IBC  DAI � lAC  BD l. These two inequalities imply the desired result. Let 20', 2,8, 2,, 28 be the measures of the arcs subtended by AB, BC, CD , DA, respectively, and take the radius of the circumcircle of ABCD to be 1. Assume without loss of generality that ,8 8. Then a + ,8 + , + 8 = 7r, and (by the Extended Law of Sines)
Third solution.
�
I
l AB  CDI = 2 1 sin a  sin ,1 = Sin
a
� , cos a �' I
42. Let I be the intersection point of the diagonals AF and BE of the rectangle
ABFE. Notice that N I is the median of the right triangle ANF with hypothenuse AF, so AF IN =  = IA 2 Likewise, BE IM =  = IE ' 2
AF IQ =  = IF ' 2
Since IA = IE = IF = IB, it follows that 1M is cyclic, as desired.
=
BE IP =  = IB . 2 IN = IP = IQ. Hence MNPQ
124
3.
A
125 3. 2. 44. Let I be the intersection point of lines BD and AF. The parallel to BD through C meets line AF at point T. First we consider AF .l BD and prove that AB .l CD. F F
GEOMETRY
SOLUTIONS
B P N
Q D E (Titu Andreescu)
< a1 :::; a2 :::; a3 :::; a4 < a1 + a2 + a3 , aI, a2 , a3 , a4 · a, b, c, d, e, j, m AB, BC, CD, DA, AC, BD, MN, b + d 2: a + c. P BC. MP NP are midlines CAB BDC, MP a2 NP c.2 Then 2m 2MN < 2MP + 2NP a+c so 2m < a + c < b + d. 43.
By Sturm's theorem, we know that if 0 then there is a cyclic quadrilateral having side lengths Denote by the lengths of the segments respectively. Without loss of generality assume that Let be the midpoint of the side The segments and in triangles and so 1 1 = and = =
=
On the other hand, if 0 is the intersection point of the diagonals, we have
b + d BC + DA < BO + OC + DO + OA AC + BD e + j, hence 2m < a + c < b + d < e + j. It suffices to prove that e + j < 2m + a + c + b + d. Note that e < c + d, j < b + c, e < a + b, j < a + d. =
A �� D
C
F
=
=
Summing up these inequalities yields
e + j < a + b + c + d < a + b + c+d+ 2m and the proof is complete. (Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1 (1978) , pp. 66, Problem 3288; Gazeta Matematidi ( GMB ) , No. 10 ( 1981 ) , pp. 402, Problem C148 )
i) Assume that we obtain
D lies on the segment CEo Then ArC
On the other hand
=
90° . Since IfAjJ : BnC,
(1)
CTI I BD so
(2)
EATC
Relations ( 1 ) and imply fiE : EaT , therefore is cyclic. It follows that BEG = 90° , hence as desired. In the right triangle ii ) Assume that lies on the segment median, so
AB .l CD, DE.
C
Because
CTF, TE is the
Ere : EaT .
CTI IB D, we have
(2) (3)
( 4)
Also,
ATEC
(5)
so from ( 3 ) , (4) , (5) , we obtain Ere : Mo. Hence is cyclic, then AEC = ArC = 90° , and .l as desired. Conversely, consider that i) If is on the segment then ME == AcE On the other hand AcE == Aci5 == ABi5 so ME == ABi5 and is cyclic. It follows that JiiF = liEF = 90° , hence as desired. ii ) If is on the segment then
AB CD,
D
C
,
BD .l AF,
AB .l CD. CE, FBIE DE,
.
,
(6)
1 26
11oreover, J.U515 = ACB15,
SO
AcE = ABl.
S
(7)
a, b, c, d
Let be the side lengths of the quadrilateral and let Because the quadrilateral is cyclic, we have
S2 = (p  a)(p  b)(P  c)(P  d). 1
S be its area. (1)
S a � b � c � d. Since S is prime number, p d=p  c = l md p  a = p b = � Summing up these equalities yields 4p2p = 2 + 2S so p = S + 1. Hence a = b = 1 and c = d = S. The required quadrilaterals are either rectangles or kites. (Titu Andreescu, Revista 11atematidi Timi§oara (R11T) , No. 2(1977), pp. 66, Problem 3067)
Without loss of generality assume that from relation we obtain
(1)
a
46. From the A11G11 inequality it follows that
S
Hence
S
1
1
(6) and (7) we obtain ME = ABl, hence FEB! is cyclic. Note that liEF = 900, so iiiF = 900 and BD ..L AF, as desired. (Titu Andreescu, Revista 11atematica Timi§oara (R11T) , No. 1(1986), pp. 106, Problem C6: 4 ) From
45.
12 7
3.2. SOLUTIONS
3. GEOMETRY
or
abcd < ( a + b +4 c+d ) 4 = p41 4 . 16abcd � p4 , 8(ac + bd)2 p4 � 8(a2 c2 + b2d2 ).
The desired inequality is now obtained from Ptolemy ' s Theorem: N �mbers
a, b, c, d are odd, hence a + b+c+d P = 2
is an integer. If is odd, then which is false. Hence p is even.
p
p  a, p  b, p  c, p  d are even and so 82 is divisible by 16,
ac+ bd = ef.
(Titu Andreescu, Revista 11atematidi Timi§oara (R11T), No. 3(1973), pp. 36, 1811; Gazeta 11atematidi (G11B) , No. 8(1980), pp. 364, Problem 18370)
Problem
47.
Let m be the length of the segment determined by the midpoints of the diagonals . From Euler 's Theorem for quadrilaterals we have
(1)
12
3.
8
and from Ptolemy's Theorem,
3. 2.
G EOMETRY
ae + bd = ef.
(1) and (2) we obtain (a  e) 2 + (b  d) 2 = (e  1) 2 + 4m 2 . Since max{ la  el , I b  dl} � 1, then 2 = 1 + 1 � (a  e)2 + (b  d)2 = (e  1) 2 + 4m2 � (e  1) 2 .
129
SOLUTIONS
D
(2)
From relations
Hence
I e  11 � V2,
AI
as desired .
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 51,
Problem 3527)
ABC DAC
AI, MI, BI AIBI, CI DI AC BI CI DIAl . Mil (Dorin Andrica, 3855;
48. The quadrilateral is cyclic so S = J(p  a) (p  b)(P  e) (p  d) 2 Since S = (�) , we have
{j(p  a) (p  b)(P  e) (p  d) = P.2 = (p  a) + (p  b) + (p  e) + (p  d)
51. We have
4 Note that this is the equality case in the AMGM inequality, hence p  a d, so the quadrilateral is a square. p  It follows that a Revista Matematica Timi§oara (RMT) , No. pp. Problem
p e = d. (Titu Andreescu, 2136)
= pb = 1(1977), 24,
=b=e=
4�. Observe that SPAB . SPCD = SPBC . SPDA, since both are equal to � PA . PB· PC· PD · sin P. The numbers SPAB , SPCD and SPBC , SPDA have the same sum and the same product, thus SPAB = SPBC and SPCD = SPDA or SPAB = SPDA and SPBC = SPCD, i.e. P is the midpoint of AC or BD, desired. as
We recall the Simpson's theorem: the projections of a point of the circumcircle of a triangle onto the sides of the triangle are collinear. Applying this result to triangles and yields that are collinear and are collinear. Hence the lines and meet at as claimed. and we deduce that (ii) From Simpson's Theorem for triangles and Since is a point of the conclusion follows. is on the lines Revista Matematidi Timi§oara (RMT), No. pp. 54, Problem Romanian Regional Mathematical Contest " Grigore MoisH" ,
(Titu Andreescu, Korean Mathematics Competitions, 2001) 50. (i) Let MI and Mil be the projections of point M onto diagonals AC and BD .
CI, MI, DI MI, ABD BDC AC,
M il
1(1979),
1995)
SABCD � 21 AC . BD,
AC BD. Since 2002 = SABCD � 21 AC . BD � 1 52 · 77 � 2 (AP + PC) . (BP +PD) =  = 2002, 2 it follows that the diagonals AC and BD are perpendicular and intersect at P. Thus, AB = .J242 + 322 = 40, BC = .J282 + 322 = 4y'fi3, CD = .J282 + 452 = 53, and DA .J452 + 242 = 51. The perimeter of ABCD is therefore 144 + 4 vTi3 = 4(36 + vTi3) . (Titu Andreescu, American Mathematics Contest 12 (AMC 12  Contest B), 2002, with equality if and only if
.1.
=
Problem 24)
52.
ABCD. Assume without loss of gener V2PA = PB +PD.
Let a be the side length of the square ality that max( We have
PA, PC) = PA.
y A(O,a) P(x, y)
a..j2PA = aPB + aPD, hence BD · PA = AD · PB +AB · PD. From the converse of the Ptolemy ' s Theorem it follows that PDAB is a cyclic quadrilateral, therefore P lies on the circumcircle of square ABCD. Conversely, using the Ptolemy's Theorem we deduce that any point of the cir cumcircle of square ABC D has the given property. P Then
B ABCD. P Alternative solution. P(x, y) y 0. A, B, C, D PC PA, V2PC = PB + PD Squaring both sides yields 2X2 + 2(y + b) 2 = (x  b) 2 + (x + b) 2 + 2y2 + 2 V(x  b) 2 + y2 + 2 V(x + b) 2 + y2 , then 2by = V(x  b) 2 + y2 + V(x + b)2 + y2 , hence (x2 + y2 + b2 2bx)(x2 + y2 + b2 + 2bx) = 4b2 y2 .
It follows that the locus of point is the circumcircle of square be a point with the given property and assume Let are considered like in the diagram. > Point so > Note that
It
B(a,O
x
(Titu Andreescu, Romanian IMO Selection Test, 1981; Revista Matematidl. Timi§oara (RMT), No. 2(1981), pp. 87, Problem 4751) 53. Let ABCD be a quadrilateral from the set P and let M, N, P, Q be the midpoints of sides AB, CD, BC, AD, respectively. The Euler's relation for the parallelogram MN PQ is MP2 + PN2 +NQ2 +QM2 = MN2 +PQ2 , or
C
Q
_
follows that
A
M
On the other hand, we have
Thus
P
D(a, 0) C(O, a)
so
x 2 + y 2 = b2
BD ABCD.
so
A.
Likewise, for that contains and so point lies on the semicircle of diameter that contains and lies on the semicircle of diameter ::; we deduce that finally we obtain the circumcircle of square
y 0
131
3.2. SOLUTIONS
3. G EOMETRY
130
P
BD
C
Hence
DB MP = AC MQ = , 2 2 AC2 + BD2 = MN2 + PQ2 . 2 AC2 � BD2 , A (P) + A� (P) = �
B
3.
132
3. 2.
GEOMETRY
In order to have equality everywhere we must have
which is, clearly, a constant.
(Dorin Andrica)
The first equality is equivalent to . rr sm
AF = 2R sin �;. 5rr
. sm 12 12 � + . rr = 4, sm sin 12 12 1 = 4 sm. 5rr12 sm. 12rr . 1 = 4 cos 12rr sm. 12rr or 1 = 2 sm. '6rr
which is clear. For the second equality, we have
AC2  4R2 . 2 AD 2  4R2 sm. 2 12 ' AE2  4R2 . 2 12 ' AF2  4R2 sm. 2 5rr12 ' It reduces to rr 2rr + sm. 2 3rr + sm. 2 4rr + sm. 2 5rr = 5 . 2sin2  + sm 12 12 12 12 12 2 ' which is also clear. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 6(1971), pp. 27, Problem 821) 55. Let A 1 A2 . . . An be the given polygon and let S be the area of AIA2 . . . An . There is a point M inside the polygon such that n L k= 1 MA% = 2S. We write n S = L SAk MAk+l ' An+l = AI. k=1 Hence n MAk MAk+1 sin AkMAk+1  � L..;n MAk MAk+ 1 S = L..; 2 2 k=l k= l 1� 1� ::; 4 L..; ( MA2k + MA2k+ 1 ) = 2 L..; MA2k = S. k=l k=l 2rr sm 12 ' 4rr sm
�
3rr
_
<
�
, no
is

Furthermore,
Ak�+l = 1 and MAk = MAk+1 , k = 1,2, . . . It follows that M is the circumcenter of a cyclic polygon and all sides subtent arcs of 90°. That is the polygon a square, desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 50, Problem 35 28) 56. In a triangle ABC with the altitude AA' and the circumradius R the following equality holds: AB · AC = 2R · AA'. (1) Let R be the circumradius of the polygon A lA2 .. . An and let P}, P2 , . . . , Pn be the projections of a point P on the circumcircle onto the sides A l A2 , A2 A3 , . . . , An Al, respecti vely. Applying (1) for triangles PAi Ai+l, where An+l = AI, yields PAi · PAi+1 = 2R · PPi , i = 1,2, sin
54. From the law of cosines we derive AB = 2R sin ;2 and
or
133
SOLUTIONS
<
as
. . . , n.
Hence
which is a constant,
as
claimed.
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 12(1980), pp. 65,
Problem 4123)
57. Applying relation (1) from the previous problem to triangles M Al A2 , MA3 A4 , MA5A6 , . . . , MA2n l A2n , we obtain MAl ' MA2 = 2R · MK}, MA3 · MA4 = 2R· MK3 , . . . , MA2n1 • MA2n = 2R · MK2n1 • MUltiplying these equalities yields MK1 · MK3 · · · MK2n  l _ MAl · MA2nR2 n. . . MA2n (2) For the triangles
relation
(1) yields MA2 · MA3 = 2R · MK2 , MA4 · MA5 = 2R · MK4 , . . . ,
134
3.
3.2.
GEOMETRY
Multiplying these equalities gives
(3) M 2 M 4 . . . M 2n = MAl ' M2AnR2 n. . . MA2n Similarly, we obtain • MA n (4) MH1 . MH2 · · · MHn  MAl ' MA 2n R2n 2 by applying relation (1) to triangles MAl An+ l, MA 2 An+ 2 , . . . , MA n A2 n. From equalities (2), (3) and (4) we draw the conclusion. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1981), pp. 68, Problem 4622) 58. Let x and y be the measures of the arcs subtended by the sides a and b, respectively. We have nx + ny 27r, or 27r x + y = . Let R be the circumradius of the polygon. Then . 2x = 2Ra and sm' 2y = 2Rb sm 27r  x so Now y = . 2'y = sm. ( ;;7r  2X ) = sm. ;;7r cos 2X  cos ;;7r sm. 2x = 2bR' sm hence b + cos 7r SI. n .X . 7r cos X2 = sm 2R n 2 Squaring b aths sides yields . 2 �7r cos2 2X 4Rb22 + 2 2Rb cos �7r sin 2X + cos2 ;;7r sin2 2X sm or b cos 7r sin X + cos2 7r sin2 X . . 2 7rn ( 1 sm. 2 X2 ) = 4Rb22 + 2sm 2R 2 2 x a . � ' a S mce sm 2 = R' we btam 2 sin2 � (4R2  a2 ) b2 + 2ab cos � + a2 cos2 � . K .
=
n
'
n
n

=

Therefore
(Dorin Andrica)

n
59. We have from the CauchySchwarz Inequality that
K
K
=
n
n
n
2 2 R = �/a 2 sin  + 2ab cos :': + b . n
n
n
135
SOLUTIONS
(tt=l �) (tt=l ai . bi) (tt=l �) 2 �
for any positive numbers Setting n
= 4,
ai , bi , i = 1, 2, . . .
yields
, n.
BCD S 2 3V " SPA where V is the volume of the tetrahedron. Then " SBCD S2 S2 = � L...J PA  3V rS r ' L...J
>
>
,
=
as desired. The equality occurs if and only if
� = a N. V b;
i = 1, 2,3,4 hence bi = a I , i 1,2,3,4. Then PAl = PBI = PC1 = PD1, so P is the incenter of tetrahedron AB CD. Remark. The inequality holds for convex polyhedra circumscribed about a sphere. (Titu Andreescu, Romanian IMO Selection Test, 1982; Revista Matematica Timi§oara (RMT) , No. 1(1982), pp. 82, Problem 4910) 60. All summations here range from i = 1 to i = 4. Let be the circumcenter and R2be the 2circumradius of A1 A2A3 A4 . By the Powerofapoint Theorem, GAi ·GA� = R  OG , for 1 � i � 4. Hence the desired inequalities are equivalent to (1) and (R2  OG2 ) L G�i � L GAi• Now (1) follows immediately from (3)
for
=
0
136
3. G EOMETRY
by the ArithmeticGeometricMean Inequality. To prove (3) , let P denote the vector from 0 to the point P. Then (4)
(4) vanishes. By CauchySchwarz 4 L GA� � (L GAi) 2 and L GAi L G�i � 16, 41 6 GAi2 6 GAi1 � 1 (6 (GAi ) 2 ) 2 6 G1Ai � 6 GAi. Hence (2) also follows from (3) . (Titu Andreescu, IMO 1995 Shortlist)
This is equivalent to (3), since the last term of Inequality, so
�
�
16
�
�
�
Chapter 4 TRIGON OMETRY
PROBLEMS
1. Prove that 2. Prove that cos3 for all x E
JR.
X

3
+ cos3
X
+ 27r

3
+ cos3
X
+ 47r

3
=
3 cos x 4

3. Evaluate the sum n1
Sn L k= l sin kx cos(n  k)x. =
4. Evaluate the sums
Sl S2
= =
+ sin 2x cos 3y + . . . + sin(n  l)x cos ny, cos x sin 2y + cos 2x sin 3y + . . + cos(n  l)x sin ny. sin x cos 2y
.
5. Evaluate the products 1) PI 2) P2
=
=
(1  tg1°)(1  tg2°) . . . (1  tg8g 0 ) j (1 tg1 °) (1 tg2°) . . . (1 tg44°).
+
6. Prove that
+
+
(4 cos2 go  3) (4 cos2 27°  3)
=
7. Let x be a real number such that sec x  tan x 8. Evaluate the product where Ixl
< 2n+7r 2 ' 139
tan g o ,
=
2. Evaluate sec x + tan x.
4.
1 40
4 .1.
TRIGONOMETRY
9. Let a, b, e, d, x be real numbers such that x f:. k7r , k E sin x a
=
sin 2x b
sin 3x
sin 4x
= e = d '
.z
and
17. Let a, (3 be real numbers with (3 � 1. Prove that
6 e = and 2 sin a  6 sin b + 7 sin e  9 sin d = 0. Prove that 3 cos(a + d) = 7 cos(b + e) . 2 cos a + cos b + 7 cos + 9 cos d
arccos a + arccos b + arccos
then
°
e = 7r,
for all a E
12. Let a, b, e be positive real numbers such that
e e = 1. 1 1 1 arctg  + arctg  + arctg  = 7r . a e b ab + b + a
13. Let x and y be real numbers from the interval (0, i) such that
7r Prove that x + y = 2 '
cos2 (x  y)
= sin 2x sin 2y
(1 + 2 sin2 a)t3 + (1 + 2 cos2 a)t3 � 213+ 1
JR.
18. Let x be a real number, x E [1,1] . Prove that for all positive integers n.
19. Prove that for all integers
Prove that
 
sin3 a cos3 a > sec(a  b) + sin b cos b 
7r for all 0 < a, b < 2"
10. Let a, b, e, d E [O , 7r] such that
11. Prove that if
16. Prove that
14 1
PROBLEMS
n
1_ < x2n + (1  x2 ) n < 1 _ 2n  1 
+ cosec2nx � 2n+1 � and for all x E (0, �) . sec2n x
°
20. Prove that for all real numbers x.
(1 + sin x) (l + cos x) :::; � + V2
21. Find the maximal value of the expression
E = sin Xl COS X2 + sin x2 cos x3 + . . . + sin xn cos Xl ,
when X l , X 2 , . . . , Xn are real numbers.
22. Find the extreme values of the function f : JR
t
1R,
= a cos 2x + b cos x + e, where a, b, e are real numbers and a, b > 0. 23. Let ao, aI , . . . , an be numbers from the interval (0, /2) such that tan ( ao  �) + tan (a1  �) + . . . + tan (an  i) �  1 . Prove that tan ao tan a1 . . . tan an � n+ f (x)
(0, i ) such that �2 (1  tga) (l  tg(3) (1  tg,) = 1  (tga + tg(3 + tg,). Prove that a + (3 + , = "4' 15. Let a, b E (0, i ) , Prove that ( Si�2 a ) 2 + ( cos2 a ) 2 = 1 sm b cos b if and only if a = b. 14. Consider the numbers a, (3" 7r
E
7r
n
n
1.
24. Find the period of the function f(x) if p , q are positive integers.
=
cospx + cos qx,
x E 1R
142
4.
TRIGONOMETRY
4. 1 .
2
25. Let aD = � + V3 + v'6 and let an+1 = 2 (aann+52) for n > O. Prove that an = cot for all n.
(
34. Prove that in any triangle B
A
C
B
C
A
� cos3 2 sin 2 sm. 2 = cos 2 cos 2 cos 2 � sm. 2 2 '
2n 3 7r  )  2
�

3

�
35. Let n be a positive integer. Prove that in any triangle
26. Let n be an odd positive integer. Solve the equation cos nx =
A
1 43
PROBLEMS
2n1 cos x.
and
27. Solve the equation
L sin nA sin nB sin nC = (_ I) n+1 + cos nA cos nB cos nC L cos nA cos nB sin nC = sin nA sin nB sin nCo
36. Consider a triangle ABC such that
A sin 2 x + B sin 2x + C = 0,
sin A sin B + sin B sin C + sin C sin A = A
where A, B, C are real parameters. and
28. Solve the equation
(1 + sin A) (1 + sin B) (1 + sin C) = 2(A + 1)
. z cos x = '23 . y cos z + sm . x cos y + sm sm
29. Prove that the equation
. .2
4
Prove that triangle AB C has a right angle.
37. Let A > 1 be a real number and let ABC be a triangle such that
3
. 3x sm . x= 4 sm x sm x sm
has no real solutions.
and
30. Solve the system of equations
{
Prove that the triangle is isosceles.
2 sin x + 3 cos y = 3 4 3 sin y + cos x = .
2
38. Prove that the triangle ABC is equilateral if and only if 1 A B C tg "2 + tg 2" + tg = 4S 2"
31. Solve the system of equations
{X
sin y +
VI 7r
x + y = 4'

x2 cos Y =
�
2
a cos A + b cos B + ccos C =
39. Let ABC be a triangle such that Prove that triangle ABC has a right angle.
37r x+y+Z= 4 tgx + tgy + tgz = 5 tgx . tgy . tgz = 1.
33. Prove that in any triangle
(a2 + b2 + c2 ).
sin2 B + sin2 C = 1 + 2 sin B sin C cos A.
32. Solve the system of equations
1
aA cos B + bA cos A = cA a2A1 cos B + b2A1 cos A = C2A1•
40. Let ABC be a triangle such that
(cot �r + (2 cot �r + (3 cot �r = ( �; r ,
abc 2R2
where s and r denote its semiperimeter and its inradius, respectively. Prove that triangle ABC is similar to a triangle whose side lengths are all positive integers with no common divisor and determine these integers.
T
1 44
1 45 51. Let ZI,Z2 ,Z3 be complex numbers, not all real, such that I Z11 I Z21 I z l 1 and 2(ZI + Z2 + Z3 )  3Z1 Z2Z3 E JR. Prove that 4. 1 .
4. 41. Prove that in any triangle . A2 cos B2 cos C2 < 89 . � sm TRIGONOMETRY
PROBLEMS
=
L..J
=
a =
42. Prove that in any triangle
A be ea ab > 4 (sm. 2 2 + sm. 2 B2 + sm. 2 C2 ) .
�
�
�
52.
++
43. Prove that in any triangle cos A
a3
+
cos B
b3
+
Let n be an even positive integer such that complex roots of unity of order n. Prove that
nII l (a + bc%) (a � + b � )2 k=O for any complex numbers a and b. 53. Let be an odd positive integer and co, C1 , . .. , cn l the complex roots of unity of order Prove that nII1 (a + bc%) an + bn k=O for all complex numbers a and b. 54. Let ZI,Z2 ,Z3 be distinct complex numbers such that I Z11 I Z21 I Z3 1 Prove that
e3 > � 16p3 .
=
cos C
44. Prove that in any triangle
2 A 2 B 2 C > _9 be ea ab .
sec  sec  sec 2 2 + ___ 2 + ___
__
45. Prove that in any triangle
!!.

n
n.
p2
=
> 3V3.
r 
46. Let ABC be a triangle. Prove that
A 2 B + cos B 2C + cos C 2A ' 47. Find the number of ordered pairs (a, b) such that (a+bi)2002 abi, a, b E 3A + sin 3B + sin 3C
sin 2
2
=
R
48. Find
and
Imz 5 Im5 and the values of z for which the minimum is reached. min
arg
:s arg
Z2 :s . . . :s arg Z2n
:s
7r.
55. Let ZI,Z2 ,Z3 be distinct complex numbers such that I Z11
Z2 f:. Z3 · Prove that
=
IZ2 1 . IZ2n l and = .: =
50. For all positive integers k define
Ak {z E I Zk I}. Prove that for any integers and with 0 < < we have Al U A2 U · · · U Am Anm+1 Anm+2 An . =
<C
n
C
=
IZ21 I Z3 1
= r
=
Iz +
56. If Z is a complex number satisfying I z3 + z 3 1 :s 2, the inequality show that
z 1 1 :s 2.
57.
m
= r.
Z
Prove that
=
=

49. Let ZI, Z2 , ... , Z2 n be complex numbers such that I Zl l
ZI
=

2 :s cos 
zE C\1R
� is odd and co , C1 ,· . . ,cn 1 the
m
U
n
U···U
(Zb Z2 ) aE
The pair of nonzero complex numbers has the following property: there is a real number [2, 2] such that z? + z� = Prove that all pairs (zf , z�) , n = 2, 3, . . . , have the same property.
aZI Z2
O.
58. Let A1 A2 . . . An be anregular polygon with the circumradius equal to 1 . Find
the maximum value of max
jII=1 PAj when P describes the circumcircle.
1 46
4.
59.
TRIGONOMETRY
n
Let be an odd positive integer and let aI , a2 , . . . , an be numbers from the interval [0, 71']. Prove that
k>
60. Let n be a positive integer. Find the real numbers ao and akl, k, l = 1, n,
l, such that
nx = ao + L akl cos 2(k  l)x 19< k 5
sin2 . sm2 x
for all real numbers x f:. m7l', m E Z.
n
SOLUTION S
1. We prove that for all integers n > O. Consider the equation
k7l' n (2n  1) � � cot 2 21 ' 3 k= 1 _
n+
sin(2n + l)x with roots
(1)
= 0,
n7l' 2 71' 71' ..., . 2n + 1 ' 2n + 1 ' 2n + 1 Expressing sin(2n + l)x in terms of sin x and cos x, we obtain
n = (2n : 1) cos2n x sin x en : 1) COs2n2 sin3 + . . = = sin2n+, ( en : 1) cat2n en: 1) cot2n2 x + . . . ) _
sin(2 + l)x
X
Set x

X
X
.
_
k7l' I 0 , we have , k 1 , 2 , . . . , n. S"Ince SIn2n+ l x r 2n + 1 2n 1 1 cat2n X _ cot 2n2 x + . . . O .


en: ) ( :) = Substituting y = cot2 x yields en: l) yn en: l) yn l + . . . = 0, 
n7l' Using the relation between coefwith roots cot2 71' , cot 2 � , . . . , cot 2 2 +1 2n + 1 2n + l ' ficients and roots, we obtain 2n + 1 n k n(2n  1) 3 7l' � 2. cot2 2n + 1 1 3
n

( ) = en: )
= 3, the desired conclusion follows. (Dorin Andrica, Revista Matematidi Timi§oara (RMT), No. 12(1979) , pp. 51,
Setting n
Problem 3831)
1 47
148
2. Applying the identity cos for
.[
4.2. SOLUTIONS
4. TRIGONOMETRY
t = x, t = x + 271", t
=
t = 4 cos3 3"t  3 cos 3"t ' t E
lR
x + 471" and summing up the three relations, we obtain + 271" + 471" 4 cos3 3' + cos3  + cos3  3 3
( X X X ) 3 cos x = ( X3 X +3 271" X + 471" ) 3 cos  + cos  + cos  . 3
On the other hand, x + 471" x 271" cos  + cos  + cos 3 3 3
+
X = 
+ =  + 
2x + 471" 471" 2 cos  cos 6 6 x 21T + 1 cos 0
= (2 cos � )
�
cos
82  81
Hence
271"
X
81
3
=
and the desired identity follows. Revista Matematica Timi§oara (RMT) , No. 2(1975), pp. 44, Problem 2124)
=
= = + . . . + sin(n  l)x cos x+ + sin(n  l)x cos x + sin(n  2)x cos 2x + . . . + cos(n  l)x sin x = = sin nx + sin nx + . . . + sin nx = (n  1) sin nx, n1 8n =  sin nx. (Dorin Andrica, Gazeta Matematica2 (GMB), No. 8(1977) , pp . 324, Problem 28n 8n + 8n sin x cos(n  l)x + sin 2x cos(n  2)x
so
1 2
[
sin
.
]
+ (n  1)h2 ) =
]
� sin [Y + (n  ll'� ] + sin � sin [Y + ( n  l) �] .
n l
1
= h . h2 2 sm sin f 2 (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 65, 82
Problem 3056)
5. We have PI = 0 because of the factor 1  tan 45° = O. On the other hand we have ( cos 1 ° + sin 1 ) . . . (cos 44° + sin 44°) P = _
0
(../2
2
../2 . ) (../2 ( )
cos 1 ° . . . cos 44 °
. 440 y'2 sm ° . . . T COS 440 + T T cos 1° + T sm 1 ../2 �
16803; Revista Matematica Timi§oara (RMT) , No. 2(1978) , pp. 30, Problem 3055)
4. Note that
44
cos 1 ° . . . cos 44 °
)
= sincos46°1 ° . .. ...cossin4489°° (v'2) = 222 . 44
= sin(x + 2y) + sin(2x + 3y) + . . . + sin[(n  l)x + nyJ and 82  8 = sin(2y  x) + sin(3y  2x) + . . . + sin[ny  (n  l)x] . Setting x + y = hI and y  x = h 2
(Titu Andreescu)
81 + 82 1
yields
81 + 82
.
+
[
.
]
hI . n hl sm T sm y + (n  1) 2 . hI sm 2 h2 ) + sin(y 2 h2 ) + . . . + sin(y . nh2 sm y + (n  1) h2 sm 2 T h2 . sm 2
. n hl sm y + ( n  1) hI sm 2 T h1 sm 2
and
(Dorin Andrica,
3. We have
=
1 2
= sin(y +
149
= sin(y + hI ) + sin(y + 2hI ) + . . . + sin(y + (n  l)hI ) =
6. We have cos 3x = 4 cos3 x  3 cos x, so 4 cos2 X  3 =
(2k + 1) . 90°, k E Z. Thus
(4 cos2 9 °  3) (4 cos2 27°  3) as
desired.
(Titu Andreescu)

cos 3x for all x cos x
sin 9° cos 81 ° cos 27° . = coscos819°° = =  = tan 9 ° , cos 9 ° cos 9 ° cos 27°
f.
15 0
4. 2 .
4. TRIGONOMETRY
7. From the identity 1 + tan2 x = sec2 x it follows that
(Titu Andreescu, Korean Mathematics Competition, 2002)
= sec2 x  tan2 x = (sec x  tan x) (sec x + tan x) = 2(sec x + tan x), so sec x + tan x = 0.5. (Titu Andreescu, American High School Mathematics Examination, 1999, Prob 1
11. From the hypothesis it follows that sin( arccos a + arccos b + arccos c)
for all Ixl
<
+
1 tan2 2 k x (1  tan2 2 k x)2
: it follows that 2 +2 '
=
L cos a cos ,B sin 'Y
cos2 2 k x cos2 2 k+ 1 X
sin(arccos x)
sin(a + ,B
=
+ 'Y) + sin a sin ,B sin 'Y
� and sin(arcsin x)
we obtains ab � + bc � + ca J1=b2 as desired.
9. Let
Problem 3054)
Because sin2 4x =
=
and the formulas
(Dorin Andrica)
= x,
x E [ 1, 1] ,
= V(I  a2 ) (1  b2) (1  c2 ) ,
(Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2(1977) , pp. 64,
sin x sin 2x sin 3x sin 4x         ;\ b a c d _
_
Then
= O.
Using the identity
lem 15)
8. Since
151
SOLUTIONS
_
_
12. The identity
.
sin 2 4x = 2 sin2 2x(l  sin2 2x). 2 2 sin 2x(1  sin2 2x) , we obtain
= 2b2 (1 ;\2 b2 ) . On the other hand, sin 3 x = AC, sin x = Aa, and since sin 3x = sin x (3  4 sin 2 x), we have c = a(3  4A2 a2 ). d2
_
Eliminating A from the relations (1) and (2) yields 2a3 (2b3  �) = b4 (3a  c),
as desired.
(Dorin Andrica)
10. Rewrite the two equalities as 2 sin a  9 sin d
+
= 6 sin b  7 sin c
arctgx + arctgy + arctgz implies
(1)
and the conclusion follows.
=
85
+ 42 cos(b + c) ,
+ + +
1 1 1 ab + bc ca  1 arctg arctg + arctg = arctg k7r. ab c  ( a + b c) a b c Because ab + bc + ca = 1, we obtain 1 1 1 arctg  + arctg + arctg = k7r, a b c where k is integer. 7r Note that 0 < arctgx <  for all real x > 0, hence 2 1 1 3 7r 1 o < arctg + arctg z; + arctg c < � 2' Therefore k 1 and 1 1 1 arctg + arctg + arctg 7r, a b c as claimed. Revista Matematidi Timi§oara (RMT) , No. 1 (1977), pp. 42, Problem 2827)
+

(2)
=

=
(Titu Andreescu,
13. The given relation is equivalent to
2 cos a 9 cos d = 6 cos b  7 cos c. By squaring the two relations and adding them up we obtain 85 + 18 cos(a + d)
= arctg 1 x+(xyy++zyz +xyzzx) + k7r
= 4 sin x sin y cos x cos y, (cos x cos y  sin x sin y) 2 = 0
(cos x cos y + sin x sin y) 2 or
152
4. Hence
cos2 (x +
=
y) = 0,
x, y E (o, �), we obtain x + y = �, as desired. Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1977) , pp. 42,
and since (
4. 2.
TRIGONOMETRY
Problem 2826)
or tan 2 a tan2 b. Because a, b E 0, we obtain a b. The converse is clear and we are done. From the given relation we deduce that there is a number such that e E 0,
( i),
Alternative solution.
sin2 a sin b
1 "2 (1  tan a  tan ,B  tan 1 + tan a tan ,B + tan ,B tan 1+ + tan 1 tan a  tan a tan ,B tan 1) 1  (tan a + tan ,B tan 1) ,
Hence
=
or
°
<
<
sin 2 a It follows that
From relation (1) we derive 1  tan a tan ,B  tan ,B tan 1  tan 1 tan a 1' tan a tan ,B + tan 1  tan a tan ,B tan 1 therefore cot(a + ,B + 1) 1. Hence a + ,B + 1 as desired. ( Revista Matematica Timi§oara (RMT) , No. 1(1973) , pp. 42, Problem 1582)
=
=
= i, Titu Andreescu,
15. The relation in the statement is equivalent to
(
sin4 a cos4 a . 2 b + cos2 b) (sm + .2 b cos2 b sm
or •
sm4 a It follows that
hence
+
cos2 b . 4  sm a cos 4 a + sin2 b
+
) = 1,
sin2 b  cos4 a cos2 b
Furthermore,
cos b . 2 sm a sin b
= cossin bb cos2 a,
= sin b sin e
and
= cos(b  e)
cos2 a cos b
= cos e cos2 a = cos b cos e.
and
cos 2a
as
Problem 2825; Gazeta Matematica (GMB), No. 11(1977) , pp. 452, Problem 16934)
=
16.
Multiplying the inequality by sin a sin b + cos a cos b cos(a  b), we obtain the equivalent form Sin3 a cos3 a . .  (sm a sm b + cos a cos b) 2: 1. . sm b cos b
(
)
+
But this follows from CauchySchwarz Inequality, because, according to this in equality, the lefthand side is greater than or equal to (sin2 a + cos2 a) 2 1. (
=
Titu Andreescu)
17. Using the inequality
x? + xr > ( 2
Xl
+ X2 2
)m
for m 2: 1 we obtain
= 1.
cos 2 b . 4 sin2 b . 2 a cos 2 a + 1  2 sm  sm a +  cos4 a . sm2 b cos2 b sin b 2 ) 2 (COSsm.bb sm. 2 a  cos a = 0 . cos b
and
°
(1)
tan a + tan ,B + tan 1 1= tan a tan ,B tan 1.
= sm. e
= cos(b + e) Since a, b, e E (0, i), we have b  e = and 2a = b + e, hence a = b, desired. ( Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1977) , pp. 41, 1
7r ,
+
=
( i)
14. Expanding yields
= + tan a + tan ,B + tan 1  tan a tan ,B tan 1 = = 1  tan a tan ,B  tan ,B tan 1  tan 1 tan a. Since a, ,B, 1 E ( 0, i ) , we have a + ,B + 1 hence
1 53
SOLUTIONS
= 1,
(1 as
+ 2 8in2 alP + (1 + 2 C082 aJII � 2 C + 2 8in2 � + 2 cos2 a ) p = 2P+1 ,
desired. (
Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1974) , pp. 30,
Problem 1942)
Because X E [ 1, 1J , there is a real number y such that x = sin y . It suffices to prove that 1 > sin2n + cos2n > __ 12n  1
18.
y
y
� 1 and I cos yl � 1, hence sin2n y + cos2n y � sin2n2 y + COS2 n 2 y � . . . � sin2 y + cos2
I
Y
= 1,
desired. For the righthand side we use the inequality x? x� > Xl X2 n 2 2
( + )
+
Hence as
as
'4
2
2
vIn
=
"2
=
"2
(Titu Andreescu, Revista Matematidl Timi§oara (RMT) , No. 2(1978) , pp. 47,
claimed.
Alternative Solution. By setting u X2 and v = 1  X2 the inequality becomes u,v E [0, 1] and u + v = 1, we have un � vn � v, n l � uunn++vvnn �� 1. +Because 2implying v 1. Also, by then power mean inequality, 1_ un + vn  2 ( u+2 V ) = 2 (!2 ) n = _ 2n  l . (Dorin Andrica) =
1
U
U,
=
>
19. We have
+ 1 2:: 2 tan x and cosec2 x = cot2 x + 1 2:: 2 cot x by the AMGM inequality. It follows that sec 2n x + cosec 2n x 2:: 2 n (tan n x + cot n x). Since tan n x + cotn x 2:: 2, we obtain sec2n x + cosec2n x 2:: 2n +l , sec2 x = tan2 x
as
+ (sin x + cos x) = 3 + V2 sin (x + 7r ) � 3 + 2, desired. Note that equality holds for x � + 2k 7r , where k is integer. Alternative Solution. Expanding the lefthand side, we see that 1 + (sin x + cos x) + � sin 2x = 1 + V2 sin (x + �) + � sin 2x � � 1 + V2 + 1 3 + V2. = 23
For the lefthand side note that sin yl as
155
4.2. SOLUTIONS
4. TRIGONOMETRY
154
desired.
Alternative Solution. Using the AMGM inequality we obtain sec2 n x
>
+ cosec2n x  2Jsec2n xcosec2nx = 2 sinn x1cosn x
(Dorin Andrica,
l_ = 2n+ l __ n
>
2n+l sin 2x Gazeta Matematidi (GMB), No. 3(1975), pp. 104, Problem

14900)
20. We have
Problem 3500)
21. We have E = sin Xl COS X2 + sin x2 cos x3 + ... + sin xn cos X l �
� sin2 Xl +2 cos2 X2 + sin2 X2 +2 cos2 X3 + . . . + sin2 Xn +2 cos2 Xl 2 Therefore the maximal value of E is and it is reached, for example, when 2 7r X l = X2 = . . . = X n = '4 ' (Dorin Andrica, Revista Matematidi Timi§oara (RMT), No. 2(1977), pp. 65, n
n
Problem 3058)
22. Because a, b
>
0, it follows that the maximal value of f is Setting y = cos x yields f (x)
=
a(2y 2
a+b+
c.
 1) + by + c = 2ay2 + by + c  a.
If  � E [ 1, 0) , then the minimal value of f is 4a � 8a(c  a) 8a 8a
b2 +
 :a E ( 00 1), then the minimal value of is f ( 1) 2a  b + c  a = a  b + c. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 1(1981) , pp. 52,
If
,
f
=
Problem 4315)
. )(1 + coo x)  (1 + sin x) 2 + (1 + COS X)2 2
(1 + ffin x _

2
<
+ 2(sin x + cos x) + (sin2 x + cos2 x) _
2
=
23. First Solution.
Let hypothesis that for each k,
bk = tan(ak  7r/4), k = 0, 1, . . . ,n.
1 bk 1 and 1 + bk 2:: 9¥k5; L n (1  bl ) . <
It
follows from the
<
0
(1)
156
157
4. TRIGONOMETRY
4.2. SOLUTIONS
Applying the ArithmeticGeometricMean Inequality to the positive numbers 1 bz , = 0, . . . , k  1, k + we obtain
other by a positive number chosen so that the sum of the pair does not change. Each such change decreases the product of the i'S. It follows that for a given sum of the bi 'S, the minimum product is attained when all of the bi ' S are equal. In this case we > for each so have bi n + 1'
l 1,
1, . . . ,n, l in (2) II n (1 bl) l=lk'5,n (1  bl) � n 09=1k'5, O'5,L From (1) and (2) it follows that l in !!n (1 + bk ) � nn+1 (Don (1  bt )n) , and hence that n 1 + bk n+l II 1  bk  n . k=O Because 1 + bk  1 + tan (ak  �) = tan ( (ak  �) + �) = tan ak , 1  b k 1  tan (ak  �) the conclusion follows. Second Solution. We first prove a short lemma: Let W,X,Y,z be real numbers with x + y w + z and Ix  yl < Iw  zl . Then xy. wz <Proof. Let x + y w + z = 2L . Then there are nonnegative numbers r, s with r < s and wz (L  s) (L + s) < (L  r) (L + r) = xy. We now use this lemma to solve the problem. For 0 ::; k ::; n, let b k = tan(ak  7r j4)
(
_
)
n1 
t
i,
( lli) n 1
tO t 1 ' " tn � : :
n+1
= C;) n+l = nn+ 1 .
+ This completes the proof. We present a solution based on calculus. We set
Third Solution.
>
=
=
=
and let
where  1 < bi < 1, and assume that
a � n  1. We then show that the product IT 1 + bk k=O 1  bk
attains its minimum when all of the b k 's are equal, that is, their common value is
aj(n + 1). The desired inequality will follow immediately. We proceed by induction . The case n 1 was established in the discussion of (1) in the previous solution. For n � 2, set n1 L k=O bk a' = a  bn n  2. The last inequality follows from a � n  1 and bn < 1. Set b = bn and c = a' jn, so b + nc a. By the induction hypothesis, (IT 11 + bkbk ) 11 + bbnn  ( 11 + cc ) n 11 + bb . k=O Thus we need to prove that ( 1 + c) n (�) � ( nn ++ 11 + aa ) n+1 (1) 1c 1b where the righthand side is obtained by substituting aj(n + 1) for each bk , k = 0, 1 , . . . ,n, in the product . Next, recall that a is fixed, and that b + nc = a. Thus we can eliminate b from (1) to obtain the equivalent inequality n+ 1 ( � (2) G ��r G �:�:�) ::� �: ) Now bring all terms in (2) to the lefthand side of the inequality, clear denomi nators, and replace c by x. Let the expression on the left define a function f with f(x) (1 + x) n (1 + a  nx)(n + 1  a) n+1  (1  x) n (l  a + nx) (n + 1 + a)n+ 1 . =
>
=
=
>
1 < bk < 1 and (1) tjtk G � ��) G � :: ) 1 + 1 + bj�k 1 bj + b + k First note that because 1 < bk < 1 and bo + b1 + . + bn � n  1, it follows that bj + bk 0 for all 0 ::; j, k ::; n with j f:. k. Next note that if bj + bk > 0 and bj f:. b k , then it follows from the lemma applied to (1) that the value of tj tk can be Then
=
=
>
.
_
made smaller by replacing bj and bk by two numbers closer together and with the same sum. In particular, if bj < 0, then replacing bj and bk by their average reduces the problem to the case where bi > 0 for all We . may now successively replace the b/s by their arithmetic mean. As long as the bi are not equal, one is greater than the mean and another one is less than the mean. We can replace one of this pair by the arithmetic mean of all the bi'S, and the
i.
=
1 58
4.
4.2.
TRIG ONOMETRY
To establish (2) it is sufficient to show that for 0 � x < 1, I(x) attains its minimum value at x = aJ (n + 1) . Towards this end we differentiate to obtain 1 ! ' (x) = n (a  (n + l)x) ( (l + x) n  1 (n + 1  a) n+  (1  x) n  1 (n + 1 + a) n+ 1 ) = = n (a  (n + l)x)g(x) , 1 n where g(x) = (1 + x)  (n + 1  a) n+ 1  (1  x) n 1 (n + 1 + a) n+ 1 . It is clear that I'
(n: 1)
=
0, so we check the second derivative. We find
( : l ) = n(n + l)g ( n : l ) > 0,
f" n
so I has a local minimum at x = aJ (n + 1). But I' (x) could have another zero, obtained by solving the equation g(x) = Because
O.
t,

g' (x) = (n  1)(1 + x) n 2 (n + 1  a) n+ 1 + (n  1) (1  x) n 2 (n + 1 + a) n+ 1 is obviously positive for all x E [0, 1), there is at most one solution to the equation g(x) = 0 in this interval. It is easy to check that g(aJ (n + 1)) < 0 and g(l) > O. Thus there is a real number aJ (n + 1) < < 1, with g(t) = For this we have
t,
t
t
O. I" (t) = n(a  (n + l)t)g' (t) < O.
t
Thus, is a local maximum for I, and no other extrema exist on the interval (0, 1). The only thing left is to check that 1 (1) � l(aJ(n + 1)). Note that the case x = 1 is also an extreme case with bo = b1 = . . . = bn  1 = 1 . This case does not arise in our problem, but we must check to be sure that on the interval 0 � x < 1, I (x) has a minimum at x = aJ (n + 1). We have 1(1) = 2n (1 + a  n) (n + 1  a) n+ 1 0,
�
since n  1 � a � n + 1, and l(aJ (n + 1)) = 0 (by design). Thus I(x) indeed attains a unique minimum at x = aJ(n + 1). USA Mathematical Olympiad, 1998, Problem 3)
(Titu Andreescu,
24. Let d be the greatest common divisor of p and q. We prove that T = 27r is d the lowest positive period of the function f. It is clear that I(x + T1 ) = I (x)
for all real x, therefore T is a period of function I. Suppose there are T1 > 0 and an integer A > 0 such that T = AT1 and l (x + T1 ) = I (x) for all real x. Then I( T1 ) = 1(0) = 2, so cos pT1 + cos qT1 = 2,
1 59
SOLUTIONS
therefore
COSpT1 = COS qT1 = 1. It follows that
T1 =
O.
2k l7r
P
=
2k2 7r q
for some integers k1 ' k2 > 27r Since T = AT1 and T1 = Ad ' 1 k1 k2 =q= and so p = k1 ( Ad) , q = k2 ( Ad) . Ad P On the other hand, d = gcd (p , q) , so A = 1, hence T = T1 as desired. Revista Matematidt Timi§oara (RMT) , No. 2(1978), Problem 3695)
(Dorin Andrica, 25. We have
pp.
75,
7r 7r 7r 1 + cos 12 2 cos2 cos 7r 24 24 = .7r = cot 7r . 24 sm sm 2 sin !!.. cos !!..12 24 24 24 \I'2 v'6 l+ + 1 + cos 4 4 7r 7r . sm 3  "4 4 4 2 4( + ( ) + + 4( v'6 \1'2) v'6 \1'2 v'6 + \1'2) + 8 + 4V3 4 62 = 2 + \1'2 + V3 + VB = ao + 2. 2n 3 7r  2 is true for n = 0. Hence an = cot 2n 3 7r 1. The It suffices to prove that bn = cot , where b n = an + 2, n recursive relation becomes ( b  2) 2  5 bn+ 1 2  n ' 2bn 2 k 3 7r b2  1 . Assuming, inductively, that bk = cot ck , where Ck = 3 ' yields or bn+ 1 = ;bn cot 2 Ck  1 = cot(2Ck) = cot Ck + 1 , bk + 1 = 2 COt ck and we are done. Korean Mathematics Competition, 2002)
(i �) ( )
_
_
( ; )
_
(; )
_
_
(Titu Andreescu,
26. If n = 1, then all real numbers x are solutions to the equation.
2::
4.
1 60
4. 2.
TRIGONOMETRY
Let n > 1 and note that cos nx
= (�) cosn  (�) cosn2 x sin2 x + . . . + n + (  1) """"2 ( _ ) cos x sin n  1 n 1 X
nl
X.
We have two cases: a) x f:. (2k + 1) � for any integer k. Then 2
l (�) cosn 1  (�) cosn3 sin2 + . . . + + (  1) n;' ( � ) sin n1 x l ::; n 1 ::; I cos xl ((�) I cosn  1 xl + (�) i cosn  3 x sin 2 xl + . . . + ( � ) I sin n  1 xl ) < n 1 I cos nxl
=
I cos x '
X
X
hence there are no solutions in this case. b) x (2k + for some integer k. Then
=
l)i {
i
cos x
= 0 and cos nx = 0, }
since n is odd, so (2k + 1) lk integer is the set of solutions.
Alternative Solution. With the substitution x = i  y the equation becomes cos (n �  ny ) = 2n  1 sin y. (1)
Because n is odd, (1) is equivalent to
= 2n1 sin y. Taking modules gives (2) I sin nyl = 2n  1 1 sin yl . But I sin nYI � nl sin yl for all y in hence nl sin yl � 2n1 I sin yl · If y f:. k7r, k E Z, then n � 2n  1 , which implies n E {1, 3}. The case n = 1 is clear and for n = 3 the original equation reduces to cos 3x = 4 cos x, that is 4 cos3  3 cos x = 4 cos x. Taking into account that cos x f:. 0, this yields cos2 x = �, which i s not possible. It follows that y k7r, which gives the solutions x = (2k + l)i, k E Z. (Titu Andreescu, Gazeta Matematidi (GMB), No. 7(1978), pp. 304, Problem ± sin ny
27. The equation is equivalent to
(A + G) sin2 x + 2B sin x cos x + G cos2 = 0 We have the following cases: i) A + G = 0 and G f:. O. Then 2B cos x = 0 or cot x =  0 ' hence 2 (2k ; 1) 71" xE { 1 k E Z } U { arcctg (  %) + h l k E Z } . ii) A + G f:. 0 and G = O. Then 2B sin x = 0 or tan x =  ' A+G hence 2 x E {k71" 1 k E Z} U { arctg (  : ) + hi k E Z }. A iii) A = B = G = O. Then any real number is a solution. iv) A = G = 0 and B f:. O. Then sin 2x = 0 and so x E {ki l k E Z } . v) A + G f:. 0 and G f:. O. The equation is equivalent to X
c
(A + G) tan2 x + 2B tan x + G = 0,
hence tan x for B 2 + G 2
=
17297; Revista Matematidl Timi§oara (RMT) , No . 12(1980), pp. 63, Problem 4107)
= B ± JB2A +_GAG + G2
� AG. It follows that
Z} U
x E {arctgY1 + k7r 1 k E {arctgY1 + k7r1 k E Z } if B 2 + G2 AG. Otherwise there are no solution. Revista Matematica Timi§oara (RMT), No. 1(1978), pp. 89, Problem 3429)
� (Dorin Andrica,
lR,
X
16 1
SOLUTIONS
28. The equation is equivalent to 2 sin x cos y + 2 sin y cos z + 2 sin z cos x or It follows that
(sin x  cos y) 2 + (sin y  COS Z ) 2 + (sin z  COS X ) 2 = O. sin x
Hence
= 3,
= cos y,
sin y
= cos z ,
sin z = cos x.
1 62
4. Z
4.2.
TRIGONOMETRY
7r + X = (4k3 + 1 ) '2
and
for some integers k1 ' k2 ' k3 and therefore
�
Y
x = [4(k1  k2 + k3 ) + 1 ] ,
= [4(k1 + k2  k3 )
�
and
z = [4(k1 + k2 + k3 ) + 1] ,
+ 1] � ,
k1 , k2 , k3 E z .
4 y = arctan 3 + 2l7r,
for some integers k and l. Revista Matematica Timi§oara (RMT ) , No. 2 ( 1978 ) , pp. 74, Problem 3694)
(Titu Andreescu,
(Titu Andreescu, Gazeta Matematica (GMB ) , No. 11 ( 1977) , pp. 451, Problem
31. Observe that x E [1, 1] and
29. Note that
From the first equation we obtain
x = sin ( arcsin x) ,
16931 ; Revista Matematica Timi§oara ( RMT ) , No. 12 ( 1979 ) , pp. 52, Problem 3835 )
sin x sin 2x sin 3x sin 4x =
� (cos 3x  cos 5x) (cos x  cos 5x) =
1 = 4 ( cos2 5x  cos 3x cos 5x  cos 5x cos x + cos x cos 3x ) =
1 6 3 =  ( 2 cos2 5x  cos 2x + cos 8x  cos 4x cos 6x ) <  = , 8 4 8 hence the equation has no solution. Revista Matematica Timi§oara (RMT ) , No. 1 ( 1977 ) , pp. 41, Problem 2923 )
+
(Titu Andreescu,
30. Squaring both equations and summing up yields 4(sin2 x
+ cos2 x) + 9 (cos2
or 13 Hence
Y
+ sin2 y ) + 12 (sin x cos y
+ sin y cos x) = 25,
+ 12 sin(x + y) = 25. sin ( x + y ) = 1,
and so X
7r + y = (4k + 1 ) '2
3
tan x = 4 '
. .j2 cos ( y  arcsm x ) = 2' then y  arcsin x = 7r
. x + y = "4 ' we get U smg for some integer k. Case x + arcsin x = have k = 0, hence
1.
Therefore
�  2k7r . Because x E [ 1, 1] and arcsin x E [�, �] , we 7r x + arcsm x = '2 •
7r arcsm x = '2  x, •
or x = cos x.
For this equation there is only one solution X o E solution 7r Y = "4  Xo X = Xo ,
(0, �). The system has the
+
arcsin x = x .
sin y = cos x.
4
tan y = 3 ·
Note that sin x, cos x, sin y, cos y are all positive, therefore 3 x = arctan 4 2k7r
+
±� + 2k7r.
� ± � + 2k7r,
x + arcsin x =
2.
Turning back to the system we obtain . x = cos y = 53 and sm . y = cos x = 54 ' sm hence
� = cos ( arcsin x) .
Case x arcsin x = 2k7r. Using similar arguments, k = 0, so
for some integer k. It follows that sin x = cos y and
1 63
SOLUTIONS
This equation has the unique solution x = ° so the system has the solution x = o,
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 ( 1977) , pp. 41,
Problem 2824)
32. Using the formula t an (x
+ y + z) = 1tan tanx +xtantanyy+tantanzytantanz xtantanyztantanzx
==

164
4.
we have 1 = tan
4. 2 .
TRIGONOMETRY
because the first column is the sum of the other two. Computing the determinant, we obtain A A B O A . B . O  sm  = cos  cos  cos  """ sin2 """ cos3  sm 2' 2 6 2 2 2 2 2 6 as desired.
51 37r = 4 1  tan x tan y  tan y tan z  tan z tan x
Hence tan x tan y + tan y tan z The equation
+ tan z tan x = 5.
t3  5t2 + 5t  1 = 0
(Dorin Andrica)
has roots tan x, tan y, tan z, from the relations between the roots and the coefficients. On the other hand, the equation has the roots 1, 2 V3, 2  V3, hence {x, y, z} ==
+
{� + k", �; + h", ;2 + p,,}, for some integers k,l,p.
35. Denote E1 =
L sin nA sin nB cos nO
Observe that
(Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 8(1971), pp. 27,
(cos A
Problem 1018)
33. Using the Extended Law of Sines we obtain
+ sin O cos 0) = R(sin 2A + sin 2B + sin 20) = R(2 sin(A + B) cos(A  B) + sin 20) = AB + 0 ABO = = 2R sin O(cos(A  B) + cos O) = 4R sin O cos cos 2 2 a cos A + b cos B + c cos O = 2R(sin A cos
= 4R sin 0 cos as desired.
(�2  B) cos (�2  A)
A
==
+ sin B cos B
4R sin A sin B sin 0
==
==
abc 2R2 '
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 65,
Problem 3060)
A B B O . O . "2 cos "2 = sm cos "2 + sm "2 cos "2 ' A 0 0 A . "2 . "2 cos "2 = sm cos "2 + sm cos "2 ' B
A B B A . "2 . "2 cos "2 = sm cos "2 + sm cos "2' O
0 A sin  cos 2 2
0 cos "2
B sin  cos 2 2
A
==
L cos nA cos nB sin nO.
+ i sin A) (cos B + i sin B) (cos 0 + i sin O) =
= cos(A + B + 0) + i sin(A + B + 0) = cos 7f By de Moivre ' s formula, (cos nA
+ i sin 7r = 1.
==
+ i sin nA) (cos nB + i sin nB) (cos nO + i sin nO) = (_ l) n .
Expanding the brackets yields E1 + Hence
iE2 + cos nA cos nB cos nO  i sin nA sin nB sin nO = (_ l)n .
E1 = ( _ 1) n+ 1
(Dorin Andrica,
+ cos nA cos nB cos nO
and E2 = sin nA sin nB sin nO. Revista Matematica Timi§oara (RMT), No. 1(1978) , pp. 65,
A 0 sin  cos 2 2 B A sin  cos 2 2
36. Subtracting from the second equality the first multiplied by 2 yields (sin A  1) (sin B  l) (sin 0  1) = O. Hence sin A, sin B or sin 0 is 1, so AB 0 is a right triangle. Revista Matematica Timi§oara (RMT), No. 2(1978), pp. 49, Problem 3514)
(Titu Andreescu, 37. Note that
A . B 0 . 0 B cos  sm "2 cos "2 sm "2 cos 2" 2 B cos "2
and &
Problem 3278)
34. Because A + B + 0 = 7r, we have
Hence
1 65
SO LUTIONS
= 0,
{ aA
a cos B + b cos A  c = O.
The system of linear equations
bA A 2a A 1 b b2 A 1 c C2 A 1 = 0 a cos B + cos A  c = 0 cos B + cos A  = 0 cos B + cos A 
166
4.
4. 2.
TRIGONOMETRY
has the solution (cos B, cos A,  1) and is homogeneous. Therefore the determinant Ll =
a aAA1 a2
b c b A C cA b2A1 2A1
From the hypothesis we have sin 2 B
+ sin2 C = 1 + 2 sin B sin C cos A,
therefore sin 2 A = 1. It follows that A = �, hence the triangle ABC is right, as 2 desired. Revista Matematica Timi§oara (RMT) , No. 12(1977), pp. 52, Problem 3838)
(Dorin Andrica,
is zero. On the other hand, 1
1
1 abc aAA1 bAA11 CcA1(a 1 )2 (b ) 2 ( A 1 ) 2 = abc(aA  1 _ b A 1 )(aA1 _ cA1 ) (bA1 _ cA1 ) . Therefore a = b, b c or c a, hence the triangle is isosceles. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1977), pp. 89, Problem 3 1 99 ) Ll
167
SOLUTIONS
=
=
=
2 + 32 + 22 = 72 and A B C r cot 2 + cot 2 + cot 2
40. Because 6
s
 =
1 2 J(P  b)(p  c) + J(P  c)(p  a) + J(P  a)(p  b) = S ( + b2 + c2 ) 4 Jp(P  a) Jp(p  b) c) Jp(p or 1 1 2 S L(P  a)(p  b) S (a + b2 + c2 ). 4 Expanding the brackets yields _p2 + ab + bc + ca = 41 (a2 + b2 + c2 ), a
=
then
4(ab + bc + ca) a2 + b2 + c2 + (a + b + C)2 . It follows that (a  b) 2 + (b  C)2 + (c  a) 2 0, hence b = c. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1972) , pp. 28, =
=
39. By the Extended Law of Sines,
a = 2R sin A, b
=
2R sin B,
c = 2R sin C.
On the other hand,
This means that we have equality in the CauchySchwarz inequality. It follows that C B A cot  2 cot  3 cot 2 2 2 2 3 6 A B 7 C 7 Plugging back into (1) gives cot 2 = 7, cot 2" = 4 ' and cot 2" = g ' Hence by 63 28 7 . . . . the Double angle formulas, sm A = , sm B = , and sm C = 1 3 0 Thus the SIde 65 25 lengths of T are 26, 40, and 45. USA Mathematical Olympiad, 2002, Problem 2)
__ 
_
__


41 . Summing up the formulas ra
A B C = 4R sin 2 cos 2" cos 2 '
rb = 4R sin 2"B cos 2"C cos 2A ' r = 4R sm. 2C cos 2B cos 2A yields B C r + rb + r . A ""'" sm ' 6 2 cos "2 cos "2 4R On the other hand + rb + 4R + hence B C 4R + r . A ""'" sm 6 2 cos "2 cos 2 = m = 1 + 4R ' c
ra
sin2 B
+ sin2 C
=
sin2 A
+ 2 sin B sin C cos A.
'
(Titu Andreescu,
=
so
(1)
= (6 cot 2A + 6 cot 2"B + 6 cot 2"C ) 2
a =
Problem 1160)
A B C cot  cot  cot  ' 2 2 2
the given relation is equivalent to
38. The relation is equivalent to ....!.=;:::::;=: .==:::;:=::
=
rc =
r,
c
a
r
168
Because
as
� � r, it follows that
hence
1 9 A B C < 1+ = '" sin  cos  cos  L...J 2 8 8' 2 2
desired.
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 49,
42. We have
b e B ecause � + b
�
a2 2 cos A + be . 2, we obtam a2 be
�
=
!!. + sin2 Q2 ) '
(Dorin Andrica)
B be ea 1 , sec2 "2 = , A = p(p  b) P (p  a) cos2 2 so it suffices to prove that 1 9 1 1  +  +  > . p  a p  b p  e  pSetting x = p  a, Y = P  b, Z = P  e in the inequality
(
(x + Y + Z) .!. + � + � x Y Z P
2be cos A 2ae cos B 2ab cos C + + b2 a2 e2 _
,
) �9
1 1 1 ( p+  +  ) > 9 , a pb pe 
45. By the AMGM inequality,
p = (p  a) + (p  b) + (p  e)
Then
_
3+
e)
(Dorin Andrica)
+ a2e2 · 2ae cos B + 1  ab22 + eb22 b2
2be cos A b2 +1=2 a2 a
2ab cos C a2 b 2 +1+ e2 e2 e2 · Summing up these equalities implies
C ab sec2 "2 = p(p 
and the solution is complete.
_
and
=
yields
43. We have
Likewise,
+)
� "2 (
A sec2 "2
b2 > 4 sin2 !i 2' ae Summing up these inequalities yields a2 b2 e2 > 4 sin2 + sin2 + + be ea ab 2 2 desired.
so
3 cos A cos B cos C >  .  +  +  a3 b3 e3 2abe By the AMG M inequality, 3 3 3 3 81 = 16p3 ' 2abe a+b e therefore 81 cos A cos B cos C > + + as ,;s �  16pS · Revista Matematica Timi§oara (RMT) , No. 2(1975), pp. 46, Problem 2134)
44. We have
A 2(1  cos A) = 4 sin2 "2
and likewise
as
 "2 ' '
(Dorin Andrica,
b e + . e b

( �
be cos A ea cos B ab cos C 3  +  > a2 + e2 b2
and moreover
Problem 3510)
or
169
4.2. SOLUTIONS
4. TRIGONOMETRY
p3
( ab22 + ab2 ) +
so p4
2
� 3 \!(P  a) (p  b) (P  e),
� 27(p  a) (p  b) (p  e)
� 27p(p  a) (p  b) (p  e) = 2782 •
� 3V38, and since 8 = !!.r � 3V3, desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 2(1982), pp. 66,
It follows that p2 Problem 4993)
pr,
as
4.
170
I"V = Q2 , Then 0° < a, (3" < 90° and a + (3 + , = 90° , We have BC 3A = sin 3a  cos((3  ,) = sin 3a  sin(a + 2,) = cos sin: 2 2 46. Let a
=
� ' (3 = � , 2 2
I
_
=
2 cos(2a + ,) sin(a  ,)
=
2 sin(a  (3) sin(a  ,).
In exactly the same way, we can show that C3B _ = 2 sin ((3  a) sin( (3  , ) cos sin 2 2
A
and
A
3C B sin "'2  cos 2 Hence it suffices to prove that
=
sin(a  (3 ) sin(a  ,) + sin(,  (3 ) [sin(,  a)  sin((3  a )] ,
0
which is positive as function y = sin x is increasing for ° < x < 90°. We keep the notation of the first solution. We have sin 3a = sin a sin 2a + sin 2a cos ai
= sin(2a + ,) sin 2a cos , + sin , cos 2ai cos((3  ,) sin(2, + a) = sin 2, cos a + sin a cos 2,i sin 3, = sin , cos 2, + sin 2, cos , . =
=
+
sin 3a + sin 3,  cos((3  a)  cos((3  ,)
�
O.
In exactly the same way, we can show that sin 3(3
+ sin 3a  cos(,  (3)  cos(,  a) � 0
and sin 3, + sin 3(3  cos(a  ,)  cos(a  (3)
�
O.
=
= z·
=
=
=
= z,
=
=
(Titu Andreescu, American Mathematics Contest 12A, 2002, Problem 24) 4 248.2 Let5 a, b be real numbers such that z = a + bi, b =f:. Then 1m z5 5a b 
10a b + b and
Imz 5 Im 5
=
5
( � ) 4  10 ( � ) 2 + 1.
O.
b z b 2 Setting x = (�) yields 1m Z5  = 5x 2  lOx + 1 = 5(x  1) 2  4. 1m 5 z The minimum value is 4 and is obtained for x = 1 i.e, for z
=
i), =f:.
a(1 ± a O . Revista Matematica Timi§oara (RMT) , No. 1(1984) , pp. 67, =
M1 , M2 , . . . ,M2n be the points with the complex coordinates Zl, Z , " " Z and let A , A '"'' A be the midpoints of segments M1 M2n , M2 M2 2n1 , •2•n" Mn Mn+1 • I 2 n M +l M2 M2n1n AAn2 Ml M2n Al 49.
=
(sin a  sin , ) (cos 2a  cos 2,) + 2(cos a  cos ,) cos(a ,) sin(a  ,). Note that sin x is increasing and cos x is decreasing for 0 < x < 90° . Since < 90° , each of the two products in the last addition is less than or o < a" , a equal to O . Hence
+,
=
Problem 5221)
=
(sin a  sin ,) (cos 2a  cos 2,) + (cos a  cos ,) (sin 2a  sin 2,)
z=
required conditions.
(Titu Andreescu,
It follows that =
=
z.
=
. (,  . . (  a ) sm  2 sm (3) ,
sin 3a + sin 3,  cos((3  a)  cos((3  ,)
a + bi, a  bi, and I z l Ja2 + b2 • The given relation becomes 2Z 00247.= LetNotez that = O.
0
=
(Titu Andreescu,
I z l (l z l 200l  1) Hence I z l 0, and (a, b) (0,0), or I z l 1. In the case I z l = 1, we have Z 2 002 which is equivalent to Z 2003 z I z l2 1. Since the equation Z2003 1 has 2003 distinct solutions, there are altogether 1 + 2003 2004 ordered pairs that meet the
Note that this inequality is symmetric with respect to a, (3 " , we can assume without loss of generality that ° < a < (3 < , < 90°. Then regrouping the terms on the lefthandside gives
cos((3  a)
Adding the last three inequalities gives the desired result. USA IMO Team Selection Test, 2002, Problem 1)
from which it follows that
sin(a  (3) sin(a  ,) + sin((3  a) sin((3  ,) + sin(,  a) sin(,  (3) � O .
Alternative Solution.
171
4.2. SOLUTIONS
TRIG ONOMETRY
Let
172 Mi, i = 1
The points , 2n lie on the upper semicircle centered in origin and with are radius 1. Moreover, the lengths of the chords in a decreasing order, hence are increasing. Thus
M M , M M , ., M M OAl , OA2 , . . . , OAnl 2n 2 2nl .. n n+l 1 Z. �Z2n 1 1 Z2 + ;2n. 1 . . . 1 Zn +2Zn+1 1 and the conclusion follows. Alternative Solution. Consider Zk = r(cos tk +i sin tk), k = 1, 2, . . . , 2n and observe that for any j = 1, 2, . . . , n, we have $
$
$
= r2 [(costj + Cos t2n j+ 2 (sin tj + sin t2n_j+l )2] = = r2 [2 + 2(cos tj cos t2nj+l + sin tj sin t2nj+l )] = = 2r2 [1 + COS(t2nj+l  tj )] = 4r2 cos2 t2n3'+21  tj . t 1  tj and the inequalities Therefore I Zj + Z2 n  j+ ll = 2r cos 2 n j� IZI + z2n l :::; IZ2 + Z2nll :::; :::; IZn + zn+ll are equivalent to t2 n tl � t2 n  l t2 � . . . � tn + l tn · Because :::; tl :::; t2 :::; ... :::; t2n :::; the last inequalities are obviously satisfied. d +
. • .
°
7r,
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1 (1984) , pp. 67, ) 50. Let p = 1, 2, . . . , m and let Z Then zP = 1. Note that n  m + l, n  m + 2, . . . , n are m consecutive integers, and, since p :::; m, there is an integer k {n  m + 1, n  m + 2, . . . , n } such that p divides k. Let k = k p. It follows that z k = (zp) k ' = 1, so Z Ak An  m+l A n  m+2 . . . U An , as claimed. Remark. An alternative solution can be obtained by using the fact that (an k 1)(anl 1 1) . . . (an k+l  1) (a  1) (a  1) ... (a  1) is an integer for all positive integers a > 1 and n > k . (Dorin Andrica, Romanian Mathematical Regional Contest " Grigore Moisil" , 1997) 51. Let Zk = cos t + i sin tk, k {I, 2, 3}. implies The condition 2 (ZI + Z2 + Z3 ) 3ZlZ2 Z3 2(sin tl + sin t2 sin t3 ) = 3 sin ( tl + t2 + t3 ). (1) Problem 5222
E Ap e
i
E
E
k
k
E
+

E
1R
C
U
U
(tl t2 , t3 ) i,[
tl, t2 , t3 < i' Let
Assume by way of contradiction that max < hence 7r 7r . E 0, '6 . The sme functIOn IS concave on 0, '6 , so
t = tl + t32 + t3
. .
( ) 1
From the relations
'
( . tl + sm. t2 + sm. t3 ) sm. tl + t32 + t3 .
'3 sm
(1) and (2) we obtain
then
)
( 2)
:::;
t. It follows that 4 sin3 t  sin t � 0, i.e. sin2 t � �. Hence sin t � �, then t � i, which contradicts that t (0, i) , Therefore max( tl , t2 , t3 ) � i, as desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1 ( 1986) , 91, Problem 5862 ) 52. Let n = 2 ( 28 + 1) and b f:. 0, otherwise the claim is obvious. Consider a complex number a such that a2 = � and the polynomial f = xn  1 = (X  co)(X cl) . . . (X cn l) ' We have f en = G) a (a  iso ) . . . (a isn . ) and f ('n = ( �1 r (a + iso ) . . . (a + ic"n ,), hence f ( ;:a ) f ( a;:) = (a2 + co2 ) . . . (a2 + 2n l ) sin
_
173
4 . 2 . SOLUTIONS
4. TRIGONOMETRY
3t
:::; 2 sin
E
pp.
C

'
Therefore
nII l (a+ bc� ) = bn nIIl � +c�) = bn nII l (a2 +c� ) = ( = bn f en f ( _ IT) = bn [(a2 )2.+. + 1]2 = bn [ m 2.+. + Ij " = b2(2s+1) ( a2s+bl2s++ bl 2S+l ) 2 = (a � + b � )2 . (Dorin Andrica, Romanian Mathematical Olympiad  second round, 2000) 53. If ab = 0, then claim is obvious, so consider the case when f:. and b f:. 0. k=O
k=O
k=O
a
We start with a useful lemma.
°
4.
174
4.2.
TRIGONOMETRY
Lemma. If an odd integer, then
,cn l are the complex roots of unity of order where is nIIl (A + B ) = An + Bn , for all complex numbers A and B. Proof. Using the identity xn  1 = II (x c ) for x = � yields l  ( BAnn + 1) = n!! ( A + ) , and the conclusion follows. Consider the equation bX2 + a = 0 with roots X l and X . Since n,
CO , C 1 , ' "
Ck
n l

k
k=O


13
The desired inequality is equivalent to
n
k=O
Ck
0
2
175
SOLUTIONS
ab1 + be1 + ea1 > R12 abc2 = 48 = a+ b + e > R R 4Rpr
..
ie or
R � 2r, which is Euler's inequality for a triangle.
(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1985), pp. 82, 5720)
Problem
55. Let AI , A2 , A3 and A be the points of complex coordinates Zl , Z2 , Z3 and let a)z3, a Hence point A lies on the line A A and the triangle ZA =A aA3Z2 +has(1its circumcenter in the origin of the complex plane. l Point 2 B is the foot of the altitude from A l in the triangle A l A2 A3 . It follows that Al A � Al B, so E
R
2
3
We have we have
nIIl (a + bc� ) = bn nII l (ck  Xt}(ck X2 ) = bn nIIl  Xl) nIIl (€k X2 ) Using the lemma for A = Xl, B = 1 and then for A = X2 , B = 1 gives nII l (ck  xt} = (Xl) n + 1 = 1 X�, nII l (ck  X2 ) = (X2 )n + 1 = 1  x� . Hence nII l (a + bcT: ) = bn (l  x�)(l  x�) = = bn [l + (XlX2 ) n (x� + x�)] = bn [1 + (�) n] = an + bn , . Xl X2 = a and Xnl + X2n = Xnl + ( Xl )n = 0. SInce (Dorin Andrica, Romanian Mathematical Olympiad  second round, 2000) 54. Consider the triangle with vertices of complex coordinates Zl , Z3 and the circumcenter in the origin of the complex plane. Then the circumradius R equals IZII = IZ2 1 = IZ3 1 = r and the side lengths are a = IZ2  z31 , b = IZI  z31 , e = IZI 1 k=O
k=O
k=O
(c k

k=O

8A1A2A3
therefore as desired.
I ZI  z 11zl  I Z2 2 z3 1 h  IZI  z2 11zl 2 z3 1 sin Al  2 h = IZ I  z22r11 zl  z31 '
_
_
_
 z3
2
1
·
IZ2 2r z31
y
k=O
k=O
k=O

o
X A(z)
b
Z2 ,

z2 ·
(Dorin Andrica, Romanian Mathematical Olympiad  final round, 1984) 56. Denote I z + 1 /z 1 by r. From the hypothesis,
176
4.
Hence r3 ::; 2 + 3r, which by factorization gives (r  2) ( r + 1) 2 ::; O. This implies r ::; 2, desired. (Titu Andreescu, Romanian Mathematical Olympiad  first round, 1987; Revista as
Matematidl Timi§oara (RMT) , No. 1(1987) , pp. 75, Problem 6191)
57. Denote t =
ZIZ , t 2
z?  aZlZ2 + z� = 0 is equivalent 2  4 ::; 0, hence t a± iJ42 a2 and It I to t2  at + 1 = O. We have a a2 4  a2 = 1. f t = cos a + i sin a, then z�n tn cos na + i sin na and we can Jwrite +zfn4a zfzg z . + z� n 0, where an 2 cos na [2, 2] . n Alternative Solution. Because a [2, 2] ' we can write a = 2 cos a. The relation z?  aZlZ2 + z� = 0 is equivalent to E
<C* .
The relation
� =
4
4. 2.
TRIGONOMETRY
1
=
=

=
E
=
E
==
=
177
SOLUTIONS
Lemma. (IMO 1973, Problem 1)
Let be a semicircle of unit radius and C
PI , P2 , . . . , Pn points on where n � 1 is an odd integer. Then I OPI + OP2 + . . . + OPn l � 1, C,
�
�
::::=t
where is the center of Proof. The key idea is to show that the orthogonal projection of the vector sum 0Pi + � + . . . + OP� onto some line has length not less than 1 (see S. Savchev, T. Andreescu, " Mathematical Miniatures" , The Mathematical Association of America, 2003, pp. 75) . Let n = 2k  1. From the considerations of symmetry, the line 0
C.
containing the middle vector 0 the fact that n is odd!) .
I
P� is a natural candidate for such a line (here we use
(1) and, by a simple inductive argument, from (1) it follows that
1...zgzn + zfzn
� =
2 cos na,
n
= 1, 2, . . .
A
(Dorin Andrica, Romanian Mathematical Olympiad  second round, 2001; Gazeta
Q����
B
o
Matematidl (GMB), No. 4(2001), pp. 166)
Al A2
58.
Rotate the polygon . . . An such that the complex coordinates of its vertices are the complex roots of unity of order n, . . . , Cn Let be the complex ' coordinate of point located on the circumcircle of the polygon and note that = 1. The equality
P
yields
Cl, C2 ,
z
Izl
zn  1 = jII=n1 (z  Cj ) Izn  l l = jII=n1 IZ  Cj l = jII=n1 pAj•
. I zn  11 ::; Iz ln + 1 = 2, it follows that the maximal value of IIn PA; is 2 j= l where and is attained for z n =  1, i.e. for middle points of arcs AjAj +l , j = 1, . . . An+ l = AI. (Dorin Andrica, Romanian Mathematical Regional Contest " Grigore MoisH" , , n,
59. We will use the following auxiliary result:
OP� . OPI, OP2 , ... , OP2k 1 OPI, OP2 , ... ,OP2k 1 l AB A B l A l B1 • OPk OP1 + OP2 + . . . + OPk1 � (k  I)OAl , OPk+1 + OPk+2 + . . . + OP2k1 2:: (k  I)OBI. This is because OPj � OAI for j = 1 , . . . , k  1 and OPj � OBI for j k + 1, . . . , 2k  1. Since OAI + OBI = 0, the proof is complete. Consider the complex numbers Zk = cos ak + i sin ak , k 1, 2, . . . , n and the points PI, P2 , . . . ,Pn with complex coordinates ZI, Z2 , . . . , Zn ' 
Since
1992)
It is technically convenient to consider I as an axis with positive direction deter As is well known, the projection of the sum of several vectors is equal mined by to the sum of their projections. Hence it suffices to prove that the sum of the signed ::::=t ::::=t � onto is lengths of the projections of greater than or equal to 1. Denote the diameter of C by and the orthogonal = 1 and also projections of and onto by and We have 

0
=
178
4.
TRIGONOMETRY
Using the above Lemma we have ...+ � 1 , or
zn l
Ik=1t k i tk=l l CO
It follows that as
1 M, + � + . . . + OP� I � 1 , hence IZI + Z2 + sin O:k � 1 .
S O: +
desired.
(Dorin Andrica, Revista Matematidi Timi§oara (RMT)
,
No. 2(1983), pp. 90,
Problem C:58)
60. Using the identities
n cos 2JX· sin nx cos(n + l)X 1 L: sm x j= l nx sin(n + l)x S2 = � sm .:... '. 2JX· = sin sm x j= l ( Sinsinnxx ) 2 S12 + S2
S = and
=
.
L.J
we obtain
2 =
On the other hand,
S� + si = (cos 2x + cos 4x + . . . + cos 2nx) 2 + +(sin 2x + sin 4x + . . . + sin 2nx) 2 = = n+ (cos 2kx cos 2lx + sin 2kx sin 2lx) =
L:
=x+2 hence
L: cos 2(k  l)x, 19<k� n
( sinsinnxx ) 2 = n +
cos 2(k  l)x. l �l<k� n = 2, 1 :::; l < k :::; n, and the problem is solved. 
Set
ao = n and akl
(Dorin Andrica, Romanian Mathematical Regional Contest " Grigore Moisil" ,
199 5 )
Chapter 5 M ATHEMATI CAL ANALYSIS
PROBLEMS
1
Let :::; a < that a < ifiii <
1.
f3.
f3 be real numbers. Prove that there are integers
m,
n 1 such >
2. Let (an ) n�O and (bn ) n �o be the sequences of integers defined by (1 + . 1o3) 2n+l = an + bn 3 , n E N. . 10 V
v
Find a recursive relation for each of the sequences
(an) n�O and (bn ) n�o,
3. Study the convergence of the sequence (xn ) n �O satisfying the following prop
erties:
1) X1n( 1, n 0,1 1,2,) . . .Xn + 1 2) 2 Xn+l  Xn+l Xn  1 , n = 0, 1,2, . . . 4. Study the convergence of the sequence (Xn ) n �l defined by Xl E (0,2) and Xn+ l = 1 + J2xn  x� for n � 1 . 5. Consider the sequence of real numbers (Xn ) n�l such that X2 + X2 + . . . + xn O. lim l n�oo n Prove that . + Xn O . Xl + X + .. 2 lim n�oo n Is the converse true? 6. Let (an ) n �l and (bn) n�l be sequences of positive numbers such that an > nbn for all n 1. Prove that if (a n ) n �l is increasing and (bn ) n �l is unbounded, then the sequence (Cn ) n�l' given by Cn an+l  an , is also unbounded. 7. Let 0 < a < a be real numbers and let (Xn ) n�l be defined by Xl = a and Xn (aXn+ ll)X+ n(a l++1)0'2 , n � 2. >
=
=
2
2
=
=
>
=
=
181
18
2
5.
5.1.
MATHEMATICAL ANALYSIS
� n�n ' q > 1; �
Prove that the sequence is convergent and find its limit.
15. Evaluate
8. Find a sequence (an ) n2:l of positive real numbers such that
(i)
nlim+oo (an+l  an ) = 00 lim (y'a n+ l �) = O. n+oo
and

9. Let (Xn ) n 2:l be an increasing sequence of positive real numbers such that
n+oo xn ( k 2:l X l  Xn k lim nk+ k+oo nk 10. Let (3 be real numbers and let (Xn ) n2:l, (Yn ) n2: l , (Zn ) n2:l be real sequences such that max { x� + O'Yn , y� + (3 xn } :::; Zn for all n ;::: l. 1 a) Prove that Zn ;:::  8 (0'2 + (32 ) for all n ;::: 1. b) If nl�� Zn = � (O'2 + (32 ), prove that the sequences (Xn ) n 2:b (Yn ) n 2:l are 1.
1m 2 = 0. n Prove that there is a sequence n h of positive integers such that
(ii)
_
convergent and find their limits .
11. The sequences (x ) 2:l and (Yn ) n>l are defined by Xl 2, Yl 1 and Xn+l = x� + 1, Yn+ l XnYnn nfor all n ;::: 1 . a) Prove that Xn / Yn < V7 for all n ;::: 1 . b) Prove that the sequence (Zn ) n 2:l, Zn = xn / Yn , is convergent and lim Zn < V7. n+oo 12. Let ;::: 0 and a f= 0 be real numbers and let (Xn ) n2:l be an increasing sequence of real numbers such that nlim+oo nC¥ (xn+ l  xn ) = a. P �ove that the sequence is bounded if and only if > 1. =
a
a
13. Evaluate 14. Evaluate
� k(n  k) ! + (k + 1) nlim+oo f::o (k + l)!(n  k) ! n (k � +l lim L n+oo k= l n2 )
(4n +
�)qn+l ' q > 1.
16. Let (Xn ) n 2:l be an increasing sequence of positive integers such that Xn+2 + Xn > 2Xn+l for all n ;::: 1. Prove that the number 00 1 n=l 10Xn fJ  "  L.J
is irrational.
17. Prove that
= o.
a,
183
PROBLEMS
is irrational for all
An = k=L00l (k!)1 n
n ;::: 1 .
18.
Let k, s be positive integers and let numbers such that
aI, a2 , · . . , ak , bl, b2 , . . . , bs be positive real
y!al + y'a2 + . . . + ifiik = yb; + yb; + . . . + \I'bs
for infinitely many integers n ;::: Prove that k = S; = b2 . . . bs·
2.
1) 2) ala2 . .. ak bl 19. Let (xn ) n 2:l be a sequence with Xl 1 and let X be a real number such that =
Prove that
II00l
n=
( 1 x_n ) = ex • _
Xn+l
_
A f= ±1 be a real number. Find all functions f : R � R and : (0,00) � R f(ln x + A ln y) = (y'X) + (..fij) for all x, Y E (0, 00). 21. Let f be a continuous realvalued function on the interval [a, b] and let ml, m2 be real numbers such that ml m 2 > O. Prove that the equation m f(x) = � a  x + b 2x has at least a solution in the interval (a, b) . 20.
9
Let such that
9
9
18 4
5.
22.
a b
5 . 1.
MATHEMATICAL ANALYSIS
(0, 1/2),
Let and be real numbers in the interval and let 9 be a continuous realvalued function such that = for all real Prove that ex for some constant e.
g(g(x)) ag(x)+bx
x.
g(x)
=
23. Find all continuous functions f : � � [0,00) such that for all real numbers
[0 ,
x, y.
f2 (X + y)  f2 (X  y) 4f(x)f(y) =
24. (i) Prove that if the continuous functions f : � � (00,0] and ) have a fixed point, then
f+
9
has a fixed point. (ii) Prove that if the continuous functions : � � a fixed point, then has a fixed point. 00
rp'ljJ
rp
Prove that the limit
f('\x)  f(f£X) 1mo "� x x exists and is finite if and only if f is differentiable at the origin. 31. The sequence (Xn ) n �l is defined by Xl 0, Xn+l eXn  1, n � 1. Prove that n�oo lim nX n 2. 32. Let Xo (0,1] and Xn+ l Xn  arcsin(sin3 xn ), � 0. Evaluate nl�� Vnxn . 33. Let f : � � � be a twice differentiable function with the second derivative
�
25. Let rp : � � � be a differentiable function at the origin and satisfying rp(O) 0 . Evaluate �� � [rp(x) + rp (�) + . . . + rp (�)] , where n is a positive integer. 26. Let a be a positive real number. Prove that there is a unique positive real number f£ such that f£X XIL  aILx for all x 0. 27. Let f : [a, b) � � be a twice differentiable function on [a, b) such that f(a) f(b) and f'(a) j'(b). Prove that for any real number ,\ the equation f"(x)  ,\(f'(X)) 2 has at least a solution in the interval (a, b). 28. Find all functions f : [0, 2] � (0, 1] that are differentiable at the origin and satisfies f(2x) 2f2 (x)  1, x E [0 , 1] >
=
=
=°
=
o
29.
Let such that
,\
be a positive integer. Prove that there is a unique positive real number
for all real number
=
=
E
nonnegati ve. Prove that
34. Let a
=
n
f(x + f'(x)) � f(x), x
=
>
1.
<
g:� [0,1] and 'ljJ : � � [1,00) have
>
30. Let f : � � � be a function continuous at the origin and let '\, f£ be two
distinct positive real numbers.
E
�.
b be positive real numbers.+ Prove that the equation ( a ; b r Y a" bY has at least a solution in the interval (a, b) . 35. Find with proof if there are differentiable functions rp : � � � such that rp( x) and rp'(x) are integers only if x is integer. 36. Let f : [a, b) � � be a differentiable function. Prove that for any positive integer n there are numbers 01 O2 On in the interval (a, b) such that f(b)  f(a) f'(OI) + f'(02 ) + . . . + f'(On ) n ba 37. Let f, : � � � be differentiable functions with continuous derivatives such that f (x) + (x) f' (x)  g' (x) for all x E �. Prove that if Xl, X are two consecutive real solutions of the equation f(x) g(x) 0, then the equation f2(x) + g(x) has at least a solution in the interval (Xl, X2 ). 38. Let f : [�, �] � (1, 1) be a differentiable function whose derivative f' is continuous and nonnegative. Prove that there exists Xo in [�, �] such that (f(XO))2 + (I' (XO)) 2 � 1. <
=
<
< ... <
_
9
9
=
=
=°
x 0.
18 5
PROBLEMS
186
5.
5. 1 .
MATHEMATICAL ANALYSIS
47. Let n > 1 be an integer and let f : [0, 1] �
39. Prove that there are no positive real numbers x and y such that
x2Y + y2X x + y. ( ) n(n+1) for some integer 40. a) Prove that if x � y �
that
=
yX +
b) Prove that
41. Let
n: 1
n +� � \IY
X l , X 2 , . . . , Xn
be positive real numbers such that X l
Prove that
f
() c
42. Let f :
n � 2, then
+ +...+
= ° for some
�
c
lR
= 1.
lR
lR
�
lR
(x), h(x), .. . , fn (x))dx
45.
46. Let 1 = (0, 00) and let f : I � I be a function with an antiderivative F that
F(x)f G) x, g(x) F(x)F (�) is a constant function and =
x in I. Prove that g : then find f.
1 + 1R,
=
t
b).
[ f(x)dx oj O.
a
<
a
<
f3 < b such that :[
b] [c, dJ be a bijective
Let a, c be nonnegative real numbers and let f a , � increasing function. Prove that there is a unique real number f£ E ( a, such that
50.
51. Let r.p :
Let p be a polynomial of odd degree such that p' has no multiple zero and let f : lR � lR be a function such that f o p is a derivative. Prove that f is a derivative.
R
=
[
44. Evaluate
�
t f(x)dx (b  a.) f ({3) ·
be continuous functions. Prove that
is a derivative and evaluate
satisfies the condition
a,
be a continuous function such that
Prove that there are numbers
E lR.
max( h
for all
has at least a solution in the interval ( a,
be a function with a noninjective antiderivative. Prove that
43. Let h , h, ... , fn :
E
X
=
49. Let f : [a, b] � lR
1
n
=
Xn
be a continuous function such
1
=
9
X2
lR
f(x)dx 1 + 2 + . . . +  . 1 1 Prove that there is a real number o (0, 1), such that f(xo ) 11  xgXo 48. Consider the continuous functions f, : [ b] Prove that the equation f(x) /." g(t)dt g(x) l f ( )dt a
+ n +V'x.
n Vn'n'H + n +� l n + 1 > 2n + 1, n _> 3.
187
PROBLEMS
lR
�
lR
b)
f(t)dt = (IJ
 a)c + (b  1J)d.
be a continuous function such that
{Jxx+Y
x, y
r.p (t)dt =
X J{x y
r.p(t)dt,
for all E lR. Prove that r.p is a constant function.
52. Let f :
lR
�
lR
!:.±lL
be a differentiable function such that
1. f(t)dt !!:.±lLy 2
x y.
5
for all real number < Prove that is a nondecreasing function.
f
53. Let f :
lR
�
lR
f (t)dt
2
be an injective and differentiable function.
188
5,
Prove that the function
MATHEMATICAL ANALYSIS
F : (0, ) � lR, F(x) = xl 1a a: f(t)dt 00

is monotone,
54. Prove that that
55. Prove that there are no Riemann integrable functions f
for all real numbers that
�
1 lim n 2 1{ x a: + 1 dx =  , ntoo 2 0
1(Ya: f(t)dt = f(x) f(y) ,
x =J y.
SOLUTIONS :
Let e =
56. Let f : [0 , 1] � lR be a differentiable function with continuous derivative such Prove that
If(l)  f (O)1 < 1 .
1. We prove that there is an integer n > 1 such that
lR � lR \ {o} such
[[/'(X}]2 dX = 1.
{r
1
1a 1 f(x)(x  f(x))dx = 12
1.

Prove that if there i s an integer m 2:: ° such that
[ fm (t}dt = (m : 1}1 '
fa has a fixed point. 5,9 . Let f : [1, 1] � lR be a differentiable function with nondecreasing derivative. Prove that 21 1(_I1 f(x)dx � f(I) + f'(I). 60. Let f, g : [a, b] � lR be continuous functions. Prove that there is a real number e E (a, b) such that {1 C f (x)dx .+ (e  a)g(e) = !. g(x)dx + (b  e)f (e). a b
n f3n ,
(Dorin Andriea, 4955)
2. Note that
f3n 
1.
f3,
1 1(1982), 90,
1,
. 1n) 2 (n+1)+1 = (l + v3. 1n) 2n+3 = ( l + v3. 1n) 2n 1 ( 1 + V3) 2 = ( l + v3 = (an + bn V3) (4 + 2V3) = 4an + 6bn + (2an + 4bn )V3.
58. Let fa : [0, 1] � lR be a continuous function and let the sequence ( fn ) n>l be
fn(x) = { fn I (t}dt, x E [0, 1].
+ e}n  an = (�) an I e + . . . + en > nanI e > ne,
because a > Take an integer > �e . Then an > n n The interval (a , f3 ) has length greater than hence there is an integer m > such that a < m < or a < rm < as desired. Revista Matematidi Timi§oara (RMT) , No. pp. Problem
 .
then the function
 an = (a
n
57. Find all continuous functions f : [0, 1] � lR such that
defined by
f3  a. Then
+
On the other hand,
(1 + V3) 2 (n+1) +1 = an+1 + bn+1 V3, and since a n , b n are integers we derive that (i) an+1 = 4an + 6 bn i (ii) bn+1 = 2 an + 4bn . an+1  4a
bn = 6 n . Substituting in relation (ii) implies an+2 6 4an+1 = 2an + 4 an+1 64an
From relation (i) we obtain
or
an+2 = 8an+1 4an . b n+ 1 2 4bn ' and the first relatI. On gI.ves O n t he ot her hand, an = bn+2 2 4bn+1 = 4 bn+1 24bn + 6bn · 189
190
5.
5.2 .
MATHEMATICAL ANALYSIS
Thus
i) If
bn+2 8bn+1  4bn· It follows that the sequences (an ) n �l ' (bn ) n �l are given by =
Xl < 1, then Xn
=
{ XlX2
1
if n = if n is even if n is odd and n >
1 X2 X3 . The equation X2 1 + J2X2  x� V2 In all other cases the sequence is divergent. . X2 2 +2' has only the solutIOn ii) Xl 1, then odd Xn { 21 ifif isis even and the sequence is divergent. iii) If Xl > 1, then odd Xn { XlX2 ifif isis even. 2 + V2 . It follows that the sequence is convergent if and only if Xl X2 i . e. Xl (Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 12(1979), pp� 56, Problem 3865) 5. We prove a more general statement: is a positive integer and (Xn ) n>l is a sequence such that . x�P + x�P + . . . + x2np 0, (1) nhm�oo then ... (2) nl'�oo Xl + X2 + + Xn 0 For this, recall the inequality ( Xl + X2 : + Xn rp ::; x�P + x�P : + X� . X3
So the sequence converges if and only if
1. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1981), pp. 71, Problem 4648) 3. Solving the quadratic equation in Xn+1 and taking into account condition 1 ) yields + 1) Xn+l  Xn + 1 +XnJ2(x; (1) l ' 0, 1, 2, . . That is Xn+ l I(xn ), 0, 1,2, . . . , where 1 : (1, 00) � is the function given by I (x) x + 1 +xJ2(xI 2 + 1) . 2 + 0. We It is not difficult to check that 1 is decreasing and 1(2 + O) distinguish three cases: Case 1. If Xo 2 + 0, then Xn 2 + J3 for all Case 2. If Xo E (1, 2 + J3), then from the monotonicity of function 1 it follows that Xo < X2 < X4 < . . , < 2 + V3 < . . . < X5 < X3 < Xl. Case 3. If Xo E (2 + V3, ) then Xl < X3 < X5 < . . . < 2 + V3 < . . . < X4 < X2 < Xo. In all cases the sequence (xn ) n �O is convergent and lim Xn 2 + 0. n�oo Regional Contest (Titu Andreescu and Dorin Andrica, Romanian Mathematical " Grigore Moisil" , 2003) for all
n
2::
_
n =
.
1R
n =
=
=
=
=
n.
00 ,
=
4.
Xn+2 1 + JXn+1 (2  Xn+l) 1 + J( 1 + J2xn  X�) (1  J2xn  X�) 1 + J1  2xn + x� 1 + /xn  1 / , hence X n 2:: 1 for all 2:: 2.
•
=
If
=
n
We study three cases.
=
=
n
n
=
n
=
=
If p
=
n
1m
•
It follows that
p
=
= .
n
. . •
. •
I Xl + X2 : Xn I ::; x�p + x�p : . . . + X� • . .
=
n
=
=
=
=
=
Note that
=
191
SOLUTIONS
+
'p
(1),
•
Using the squeeze theorem and the hypothesis the conclusion we obtain the initial problem. The converse is not true. Take = and observe that
1
Xn (l) n Xl + X2 + . . . + Xn 1 n
=
{o
n
if n is even if n is odd
(2) follows. For
12
5.
9
5 . 2.
MATHEMATICAL ANALYSIS
Hence
... n�oo Xl + X2 +n + Xn = 0 but XI + X� + . . . + X2n = 1 . lim n�oo n (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp . 47, Problem 2570) 6. Assume by way of contradiction that there is M > 0 such that an+1 an M
First,
lim
<
for all n . Summing up these inequalities from 1 to n yields
193
SOLUTIONS
an+ !  an
= (n + 1) In(n + 1)  n ln n = In(n + 1) + In
so
(1 + �r
Second,
v'an+1  va;; J(n + 1) In(n + 1)  v'n In n (n + 1) In(n + 1)  n ln n  (n + 1) In(n + 1) + v'n ln n J ln In ( 1 +  ) In(n + 1) n  (n + 1) In(n + 1) + v'n ln n + (n + 1) In(n + 1) + v'n ln n = J J ln In ( 1 +  ) 1 n In(n + 1) + n+1 / n In n (n + 1) In (n + 1) + n In n J 1+ V (n + 1) In (n + 1) Because n = 1, lim In ( 1 + .!. ) ntoo n we have In ( l + � r .n�oo = O. hm n ln n J(n + l) ln(n + 1) + Jn1Iln In(n + 1) n ln n . Because lim = 1, It follows that lim n�oo n + 1 0 and n�oo (n + 1) n (n + 1) nlim�oo (v'an+ 1  va;;) = 0, as desired. Remark. Another such sequence is given by an = ny'n, n � 1. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 70, Problem 3087) 9. Assume by way of contradiction that there is no sequence (nkh�l with the desired property. Then there is a > 0 such that xn+1  xn  a 0 n hence Xn+ l  xn ;::: an, for all n � 1. It follows that Xn  Xl ;::: a(l + 2 + . . . + (n  1)) = a (n 2 l) n ' =
=
_
_

_
or
an+1 a1 + M. (1) n n From an ;::: nbn it follows that an+ln ;::: n n+ 1 bn+1 Since the sequence (b n ) n>l is not bounded from above we obtain that the sequence ( an n+ 1 ) n>l is not bou:ded from above, which contradicts (1). (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1(1978), pp. 91, <

Problem 3441)
. lty X2 = (aXl++1)XI(a ++1a) 2 a. The eft mequa and b ' the right inequality is equivalent to Xl a. Since 0 a we obtain  likewise X3 a and then Xn E (0, a) by inducting on n. On the other hand, Xn  Xn 1 (aXn+ ll)X+ n(a1++1a) 2  Xn l = Xn1a2 + X�(a _+1 1) 0 therefore the sequence is increasing and bounded. It follows that the sequence is convergent . and let l nlim�oo Xn. Then (a + l)l + a2
7. Note that 0
<
<
<
o
<
1
1"
IS 0
VlOUS
< X2 <
<
=
>
,
=
1 =
so 1 = a.
l + (a + 1)
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1976) , pp. 53, 2567) 8. We prove that an = n In n, n � 1, satisfies the conditions.
Problem
;:= : ====;�==
. �. v
.
=
1
>
>
194
5.
then
5. 2.
MATHEMATICAL ANALYSIS
Xn  n  1 2l a n 2n n 3 for all n � 1. This is in contradiction with l' n�� nn2 = 0, and we are done. (Dorin Andrica, Romanian Mathematical Olympiad  final round, 1984) 10. a) Summing the inequalities x� O'Yn � Zn and y� f3xn � Zn we obtain 0 ::; ( �) 2 (Yn � r ::; 2 (zn + a2 � ,12 ) , for all "= 1 . The conclusion follows immediately. b ) Notice that f3 � and Yn a � J2 ( f32 ) � � J2 Zn  ) 8 I I I f31 ( 8 From the squeeze theorem follows that n+limoo Xn �2 and n+1moo Yn =  a2 . (Dorin Andrica, Romanian Mathematical Regional Contest " Grigore MoisH" , 1997) and 11. By induction on n we obtain Yn = l 2 1 1 1 ... Zn = Xl for all � 1. Since Xn+l = x�  Xn 1 0, the sequence is increasing. From 5
On the other hand,
X 0' 2 >  +  >
Xn+1 Z; n+l ;:  Yn+l Xn  XXn+lYn  Xn+lYn l nYn+ Yn Xn+l x� 2 1  1 �2 > 1 _
+
Xn +
Xn + "2
.
+
+
+
n
0'2 + 2
=
+ "2
Zn +
0'2 +
1.

X
+  +  + Xl Xl X2
X
. . . X n l
+

Xl X2 · . . Xn l
n
X2 =
>
+
Xn
(Xn ) n �l
we obtain
hence It
1  (5) Zn < 2 "21 21. 5 + . . 2. 51n2 = 2 "21 ''": 1.: 1  5 5 [ 1 ] < 2 + 85 = "821 < ..[7, = 2 8 1  ( 5) l l n
+
+
.
+
+
n l
+
as
desired.
_
X2n
.

+ Xn
+
'
Xn
is convergent and (Zn ) n�l is an increasing and bounded sequence. Then 2 1 nlim+oo Zn � 8 < ..[7, claimed. (Dorin Andrica and $erban Buzeteanu, Romanian Mathematical Regional Contest " Grigore MoisH" , 1992) 12. Let 0 such that o . From lim n C¥ (x n+ l  xn ) = n+oo it follows that there is an integer nl such that 1 1 (1) nC¥ < Xn+l  Xn < for all � nl. Summing up inequality (1) from n = nl to n  1, 0 implies (2) L 1 L n 1 < xnl+p  xnl 1 The series f ( l + ) converges if and only if a 1, therefore applying the squeeze theorem to the inequality (2) leads to the conclusion. (Dorin Andrica, Gazeta Matematidi (GMB ) , No. 11(1979), pp. 422, Problem 18011) 13. Note that n en n 1 1) n � l (n � � n =L L ( ki1  (k 1 ) + 1 L (k) = nL + = 1  (n 1 + 2n . Since 1 n+1moo (n = n+1moo n! 0 so
(Zn ) n �l
as
c>
ac >
a
(a + c) ncx
(a  c)
n
=
p l
(a  c)
k=O
k
follows that
_
_
_
Xn X nYn
X
+
195
SOLUTIONS
k
k=O (k
I)!
k
n
k + (k + I) !
=
n! n  k)! ! k ! ( k=O
=
>
( cx n l + k)
+ I) !
=
n
+ I)!
k=O
+ I) !
.
k!(n  k) !
n

1
P
>
_
I
k=O
c¥
k)! + (k + ( k + ) !  k)! +
pl
« a + c)
+ c¥ k=O ( l k) n
nl + P
1.
n!
2
n

=
,
n!
k=O
n
196
5.
it follows that
� n�� � k(n(k + k)!1)!(n+ (k k)!+ 1) r
_
= 1.
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1975), pp. 52, 2281)
Problem
14. Note that Z lim z+O X
z>O
Indeed, Z lim ",+0 X
z >O
=
e z 1n z ",lim +0 ",>0
after applying L'Hospital rule. Then
=
= 1.
!�o ¥ e "'>O z
(z) z>O = e l� = 1,
x +1 1 . lim z_o x z>o Let e > O . There is an integer n ee) > 0 such that for any integer n � nee), we have ( � +1 1  e < :2 )k < 1 + e, k = 1, 2, . . . , n. n2 Summing up from k = 1 to k = n and using algebraic manipulations yields n (�) � +1 L 1  e < k= 1 nn2 k < 1 + e, n � n (e), L n2 k=1 or n 21  21 (e  n1 + ne ) < k=�l ( nk2 ) � +1 < 21 + 12 (e+ n1 + ne ) ' for any integer n � nee) . Therefore n ( k � +1 1 lim n+oo t; n2 ) = .2 (Dorin Andrica) z

=
L....J
15. (i) We have
1 +x+ x + · .. +xn + " ' = I 1 x , I x l < 1. 2
5.2.
MATHEMATICAL ANALYSIS

197
SOLUTIONS
Hence
r1/q (1 + x + x2 + . . . )dx = r1 /q dx ' Jo 1  x Jo so 00 1 q q > 1. In = n I: 1 nq q n=1 (ii ) For any I x l < 1 we have 1 1 + x4 + x 8 + . . . + x4 n + . . . = _ 1 _x4 . Hence 1 /q (1 + x4 + x8 + . . . )dx = 1 /q dx 1 x4 ' 1 1 o so 1 4n+l  q1 + 41 ln (1  12 ) + '21 arctgq1 ' q � (4n + l)q (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 70, Problem 3091; Gazeta Matematica ( GMB ) , No. 1(1981), pp. 40, Problem 18608) _
'
0
00
=
16. The number () has the decimal representation
() = 0 .0 . . . 010'.;' . . . 01 . . . 01 . . . 10'.;' . . . 010'.;'
k1 , k '
where 2 Because
. . . ,
k
n
kn
k2
kl
are the number of zeros between two consecutive ones.
we have
kl < k2 < k3 < . . . < kn < . . . ,
hence () does not have a periodical decimal representation. It follows that () is irrational, as claimed. Revista Matematica Timi§oara (RMT ) , No. Problem
(Dorin Andrica, 4661)
17. Note that
An = 1 + L00 (k!1) > 1 k=2
n
and that
An = 1 + L00 (k!1) n < 1 + L00 k!1 = 1 + e < 2 . Hence 1 < An < 2 for all n � 1. So An is not an integer. k=O
k=O
2(1981), pp. 73,
198
5.
5. 2.
MATHEMATICAL ANALYSIS
n,p, q with q :f 1 A . n q pqn l (q  1) 1 = A + (q +1 l )n + (q + l )n1(q + 2)n + . < 1 1 <A+ (q + l) n + (q + l)n (q + l )n + . . . = 1 ( 1 1 ) 1 =A+ (q + 1 ) n 1 + (q + 1 ) n + (q + 1) 2n + . . . = A + (q + 1) n 1 , for some integer A � O. It follows that 1 1 B= (q + l) n + (q + l )n (q + 2) n + . . · < 1 ' so A + B is not integer, which is false. Therefore An is irrational for all � 1 . (Dorin Andrica) Assume by way of contradiction that there are positive integers such that P. = Then
199
SOLUTIONS
19. Let Pn = IT (1 �) k=1 Xk+ l . Then 
. .

n
18. 1 ) Using the limit
nlim+oo � = 1
and taking the limits in both sides of the equality, we obtain
� = .I + l +...
k
so
k=
.
. . + 1"
8.
2) Using the limit
and the relation n
n+oo n ( � lim
( \Ial  1 ) + ( \Ia2 n


as
9
a
1) + . . . +
n ( ifiik 1) = 
This implies desired .
e
9
1 ) = In a
we obt�in after taking limits:
as
X X  (n + l) Xn+l = xn+l 1 X = n+2  n+ l = n+2 (n + 1) 1 (n + 1) r Pn+l Pn (n + I)! � It follows that 1 1 x2 x3 x n+ l =  +  +  + . . . +  = Pn+ l PI 2! 3! (n + 1)1 n+ 2 x x +...+ X l ' =1+ + 11 2f (n + 1)1 Because ( xn+ l ) X x2 lim 1 +  +  + · · · +  = z , (n + 1) 1 n+oo x I! 2! it follows that lim Pn+ 1 = e , desired. n+oo and Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. (Titu Andreescu 1 ( 1 977), pp. 49, Problem 2843) 20. Interchanging x and y we obtain f(lnx + Alny) = (v'X) + g (.JY) , x,y E (0,00) so f(ln y + Alnx) = (v'X) + (.JY) , X,y > o. Let = ln x + Alny and b = Iny + Alnx. Then Ab a and y = e�, x = e� hence f ( ) = f(b) = (e2(�t_al) ) + ( (�2=b ) , b E It follows that f is a constant and let f (x) = C. Then for x = y we have g(.jX) = � , so is a constant and g(x) = � . (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 12(1979), 51, Problem 3827) 21. Consider the function F I F(x) = (x  a)(x  b)f(x) + ml (x  b) + m2 (x ) 1
9
s tim es
times
hence
(Dorin Andrica, Romanian Mathematical Regional Contest "Grigore Moisil" , 1999)
A a.  b
a
9
9
e2
l)
R
a,
9
pp.
:
t lR,

a .
200
5.
Note that
5.2.
MATHEMATICAL ANALYSIS
F is continuous and
hence
E
=
9
9
l
E
9
9
9
+
9
+
 00 .
E
9
= =
2001, Problem B5)
x y f(O) O. For x = y we obtain f2 (2x) = 4f2 (X) and f( ) 2f(x), f(x) � O. f(nx) = nf(x), n � 1. Assume that f(kx) = kf(x) for all k = 1,2, . . . , n. We have f2 ((n + l)x)  f2 ((n  l)x) = 4f(nx)f(x),
Setting = = 0 yields then 2;C = since We prove that
23.
then
=
as desired. It follows that if p, are positive integers then
q
qf (�) f(P) = pf(l), so f (� ) � f(l) and f(r) r f(l) for any positive rational r. Setting x 0 in the initial condition gives =
=
=
=
00 ,
=
9
f((n + l)x) (n + l)f(x), =
c (a, b) such that F(c) = O. It follows that m f(c) � a  c + b 2c and the solution is complete. (Dorin Andrica) 22. Note that g(x) = g(y) implies that g(g(x)) = g(g(y)) and hence x = y from the given equation. That is, is injective. Since is also continuous, is either strictly increasing or strictly decreasing. Moreover, cannot tend to a finite limit L as x or else we'd have g(g(x))  ag(x) = bx, with the left side bounded and the right side unbounded. Similarly, cannot tend to a finite limit as x Together with monotonicity, this yields that is also surjective. Pick Xo arbitrary, and define Xn for all n Z recursively by Xn+ l = g(xn ) for n > 0, and Xn  l = g  1 (xn) for n < O. Let rl = (1+Ja 2 + 4b)j2 and r2 = (aJa2 + 4b)j2 be the roots of x2  ax  b = 0, so that rl > 0 > r2 and 1 > I r ll > I r2 1 . Then there exist C , C2 lR such that Xn clr� + c2 r� for all n E Z. Suppose is strictly increasing. If C2 f:. 0 for some choice of xo, then Xn is dominated by r� for n sufficiently negative. But taking Xn and X n+ 2 for n sufficiently negative of the right parity, we get 0 < Xn < Xn+ 2 but g(xn) > g(Xn+ 2 ), contradiction. Thus C2 = O. Since Xo = Cl and Xl Clrl, we have g(x) = rl X for all x. Analogously, if is strictly decreasing, then C2 0 or else Xn is dominated by r� for n sufficiently positive. But taking Xn and Xn+ 2 for n sufficiently positive of the right parity, we get o < X n+ 2 < Xn but g(Xn+ 2 ) < g(xn), contradiction. Thus in that case, g(x) = r2 X for all x. (Titu Andreescu, The " William Lowell Putnam" Mathematical Competition, hence there is
201
SOLUTIONS
then for all real
f(y) f ( y), f(r) Ir l f (l), =
y, hence r. f(x) I x l f (l) (rn ) n>l
=
for all rational numbers = We prove that for all real numbers Let be an arbitrary real = Because number and let be a sequence of rational numbers with lim
x. x �oo rn x.
f is a continuous function, it follows that nlim�oo f(rn) nlim�oo I rnl f (l) = f (nlim�oo rn) , hence f(x) = f(l)lxl · Note that a = f(l) � 0, therefore the desired functions are f(x) = al x l for some a � O. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 90,
and
=
Problem 3203)
24. (i) Let a and b be fixed points for f and g, respectively. We have a  f(a) � 0, b = g(b) � 0 so a � b.
<p <p(x) f(x) + g(x)  x. f and are continuous. Moreover, <p(a) f(a) + g(a)  a g(a) � 0
= Consider the function : lR + lR, The function is continuous because functions
<p
=
=
9
202
5. MATHEMATICAL ANALYSIS
and
cp(b) = f(b) + g(b)  b = f(b) ::; O. By the Intermediate Value Theorem, there is an E (a, b) such that cp(xo) = 0, hence (f + g)(xo ) = Xo, desired. (ii) Let and (3 be fixed points of the functions cp and 'l/J, respectively. We have o ::; 0: = cp (0:) ::; 1, (3 = 'l/J ((3) � 1 so ::; (3. Consider the function w : lR lR, w(x) = cp(x)'l/J(x)  x. The function w is continuous because functions cp and 'l/J are continuous. Moreover, w(o:) = cp(o: )'l/J (o:)  0: = O:('l/J(o:)  1) � 0, w ((3) cp((3)'l/J ((3)  (3 = (3(cp((3)  1) ::; O . Likewise, there is an E [0:, (3] such that w(,o) = 0, hence (cp'l/J)(,o) = desired. (Titu Andreescu, "Asupra unor functii cu punct fix" , Revista Matematica Timi§oara (RMT), No. 1(1977), pp. 5 10) Xo
as
0:
0:
+
=
,0 ,
,0
25. We have
�� � ( cp(x) + cp (�) + . . . + cp (�)) =
(<p(X)x  <p0 (O)
)
as
1 <p (�)  <p(O) 1 <p (�)  <P(O) 2 �2 0 + . . . + n �n  0 = = cp'(O) ( 1 + � + . . . + �) ' 2 n since cp(O) = 0 and cp is differentiable at the origin. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 71, Problem 3095; Gazeta Matematica (GMB) , No. 3(1979), pp. 111, Problem 17671) 26. Consider the function f : (0, 00) lR, f(x) = Inxax . We have !' (x) = 1 xIn so !,(x) = 0 if and only if x = �a . e It follows that = is the maximum point of the function I so is the only point such that I ( x) � f (l") for all positive real numbers x. z+o
= lim
+
_
+
ax
2
I"
Hence

a
203
5.2. SOLUTIONS
I"
x > 0, desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 54, Problem 3 54 7) 27. Consider the function F : [a, b] lR, F(x) = f (x)e>'f(z) , E Function F is differentiable, since I and I ' are differentiable, and F(a) = F(b). By Rolle ' s theorem it follows that there is c E ( b) such that F' (c) = O . On the other hand, F' (x) = e >'f(z) (I" (x)  >'(/' ( )) ) hence II/ (c)  >'(/' (c)) 2 = 0, desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1981), pp. 76, Problem 4677) 28. Let g : [0 , 2] [0, �) , g(x) = arccos /(x) . Then f(x) = cosg(x) for all x E [0 , 2] and the condition is equivalent to cos g(2x) = cos 2g(x) . It follows that g(2x) = 2g(x) + k (x) 7I' for all x E [0 , 1], where k [0 , 1] Z . On the other hand, 0 � g(x) < 2' for all x E [0 ,2] , hence k (x) = 0 and g(2x) = 2g(x), x E [0,1] By induction on we obtain g(x) = 2ng ( ; ) . Because f(O) = 1, g(O) = arccos 1 = 0, so g(x) = ( � )x g(O) ;2n for all x E (0,2]. Since I is differentiable at the origin is differentiable at the origin and g(x) = nlim g ( �)  g(O) g' (O) +oo � for all x E (0,2].
for all
as
+
>.
R
a,
X
as
+
+
:
71'
n
9
9
X
=
2
,
204
5. MATHEMATICAL ANALYSIS
It follows that g(x ) = I"x, where I" = g' (O) E [O, �) , because 0 � g(x) < � for all x E [0 ,2]. Therefore the desired functions are fJ1.(x) = cOSI"X, I" E [O , �) . (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1977), pp. 45, Problem 2852) 29. Consider the function f : (0, 00) f(x) = Inx:. Then f' (x) = 1 X>.A+Inl x ' and f' (x) = 0 for x = et.l It follows that () = e l>' is the only maximum point the function, so t R,
f ( ) � feet), x
Hence
x > O.
of
xl" .
_
I"
<5,
a

0:
1(
_
a2
a
_
y f ( �0: )  f (JL) 1 )< y 0:2 < a2 (a + c) , a
'"'co
1 1 (  c) < f ( 0:!l)  f ( 0:Yn ) < (a no: an + c). y Summing up these inequalities yields 1  � (  c) < f (y)  f ( JLo:n ) < 1 � 1  � (a +c). 0:1 � y �  a
1_ a
a l_ �a
a
Because f is continuous at the origin nlim+oo f ( JLo:n ) = f(O) , and so 1 fey)  f(O) 1
0:  1 ( c) � Y � _ 0: _1 (a + c). It follows that )1 ' (0) = limO y y+ fe y f(O) = 0: a 1 = � A  J.£ , _ a
_
_
__
only for () = e t . (Dorin Andrica, Gazeta Matematica (GMB), No. 3(1976), pp. 104, Problem 15768; Revista Matematica Timi§oara (RMT), No. 12(1979), pp. 57, Problem 3870) 30. a) Let A > I" and let y = Then f G Y)  f ly) = A, . . f lAx)  f(p,x) = hm hm y+O I"y z+o x so ( y) y+limO f o: Y fey) = � = a, where 0: = I"A > 1. Let c > O. Then there is a 0 such that for any I y l < we have f (ay)  f (y) < a + c.  c <. (1) y Substituting y with 0:Yk ' k = 1,2, . . . , n, in the relation (1) we obtain 1  c) < fey) f ( �0: ) < 1 (a+c), (a �>
205
5.2. SOLUTIONS
so the function f is differentiable at the origin. Conversely, if f is differentiable at the origin, then f( x) f(O ) f (AX)  f(O) z � AX = 1' (0) and zlim+o l" J.£X = f' (O) . Hence f(AX)  f(l"x) = (A  1") f'(O), lim o + x z as desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 76, Problem 3708) 31. Since eZ > 1 + x for all real x ::j::. 0, we have Xn+l  Xn = eZn  1  Xn > 0 so the sequence (Xn ) n �l is increasing. By induction on n we obtain Xn < 0 for all n � 1, so the sequence (Xn ) n �l is bounded. Therefore the sequence converges and let l be its limit. The equation el = l + 1 has the unique solution l = 0, hence nlim Xn = O. +oo Using zlim +o (�x  � eZ  1 ) = 21 , it follows that 1 1 1 1 X X eZn n+l n nlim+oo (n + 1)  n nlim+oo  11 CesaroStolz's Theorem implies 1 nlim+oo nXn hence nlim +oo nXn = 2, as desired. r
206
5. 2.
5 . MATHEMATICAL ANALYSIS
(Dorin Andrica, Revista Matematidl. Timi§oara (RMT) , No. 2 (1982), pp . 68,
Problem 5004)
32. We first prove by induction that Xn > O. This is true for n = 0 and assuming
Xk > 0 for some positive integer k yields 0 < sin xk < 1, hence sin3 Xk implies Xk+ l > Xk  arcsin(sin xk ) It is not difficult to see that (xn )n> O is convergent and lim X n n �oo
<
sin xk, which
= o.
= O. We have
(vn sin x n ) 2
=+ sin2 X n
and
=
(
=
)
+
(
)
•
+
X�+ l
[
1
_
0,
<
=
e
Problem 6143)
=(xn  arcsi n (s in3 xn ) ) 2
1 ]=
<
+ f'(x)) 0, as desired. Let x be a real number such that f' (x) > O. Likewise, f(x + ! ' (x) )  f(x) ! '(x) !' (c) , for some E (x, x + f' (x)) and f' (e) > f' ex) > O. Hence f(x + f' (x)) � f(x) for all real numbers x, as claimed. (Dorin Andrica, Revista Matematidl Timi§oara (RMT) , No. 12(1989), pp. 67,
sin2 Xn
It is clear that the last limit is obtained from (arcsin t3 ) (2 arcsin t  arcsin t3 ) lim lim t+ o (arcsin t  arcsin t3 ) 2 arcsin2 t t+ O (arcsin t  arcsin t3) 2 arcsin2 t
! , (x)
f(x)  f(x
x n ) 2 1 · . n hm . ( =2 = nhm+oo . 1 n+oo sm Xn
x�
<
and
= J22 . Alt�rnative Solution. We have seen above that (x n ) n >O is convergent, lim Xn = 0 n �oo and X n > 0 for all n = 0, 1,2, . . . We will calculate 1 . n + 1  n1 = nhm+. oo hm (1) =1""""1:: · +oo 1 
=
o.
+ f' ex)) =  !, (X)f' (c) ,
! ,(e)
Thus nlim foxn +oo
n
+
+
. n1 =
hm (vnX n ) 2 n +oo
(Titu Andreescu,
for some c E (x f' ex) , x). Because the second derivative is nonnegative, f' is non decreasing, hence
+
+
hence
(1)
f(x)  f(x
•
=
t3
is � and the conclusion is obtained via CesaroStoltz It follows that the limit 2 Theorem. Gazeta Matematidl. (GMB) , No. 10(2002) , pp. 409, Problem C:2557)
+
•
•

t
33.
n+1n 1 = 
(2 arcsin t  t2 arcsin t3 ) t = tlim+O ( arcsin t arcsint3 t3 ) 2 = 2.   t2
207
Let x be a zero of 1 '. Then f(x f' (x) ) f(x) and the conclusion follows. Let x be a real number such that f' ex) < Applying the Mean Value Theorem on the interval [x f' (x) , x) we obtain
. 2 X + sm2 X sm 1. n n l lim :1 :: :;1m . . 2X + oo sm2 Xn  SIn + nn �oo n l sin2 X n+ l sin 2 Xn . 2 Xn sm2 X n+l sm2 X n sm2 Xn+ l sm lim lim n + oo sin(xn  xn+ d sin(xn Xn+ l ) n + oo x� sin(xn xn+d . Xn . sm + l 2 lim sm Xn 2 lim X n Xn+ l lim n+ oo Xn Xn+ l n +oo sin(x n + Xn + l ) n +oo Xn 1. 1 1 . 1.1m 1m Xn+ l  2 · n + oo Xn + l n�oo 1 Xn From CesaroStolz Theorem we obtain 1 2 ' nhm +oo •
=
t(2 arcsin t  arcsin t3 ) lim t+ O (arcsin t  arcsin t3 ) 2
SOLUTIONS
34. Applying the Mean Value Theorem a [a, � b] yields a + b  In a 1 In 2b  a x ' 2 
for the function f(t)
= In t on the interval
( a + b)
x E a '  . 2
Hence
=
(1)
208
5. 2.
5. MATHEMATICAL ANALYSIS
Using the same argument for the interval a+b ln b  In 2 ba 2 hence In From the equalities
1
y'
[ a ; b , b] gives yE
Summing up these equalities implies
( a + b , b) , 
or
2
Y = b  a. (� a + b) 2
(2)
(1) and (2) we obtain ( a + b ) :I: = ( � ) Y 2a a+b
35. We prove that no such function exists. Assume the contrary and let k be an
integer. From the Mean Value Theorem we obtain
cp(k + 1)  cp(k) = (e) , e E (k, k + 1). Since cp(k) and cp(k + 1) are integers, cp(k + 1)  cp(k) is also an integer and so is cp' (e) · On the other hand, e is not an integer, hence cp' (e) is not integer, a contradiction. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 67, Problem 3618) 36. Define Xk = a + k (b  a) , k = 0, 1, and note that ba , k = 0, 1, Xk+ l  Xk = ;;:'
. . . ,n
. . . , n.
The Mean Value Theorem yields
desired.
(Dorin Andrica, Revista Matematica Timi§oara (RMT), No. 12(1979), pp. 58, 3878) 37. Let F lR lR, F(x) f(x)  g(x) and note that F is differentiable. Since F(x ) = F (X2 ) = 0 from Rolle's Theorem there is E (X l , X2 ) such that F'(c) = O. On the other hand, F'(x) = f '(x)  g' ( X ) = f (x) + g(x), and therefore f(c) + g(c) = 0, desired. (Dorin Andrica, Revista Matematid1 Timi§oara (RMT) , No. 2(1977), pp. 74, Problem 3113) 38. Assume that ( f (X)) + ( f' (X)) 2 1 for all X in [i, i]' Then f'(x) Jl  (f(X))2 1 for all X E [�, �] . Integrating from  i to i yields arcsin (i )  arcsin f ( i ) 2 2 7r. On the other hand, it is clear that arcsin f ( 2"7r )  arcsin f (  2"7r ) � 2"7r + 2"7r = 7r. contradicting the previous inequality. ( Titu Andreescu, Mathematical Horizons, 2000) 39. If X = y, then x = y = 0, which is impossible, because x and y are positive. Assume that there are x f:. y 0 such that x . 2Y + y 2:1: = X + y, and let y = Xl  X2 , X = X2  X3 for some Xl X2 X3 O. Then 2:1:1 :1:2  1 1  2:1:3 :1:2 or 2:1:1 2:1:2 2:1:2 2:1:3 Xl  X2 X2  X3 By the Mean Value Theorem there are (} l E (X2 ' xI ) and (}2 E (X3,X 2 ) such that 2:1:1 2:1:2  = 2()1 n 2, (} l E (X 2 ,X I) and Xl  X2 I
(Dorin Andrica, Revista Matematica Timi§oara (RMT), No. 2(1978), pp. 54, Problem 3548)
n
n
:
desired.
cp
as
f (b)  f (a) = � t!'((}i ), ba i=l
Problem
Then as
209
SOLUTIONS
�
=
c
as
2
>
>
f
>
.
>
_
_
1
>
_
>
>
5.
21 0
5.2.
MATHEMATICAL ANALYSIS
(}2 ,
Hence 291 In 2 = 29 2 In 2, that implies (h = a contradiction. Revista Matematidi Timi§oara (RMT), No. 12(1980), pp. 70, Problem 4152)
(Dorin Andrica,
40. a) Consider the differentiable function f
:
Consider f(x) = ln x, have
211
SOLUTIONS
Yi = �Xi , YlY2 . · ' Yn and Ai = !!..Yi for i = 1 , 2, . . . , n. We P=
(0, 00) � lR,
f(t) = ifi n+0. We have !,(t) = .!.n t n�l  1 (t ntn\l) n n+_1 ) ' n(n+ l) then f'(t) 2:: O. Hence f is increasing if t 2:: ( on ) n(n+l ) n :) 1and for x � y � r (n+1) , we have /(x) � /(y). There (n: 1 [(n: 1) 
_
'
SO,
, 00
fore as
Hence
yr yr . . . y:£; :s In
desired. b) We prove that
n 2:: 3.
for all The inequality is equivalent to
which is clearly true, because
and since
(n
\Iii
Yi = �Xi , i = 1 ,
(1 + � r $ n,
(1 + �r < e < 3 for all n 2:: 1. Setting x = n n+ l and Y = + l) n , x 2:: Y in the inequality from a) yields + 1 2:: 2n + 1, n 2:: 3 n + n+mn
as
so
_
n
desired.
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1981), pp. 74,
Problem 4668)
41. Recall Jensen's inequality for a concave function f:
,
2 . . . n,
(t ) i= l �Yl
i�l Ji
it follows that
1 X�l X�2 x�n
, :s • • .
thus as desired.
( ) nnXi L i= l
:E Zi
0=1
=
n,
(Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1977) , pp. 74,
Problem
3111)
F Xl < X2 F(xI) F(X2 ) ' [X l,X2 ] f(c) F'(c) (Titu Andreescu)
42.
Let be an antiderivative of the function f. Since F is noninjective, there = are real numbers such that Applying Rolle's Theorem on the interval we obtain = = 0 for some c E as desired.
43. Because max( h (x) , h (x))
=
h (x)
(X l ,X2 ),
+ h (x) + I h (x) 2

h (x) 1
212
5.
5 .2.
MATHEMATICAL ANALYSIS
and iI , h are continuous, ma:x.(fI , h) is also continuous. Assume that if iI , are continuous, then ma:x.(iI , is continuous. It follows that ma:x.( ma:x.(iI , . ma:x.( iI , I),
12, ·· ., Ikl
12, ... ,Ik  I) . . , Ik Ik)
12, . . . Ik) =
. , Ik)
and according to the 'first step, the function ma:x.(iI , . . is continuous. Hence ma:x.(iI , is continuous and furthermore a derivative function. Note that if n is even, then
. . . , In)
n (oo,I) x, .. . ,xn ) = { x1,n , XX EE [I, x , x E (1,00),I] and xn++l 1 + C, X E (oo, I) J max(l, x, ... ,xn)dx xxn++lC,+ X E [I, I] + C, x E (1,00). n + 1 On the other hand, if n is odd, then x l, n { x E (00, 1) n ma:x.(I, x, ... ,x ) = X [I, I] xn , x EE (1,00), and xn + 1 + C, x E (00,1) [ I,I] J max(l, x, . ,xn)dx xxn++lC,+ n + C, XX EE (I,oo). + 1 (RMT), (Dorin Andrica, Revista Matematidin Timi§oara ma:x.(I,
=
{
n
n
n
1,
..
Problem
5185)
=
1
n n
No .
2(1983), pp. 62,
We have
earctg:c dx VI x2earctg:c dx = II J VI + x2 J 1 x2 J V I + x2d(earctg:c ) = .)1 + x2earctg:c J VIxearctg:c + x2 dx = =
=
hence
On the other hand,
I
12 = J VIxearctg:cx2 dx J xl +1 x+2 exarctg:c 2 dx = J xV 1 x2 d(earctg:c ) = dx' = x yii+x2earctg:c  I  2 J xVI2 earctg:c 2 + x so :c dx = xvf1+X2earctg:c  (I 12 ) = J XVI2 earctg 2 2 +x arctg:c 2 e = (x  I) VI 2+ x + C. (Dorin Andrica, Romanian Mathematical Olympiad  final round, 1975; Revista Matematica Timi§oara (RMT), No. 2(1978), pp. 35, Problem 2125)
�
=
+
+
_
+
=
�
+
1
I
Let g, : be functions such that: is a derivative; is differentiable with a continuous derivative. . Then h is a derivative function. Proof. Let H be an antiderivative of h and define u : Then u' (x) = g(x)h(x) + g' (x)H(x), 45.
+
We start with the following lemma. h lR t lR
Lemma.
1) h 2) 9 9
lR
t
R,
u(x) g(x)H(x). =
g(x) h(x) = u'(x) g'(x)H(x). The function u' is a derivative and g' . H is continuous, hence the function g . h is a derivative, as claimed. Applying the lemma for h l o p and g(x) = xk for a nonnegative in k teger k, it follows that x (f p)(x) is a derivative, hence p'( 1 p) is a deriva tive. Since the degree of p is odd, p( ) = llt Assuming that lim p x = 00, :c+oo there are real numbers Xl, x� , x�, ... , xm , x� such that p is increasing on each of the interval ( 00 Xl , [X , X2 ] , ... , [X�_ 1 ,X ],[x �,00) and p(xI) = p (x = Mil P(X2 ) = p(x�) = M2 , ... , p(xm ) = p(x�) m= Mm . Let Fl ,Hl , ... ,Hm be an anti derivative of p( 1 p) on the interval (00, X l ], [x�, X2 ] ' " . , [x�, 00), respectively. It follows that Fl p  l, HI p  l, .. . , Hm p l are antiderivative of I on the intervals or
0
=
0
,
44. Denote
21 3
SOLUTIONS
0
0
J
0
lR
D
�
0
( )
0

(00, Ml], [Ml , M2], . . . , [Mm , 00), respectively, hence I is a derivative function on all llt (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1985), pp. 76, Problem 2) 46. Substituting x x1 in the given condition yields F ( ; ) f(x) = � , (1) t
21 4
5. MATHEMATICAL ANALYSIS
for all x in I. We have g' (x) = F'(x)F
G)
+ F(x)F'
(D (  :2 )
= f (x)F
1 1
G) : f G) F(x) � 
0
for all x in I, so 9 is constant. Then there is a constant c > such that
(1) we obtain
1
f (x) F(x)
ex '
48. Consider the function F : [a, b) F(x) =
2
=   2 · x = o X x
for all x in I, and from
F(x)F (�) = c
x E I.
F
�e ln x + ln d, where d > 0. It follows that F(x) = dx � , for all x in I. The relation F( x)F ( � ) = c becomes tP = c, so d = ..;c. Finally, 1 = 1 x c , x E I, f (x) = 1 ...;c xF (  ) x where e is any positive real constant . (Titu Andreeseu, Romanian Mathematical Olympiad  final round, 1987; Revista Matematica Timi§oara (RMT) , No. 2(1987), pp. 86, Problem 6307) 47. Consider the function [0 , 1) lR, g(x) = f (x)  (1 + x + . . . + xn  ) 1
and note that
9
11
is continuous. We have
1
1
g(x)dx = =
[
g(x)dx
f (x)dx 
 11 (1 + . . . + n (1 + � + . . . + � ) = O. x+
x
1
) dx =
(0, 1) such that g(xo ) = [ g(x)dx = 0, 1  x� f( Xo ) = l + xo + · · · + x o =  , 1  Xo n l
as desired.
[ t dt l f(t)dt. g( )
[
0
g(t)dt = g(c)
(Titu Andreeseu, Revista Matematica Timi§oara (RMT) , No. 12(1979), pp. 33, Problem 3444)
O. Applying Rolle's Theorem,
t f(t)dt,
(Dorin Andriea)
49. Consider the function F : [a, b) F(t) =
1.'
+
1R,
f(x)dx
[
and observe that F is continuous and F(a)F(b) that F(O') = so
0
['
f (x)dx =
l
f (x)dx
<
O. Then there is
a
E (a,
b) such
f(x)dx
b) such that l f(x)dx = (b  )f ({3) , therefore there are numbers f3 E (a, b), f3, with { f(x)dx = (b  a)f({3) , From the Mean Value Theorem there is f3 E ( a , a
a,
From the Mean Value Theorem there is Xo E
hence
b)
f t c)
1.  1
+
lR,
+
Note that is differentiable and F(a) = F(b) = we obtain e E (a, such that F' (e) = hence
as desired.
Integrating gives In F(x) =
9 :
215
5.2. SOLUTIONS
a <
as claimed.
(Dorin Andrica, Revista Matematica Timi§oara (RMT), No. 12(1979), pp. 61, 3897)
Problem
50.
Since f is an increasing bijective function, f is continuous. Denote 81 =
l
f (x)dx,
and note from the diagram below that
82 = ld r 1 (y)dy,
(1)
216
y . . . . . . . . . . . . . . .(
x, y f (I). (Titu Andreescu, 2349;
c a
=
'"
+
=
lit
=
=
=
:
R
+
R,
=
=
x, f 1(1976), 56, 2(1980), 68, 18154)
+
�,
h(x) xf(x)  10" f(t)dt. Because f is differentiable, h is differentiable and (1) h(x) = xf'(x), x � O. Since f is injective and continuous, f is either increasing or decreasing, so f' (x) � o for all x or f'(x) � 0 for all real numbers x. Case 1. If f'(x) � 0 for all x, then from (1) we deduce that h'(x) � 0, x � 0, hence h is nondecreasing. It follows that h(x) � h(O) = 0, x � O. Case 2. If f'(x) � 0 for all x, then h'(x) � 0, x � 0, and h is nonincreasing. It follows that h(x) � h(O) 0, x � O. Since F is differentiable and xf(x)  1(0X f(t)dt h(x) F'(x) = x """'X2 ' we derive that F'(x) � 0 for all x 0 or F'(x) 2:: 0 for all x 0, hence F is a monotonic function, as desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 76, Problem 3709; Gazeta Matematica (GMB), No. 1(1980), pp. 38, Problem 18115) 54. Recall from Problem 14 that =
(c, d) s,uch that { f l (y)dy (d  c) rl (e). Observe that e is unique and let fJ; = f  1 (e) . The relation (1) gives t f(t)dt (  a)e + (b  ",)d, E (a, b) , as desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 56, Problem 3556) 51. Consider the function F � �, F(s) f <p(t)dt, and note that F is differentiable. From the hypothesis we obtain F(x + y)  F(x) = F(x)  F(x  y), so F(x + y) + F(x  y) = 2F(x), x,y E �. Differentiating with respect to y it follows that F' (x + y) = F' (x  y), x, y E Sett �ng x y �, z E �, we obtain F'(z) F'(O), so <p(z) F'(O) for all z E Hence cp is a constant function, as desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1977), pp. 47, Problem 2865) . 52. Consider the function <p <p(s) fo8 f(t)dt. The condition is equivalent to 10" f (t)dt + 10" f(t)dt ?: 2 fo"¥ f(t)dt, From the Mean Value Theorem there is e E
:
0 f'(x) 0
53. Consider the function h : [0,00)
x
b
= '"
(1)
>
2
for all E �. Since is differentiable, cp is twice differentiable and moreover is concave up, from relation Hence cp" (X) 2:: or 2:: for all so is a nondecreasing function. Revista Matematica Timi§oara (RMT) , No. pp. Problem Gazeta Matematica (GMB), No. pp. Problem
c
o
cp(x) + cp(y)  cp ( x + y ) 2
hence
d
217
5.2. SOLUTIONS
5 . MATHEMATICAL ANALYSIS
R
=
=
2
>
>
O. 1
0. x I xx  11 fo� (x·+1  X)dxl :0; fo ;; Ix'+!  xldx = 10 ;' xlxX  ll dx 10 ;' xdx = 2 .
and let c > There is 0 > such that for all Then for n > "8 we obtam
I
' n
=n
n
2
< 0,
< c.
'
< cn
2
c
218
5. It follows that
5. 2 .
MATHEMATICAL ANALYSIS
57. The relation is equivalent to
* (XX+l  x)dx = 0, r 2 lim n n+oo io * nlim+oo n2 iro xx+l dx 21 ,
hence
ri l (xf(x)  f2 (X))dx = irl 4x2 dx. o o Hence [ (f2 (X)  xf(x) + x: ) = 0, or [ (f(x)  � r = O. Because f is continuous, f(x)  i = 0, for all x E [0, 1], so f (x) = �, x E [0, 1]. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1978), pp. 72, Problem 3319)
=
as
desired .
dx
Alternative Solution. Consider the function F(t) = f.' x·+1 dx. Then F(O) = 0 and we can write 1 * xx+l dx nlim n2 F ( 1 ) limO F(u)2 limO F'(u) = lim n 2 +oo n u+ u u+ 2u n+oo 1 uu+ l 1 u+limo 2u = 2 u+limo UU = 2 . (Dorin Andrica, Gazeta Matematidi (GMB) , No. 11(1979), pp. 424, Problem 18025; Revista Matematica Timi§oara (RMT) , No. 12(1980), pp. 71, Problem 4160) 55. Assume the contrary and let x =f. y be real numbers. Then rY f(t)dt = f(x)(y) and 1x f(t)dt = f(fexy)) , f Jx hence f (x) f(y) It follows that f 2 (X) + f 2 (y) 0, so f(x) f(y) = 0, which is absurd since f(x) =f. 0 for all x. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 35, Problem 3126) =
0
=

=

=
dx

58. The relation

is equivalent to Since
y
=
and
as
desired.
[ (fm (t)  : ) dt = O.
fm is continuous by the Mean Value Theorem there is Xo E (0,1) such that
xo m l iro (fm l(t)  (mt I)! ) dt O. Using the same argument, we obtain Xl E (O,xo) such that fm l (xt } = (mX�I)!l ' Continuing this procedure, we obtain Xm E (0, x m  d such that fo (xm ) = x m, desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1978), pp. 72, Problem 3320) 59. From the Mean Value Theorem we deduce that for any x E [1, 1] there is Cx E (1, x) such that f (x) f ( 1) (x + 1) f' (cx ) . Since f'( cx ) � f'(I), f(x)  f (  1) � ( x + 1) 1'(1), hence L f(x)dx  2f( 1) � (x � 1) 2 1�J' (1) =
_
=
_
[(I!' (X)I  l)2dx ;::: O. f. (f'(x)) 2 dx  2 irol I f'(x)ldx + irol dx � 0,
as
I
1 ;::: [ 1 !,(x)l dx ;::: I [ !'(X)dx l ,
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 77, 3130)
Problem
[ fm (t)dt = (m : 1)1 '
or
56. It is clear that Hence
219
SOLUTIONS

j
J
=
5 . MATHEMATICAL ANALYSIS
220
and the conclusion follows. Revista Matematica Timi§oara (RMT) , No. 2(1981) , Problem 4686)
(Titu Andreescu,
=
�

and note that h is differentiable and h' lt)
=
(a (
77,
(a, b] �, (b t) [ f (x)dx + (t  a) [ g(x)dx
60. Consider the function h : h( t)
pp.
[ f(x)dx
+ (b 
t)f(t) + [ g(x)dx  (t  )g(t) . a
Since h ) = h b) = 0, from Rolle's Theorem it follows that there is a real number b) such that h' (c) = Then cE
(a,
 /.' f(x)dx
O.

a g( ) + (b  e)f(e) + t g(x)dx = 0,
(e  )
(Titu Andreescu,
C O M P REHEN SIVE P RO B LEMS
e
and the conclusion follows. Revista Matematica Timi§oara (RMT) , No. 2(1985) , Problem 5628)
Chapter 6
pp.
56,
PROBLEMS
1. Let A be a set with
A
B
elements and let be a subset of with m � 1 elements. � B. Find the number of functions f : t such that n
A
A A
f(B) elements and let X, Y be subsets of A with p � 1
2. Let be a set with n and q 1 elements respectively. Find the number of functions f : YC
�
f(X).
3.
A
X
Let be a set with n elements and let be a subset of Find the number of functions f : t such that =
A A
4. Consider the sets A = {I,
2,
...,n
A A such that +
A with k � 1 elements.
f(X) X. } B = {I, 2, . , m} and let k ,
. .
B having exactly k fixed points. 5. Let aI, a2 , ' . . , an be positive real numbers and let m � 1. Prove that (I al )m (1 a2 )m . . + (I + an ) � . 2 .
Find the number of functions f :
A
t
+ .
m
}
:::; mini n, m .
m
a2 a3 al 6. Let aI, a2 , . . . ,a n be positive real numbers and let k � O. Prove that +
+
+
n
7. Let a, b, c be positive real numbers such that abc = 1. Prove that 8.
Let numbers x,
(3" be positive real numbers and let [a, b] be an interval. Find the y, z E [a, b) such that E(x , y, z) = o (x  y) 2 + {3(y  Z) 2 + ,(z  X) 2 0,
is maximum.
223
224
6.
6.1 .
CO MPREHENSIVE PROBLEMS
9. Find the maximum number of nonzero terms of the sum
n L If (i)  f (j) 1 i,j= l where f : {I, 2, . .. , n } {a, b, c} is one of the 3n possible functions. 10. Let aI, a2 , . . . , an , bl :::; b2 :::; . . . :::; bn be positive numbers such that al :::; bl , al + a2 :::; b1 + b2 , . . . , al + a2 + . . . + an :::; b1 + b2 + . . . + bn · +
Prove that
+ ..fo2 + . . . + � ::; .jb; + .jb; + . . . + A· n 11. Define S (n, p) = L (n + 1 2i ) 2p for all positive integers n and p. Prove that i= l for all positive real numbers ai , i r,n the following inequality holds 4P � p �l� (ai  aj) 2p :::; S (n,p) L....t a2i . 1 ��<J�n i= l 0il
=
•
12. Prove that for all integers n 2::
k 2::
2.
13. a) Consider the real numbers aij , i = 1,2, . . . , n  2, j = 1,2, . . . , n, n 2:: 3,
and the determinants
Prove that
Ak , k = 1, 2, 1 1 au
. . . ,n
1
Al + Ag + As + . . . = A2 + A4 + A6 + . . . ' b) Let X l , X2 , . . . ,Xn be distinct real numbers and let n( k l ) kl Pk = iII= O+ (Xni  Xk ), qk = i=lII (Xk  Xi ), for k = 1,2, . . . , n. Prove that
1
PROBLEMS
225
c) Prove that
n (l) k k2 + = 0, k=l for all integers n 2:: 3. 14. Let P(x) xn + alxn l + . . . + an be a polynomial with all zeros positive real numbers. Prove that if there are m =I{I, 2, . . . , n} such that am = (W (:) and ap = (1JP (;) , then P(x) (x  l) n . 15. Define the polynomials PO,P1 , ,Pn by PO(x) = 1 and Pk+1 (x) = (n  k + x)Pk (x) + xPk(x) for all k = 0, 1, . . . , n  1. Prove that deg Pk = k for all k and find the polynomial Pn · 16. Let (Pn ) n �O be a sequence of polynomials defined by Pn+l (x) = 2xP�(x) + Pn (x) and Po(x) = 1 . . Find Pn (O ) . L (n  k)!(n k) !
=
pE
=
• • •
17. Consider the polynomial
n P(x) k=LO n + 1 + 1 x k . Prove that the equation P(x 2 ) = p2 (x) has no real roots. 18. Let P be a polynomial with real coefficients. Find all functions f : 1R such that there is a real number t such that f (x + t)  f ( ) = P(x) =
k
x
for all x E
lit
19. Find the real numbers a, b, c, d, e
[2, 2] such that a + b + c+d+ e = O
and
E
+
1R
226
al
6.
COMPREHENSIVE PROBLEMS
20. Let p be an odd prime. The sequence (an ) n?O is defined as follows: ao
a 2
an
==
0,
anl
= 1, . . . , p = p 2 and, for all n � p  1, is the least integer greater than that does not form an arithmetic progression of length with any of the preceding terms. Prove that, for all n, is the number obtained by writing n in base  1 and reading it in base
p
an
p.
a, C
21.
i) Let be nonnegative real numbers and let 1 : increasing function. Prove that
[! (k)] +
L
k
L
p
[a, b) [c, d] be a bijective +
[I l (k)]  n(GJ ) = [b][d]  a(a)a(c)
a ii) Evaluate
(x)
=
{
[x] xI 0
if if if
E
is a perfect square for all
E
[a, b] [c, d] be a bijective +
==
Ak(k,
26.
< n l < n2 < . . . < nk <
24.
l
(Xn ) n?l be a sequence defined by Xl = 3 and Xn+l = x�  2 for all
Let positive integers n.
=
{
n � 1.
if n = O (mod if n = 2 (mod if n = l (mod
3) 3) 3)
t (�) ak �:�, k=O cn�' and find similar relations for (bn)n2:1 and (Cn ) n ?l ' 28. Consider the sequences (an)n?l, (bn ) n? l (Cn ) n?l, (dn ) n2: 1 defined by al = 0, bl 1, Cl 1, dl = 0 and an+l 2bn + 3cn , 'bn+ l an + 3dn , Cn+l = an + 2dn , dn+l bn + Cn , n � 1. =
=
=
=
==
N* N* N* be function such that 1(1, 1) = 2, I(m + 1, n) I ( , n) + and I( m , n + 1) = I(m, n)  n for all n N*. Find all pairs (p, q) such that I(p, q) = 2001. 30. Determine all functions f Z Z satisfying 29. Let 1
are consecutive. Prove that for all positive integers m between n + n 2 + . . . + nm and n + n2 + . . . + nm there is a perfect square.
+l
k
Find a closed formula for the general term of these sequences.
. . . be a sequence of integers such that no two
l
� + .ij4) n an + bn� + cn �'
TK
=
for all i!ltegers n � 1.
23. Let 1
.
27. Let (an ) n?l, (bn)n2:1' (Cn)n2: 1 be sequences of positive integers defined by
[J (k)]  L [J  l (k)] [b]a(c)  [dJ a(a), where k is integer and a is the function defined in previous problem. ii) Prove that n [ 2 ] n2 [ ] L � VIi k= l k2 = k=Ll � L
2n  1
In a coordinate system xOy consider the points n  1) , = 0, 1, . . . , n  1 for a given positive integer n. Find the number of open segments that do not contain a point with integral coordinates.
Prove that decreasing function. Prove that
� <p(n)
n= l
(1 +
:
n � 1.
OAk
1R
=
22. i) Let a, c be nonnegative real numbers and let 1
=
�
x \Z x 0 x Z \ {O}
227 1 for all n � 1, there are sequences (u n ) n? 1 ,
PROBLEMS
Prove that
25. Find the sum of the series
where is integer, n (G J ) is the number of points with nonnegative integer coordinates on the graph of 1 and a : 1R + is defined by
Z
6.1. Taking into account that (x n , Xn+ 1 ) (Vn ) n?l of positive integers such that
:
x
+
=
m,
a
m
m
E
:
for all integers
x,
y,
z.
+
228
6.
31. 1981
CO MPREHENSIVE PROBLEMS
points lie inside a cube of side a distance less than
1.
9. Prove that there are two points within
32. The squares of a chessboard are randomly labeled from 1 to 64. On the first 63 there is a knight. After some moves, the 64's square, initially unoccupied, is also unoccupied. Let n k be the square number of the knight who was initially on the k 's square. Prove that 63 L k=l i nk  kl � 1984. 33. The Fibonacci sequence (Fn ) n2:: l is given by Fl F2 1, Fn+2 = Fn+l +Fn ' n � 1. Prove that L'2n  F?n+2 +9 F?n2 2 23n for all n � 2. 34. Let n be a positive integer and let Nk be the number of increasing arithmetic progressions with k terms from the set {I, 2, . . , n}. =
=
D
_
F.
_
.
Prove that
q = [ nk  11 ] . 35. Let Fn be the nth Fibonacci number (that is, FI = F2 = 1, Fn+l = Fn +Fn  l for n � 2), and let P(x) be the polynomial of degree 990 such that P(k) = Fk for k 992,993, . . . ,1982. Prove that P(1983) = Fl 9 83 1. 36. Let Xl, X2 , f3 be real numbers and let the sequence (xn ) n2::1 be given by where
_
=

0.,
x� Xm+1 Xm l f:. 0 for all
If such that

and for all
m
>
1, prove that there are real numbers AI , A2
'1".·,' ,
6. 1 .
229 37. Consider b E [0 ,1) and the sequence (an ) n2:: 1 defined by al = a2 = . . . = ak l = 0, k � 3, and an+l � (b + an + a�_l + . + a�=�+2 ) for all k. =
. .
Prove that the sequence is convergent.
38. Let (an ) n2:: l and (bn ) n2:: 1 be sequences such that
(bn ) n> l an ; n �oo bn ... ) aan+l + bn+l 2 nthat limbn an , o.n  1 . Prove n�oo bn 39. Let (un ) n 2:: l , be a sequence defined by UI E 1R \ {O, I} and Un U n Un+l Ul U2 U3 0 i) is strictly monotonic and unbounded; ii) there exists lim 
11l
>
=
=
n
If the sequence converges, evaluate
1 lim II (1 + Ul . . . U k ). n+oo n k=2 40. Let (an ) n2:: 1 and (bn ) n2:: 1 be sequences such that i) 0 b l b2 bn b I + ii) � � k 1, n � 1; n an . iii) there exists n� lim oo . and IS. equa to n�oo . oo abn+n+I1bn abnn eXIsts Prove that n� hm 1m abnn 
<
< .. . <
<
< ... ;
>
1
1·
•
41. Let k be a positive integer and let
n�O Prove that
42. Let f
n > 2.
PROBLEMS
:
1 8k2 1R
�
1R
n f(x) k=L1 sinak x, =
230
6.
43. Let f : lR
f(xo ) I O.
t
1
CO MPREHENSIVE P ROBLEMS
where a k are real numbers. Prove that if number Xo such that that
.,
lai l I lajl for i I j, then there is a real
lim t(x) = 00. f(x) = x*oo x*oo Prove that the function : lR lR, g(x) = sinf(x), lim
for any � y > O. ii) Prove that
x
is not periodical.
an.
t
[0, 1), f(n) = {an } i.e. the fractional
part of the number i) Prove that is injective if and only if a is irrational. ii) If is rational, find the number of elements of the set
f
a
=
f: f f(x) M f(x) = M
Let lR t lR be a function such that i) has a period T > ii) � for all iii) if and only if kT, for some integer k. Prove that, for any irrational () the function 9 lR t 1R,
x;
0;
x= : g(x) = f(x) + f(()x),
t
lR be a continuous function with a period T > O.
[ f(x), max f(X)] there is a
a) Prove that if T is irrational, then for any A E min sequence
(Xn) n� 1 of integers such that lim
n* oo
xER
f(xn ) = A.
[
xER
]
b) "Prove that if T is rational, then for any A E min f(X) , max f(X) and for any irrational number (), there is a sequence
47.
x,
xER
1
xER
(Xn) n� 1 of integers such that
i) Let y, z , v be distinct positive integers such that that there is no A > such that
d1' . p
(n +enl) n n! (n +en1)n+ , 
1. 49. Let a, c be nonnegative real numbers and let f : [a, b) [c, d), be a bijective function. i) If f is increasing, prove that /. 1 f(t)dt + ld r' (t)dt = bd  ac. ii) If f is decreasing, prove that /.1 f(t)dt  ld r' (t)dt = bc  ad. 50. i) Let J.L : (0,00) lR be a continuous function such that x*"oo lim J.L(x) = Prove that lim e  x ifX e t J.L(t)dt = 0 x*oo o ii) Let f : [0, 00) lR be an ntime differentiable function with the nderivative continuous such that there exists n k lim L x*oo k=O C� f ( )(x) = A. Prove that lim f (x) exists and x*"oo lim f(x) = A. x *oo 51. Let f : [a, b) [c, d] be continuous function such that a � b /.1 J' (x)dx = cd. Prove that if c + d I 0, then 1 I.b b  a ( c  d) 2 �  ° � f(x)dx c+d 4 c+d 1
<
<
n�
o.
t
is not periodical.
46. Let f : lR
=
t
M { f (n)1 n E N} .
45.
b, c, d be distinct positive integers such that
+ bP cP + la  c l + I b  d l �
Prove that
23 1
PROBLEMS
ii) Let p be a prime number and let a,
48. i) Prove that
t
44. Let a be a real number and let f : N
6. 1 .
aP
lR be a differentiable function with continuous derivative such
9
1
x+
y
= z
t
t
+
v.
Prove
a
a
232
6. 1 .
6. COM PREHENSIVE PROBLEMS
52. Let I : [a, b]
lR be a continuous monotonic function and let F : [a, b] lR, (x ) = (x  a) I.' f (t)dt + (x  b) [ f (t)dt.
t
F
t
Prove that all values of F have the same sign.
lR and [1, 00) lR such that 1) is differentiable with continuous derivative; 2) I is continuous and the function h : [1,00) lR, h(x) = gl(X)  I(g(x)) is 53. a) Consider the functions I : (0, 00) 9
t
t
9:
t
nonincreasing. Denote
qn  1 n (X)  { 'acos(q)�X ,cos X . . cos X , XX = 00 n E W , n is a derivative function for any integer O. 58. 1) Let I : lR lR be a continuous function such that 1 1 I(x)dx = M(/) . lim 
57. Let q be a positive integer. Find the number an (q) such that In : lR f
�
an = kL h(x) . an+1  h(l) � 1.9(n+1) I (x)dx ::; an , n 1 1 lim L cot = 00.
b) Prove that
2) I : lR lR
n roo
k=l
M(/)
k
k2
54. Let a be a positive real number and let I : [0 ,1]
function. Evaluate
t
lR+ be an integrable
() (fa�f k �  n) . 55. Consider the functions In : lR lR, (s�n r) X 0, X k17r ' k E Z* sm X X = k1 ' E Z * lim n
roo
t
2
IT
7r x=o
� O.
=f. k
for all integers n Find the numbers Find the number
1)
2)
56. Let
p,
a ,k such that In is continuous on lR*. that In is a derivative function. annsuch q � 0 be integers. Find the numbers such that I : lR lR, 1 1 f (x) = { sm X cos x ' ifif xx #= 0O •.•
is a deri vati ve function.
.
p
_
.
q
Cp,q
{
0
G) , g(X) = fM(/), x =O
p,q
t
= Itllimroo lt iot I (x)dx = l iTo I(x)dx
O. Prove that
T
I : lR lR : lR lR, g(x) = I /(x) 1 also a 60. If /l , /2 : lR lR are derivative functions then I = max{/l , /2 } is also a
59.
If t is a derivative function, then is derivative function? derivative function?
=l
,
Y
I Y I r oo Y
is a derivative function. Let t be a continuous function with a period T >
g(l)
lR,
=f.
.
t
=l
Prove that
t
n>
Prove that the function
n
233
PROBLEMS
t
9
t
SOLUTIONS
m functions h B B. Each of them can be extended in nnm ways to a function f : A A which satisfies f(B) � B. nn m . Hence the required number is (Dorin Andrica, Revista Matematidl Timi§oara (RMT) , No. 12(1981), 81, Problem Cl: l ) 2. Let Xq be a subset with q elements of X. Since Y has elements, it follows that there are q! bijective functions Xq Y. Each of them can be extended in nn q ways to a function f A A which satisfies Y � f(X). The number of subsets Xq of X is (:) , hence the requested number is (:) nn  p . Remark. If q consider (:) 1. There are
:
m
t
t
mm .
pp.
:
9
t
> p,
:
q
t
= O.
(Dorin Andrica, Romanian Mathematical Regional Contest " Grigore MoisH" , 2000)
f(X) X, k!nn  k .
3.
X.
k! such bijections : A A. Hence the
Since = it follows that f is bijective on There are ways to a function f and each of them can be extended in desired number is
nnk
t
(Dorin Andrica)
4.
We consider two cases. i) � m. Let There is only one be a subset with elements of the set that has the property h (i) = i for all i E This function can t function h : be extended in (m ways to a function t B such that f= i for all iE
n
A \ Ak•
k
A Ak Akk n k  l)
A.
f:A
Ak .
f(i)
Ak of A is (�), hence the desired number is (�) l) nk . ii) n. Let Bk be a subset with k elements of the set There is only one function h : Bk Bk such that h(i) i for all i Bk . This function can be extended in  l) m k ways to a function B B such that g(i) f= i for all i \ (m
The number of the subsets m �
(m
t
=
9
:
t
235
E
B.
E
B
Bk •
23 6
mnm ways to a function f : A
Moreover, each function 9 can be extended in that clearly has exactly fixed points.
k
The number of the subsets
(7) mn m (m _ l)mk .
Bk
of
Therefore the number of functions f
B
is
:A
4
B with
k fixed points is (�) (m
l)nk if n ::; m and (7) mnm (m _ l) mk if m ::; n. (Dorin Andrica, Romanian Mathematical Regional Contest " Marian 'farina,
2001) 5. Using the inequality xr for
+ x� + . . .
+
x�� �
From the previous relation we deduce
B,
ll (k + ail
(7) hence the desired number is
t
n� l (Xl + X2 + . . . + Xn ) m
kn + (�) (ll air kn 1 + (;) (ll a;f kn2 + n .+ . . + (�) k+
�
Thus
(llf ( � k + � II a; ::; II (k + a; ). n
so
1 Ln ai (k + ai ) ::; k + II ni l i n
(2) (Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 2(1977), pp. 63, Problem 3045)
we obtain
First Solution.
On the other hand,
al a a a al a a2 + a23 + . . . + aln > n a2 . a23 . . . anl = n from the AMGM inequality. Therefore (l + aa2l ) m + (1 + aa23 ) m + " ' + (l + aaln ) m � � nm (2n)m = n ' 2m, as desired. (Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1(1974), pp. 7, n
I
1564; Gazeta Matematidi (GMB), No. 2(1976), pp. 65)
6. We have
n + ai) = kn + a kn l + L a a kn2 + . . . + II(k l2 Ll i= l
Using the AMGM inequality gives
(1)
Using again the AMGM inequality, we obtain
n
Problem
237
6.2. SOLUTIONS
6. COMPREHENSIVE PROBLEMS
a l a2
· · . an ·
=
=l
'
abc = 1,
7. Since this nonhomogeneous inequality can be trans formed into a homogeneous one by a suitable change of variables. In fact, there exist positive real numbers such that
p, q, r
a = pq , b = ,qr c = .pr Rewriting the inequality in terms of p, q, r, we obtain (1) (p  q + r)(q  r + p)(r p + q) ::; pqr, where p, q, r > o. At most one of the numbers = pq+r , = qr+p, = rp+ q is negative, because any two of them have a positive sum. If exactly one of the numbers is negative, then ::; 0 pqr. If they are all nonnegative, then by the AMGM inequality, 1 ..;uv ::; "2 ( + ) = p. Likewise, ..;vw ::; q and ..jWu ::; r. Hence ::; pqr, as desired. Second Solution. Expanding out the lefthand side of (1) gives (pq +r)(q r +p)(r p+q) = [p(p r) + (r q) (P q) + q(r  q) +pq][r + (q p)] = = pr(p  r) + r(r  q)(p  r) + rq(r  q) + pqr + p(p  r)(q p)+ +(r  q)(p  r)(q p) + q(r  q)(q p) + pq(q p) . u
uvw
v
<
u
v
uvw
W
u, v , W
6.
238
,
Note that pr (p
 r) + rq(r  q) + pq(q  p) + (r  q) (P  r) (q  p)
=
O.
=
+
<
=
=
+
ca .
=
we
=
239
=
(1) is equivalent to o � p (p  q)(p  r) + q (q  r)(q  p) + r(r  p) (r  q), which is a special case of Schur's inequality. Third Solution. Denoting the lefthand side of the desired inequality by L, we have L = abeL = b (a  1 + D c ( b  l + D a (c 1 + D = (ab  b + 1)(be  e + 1)(ea  a 1) = L1 • Also, since lIb = ae, lIe = ab, 11a = be, L = (a  1 + D (b  1 + D (C 1 + D = = (a  1 + ae) (b  1 + ab) (e  1 + be) = L2 . If U = a  1 + lIb � 0, then a 1 and b > 1, implying that v = b  1 + lIe > 0 and W = e  1 + 1/a > o . Then L uvw � 0, as desired. Similarly, either u � 0 or v � 0 yields the same result. If u, v, W > 0, then all factors of Ll and L2 are positive. The AMGM inequality gives J(ab  b + l)(b  1 + ab) � 21 [(ab  b + 1) + (b  1 + ab)] abo Likewise, J(be  e + 1) (e  1 be) � be, J(ea  a + 1) (a  1 + ae) � Hence L = � � (ab)(be)(ea) = (abe) 2 = 1. Fourth Solution. Using the notations established in the third solution, it is easy to verify the equalities beu + ve = 2, eav + aw = 2, abw + bu 2. As in the third solution, only_ need to consider the case when u, v, W > O . The AMGM inequality gives 2 2:: 2 e�, 2 2:: 2aJcvw, and 2 2:: 2Jawu, from which uvw � 1 . Fifth Solution. Let Ul = abb+ 1, VI = bee+ 1, WI 1; U2 1be+c, = 1 ea + a, and W = 1ab + b. As in the third solution, we only need to consider V2 2 the case in which U i , Vi, Wi > 0 for i = 1 , 2 . Again, we have Thus
6.2. Let X = a + b + e and Y = ab + be + ca. Then Ul + VI + WI = Y  X + 3 and U2 + V2 + W2 X  Y + 3. Hence either U l + VI + WI � 3 or U2 + V2 + W2 � 3. either case L � 1 follows from the AMGM inequality. (Titu Andreescu, IMO 2000, Problem 2) 8. Without loss of generality can assume that a � (3 � 'Y. Let x, y , be three arbitrary numbers from the interval [a, b] such that � y � Then E( x , y, z )  E(x ,z, y ) = ('Y  a)(( z  X) 2  (y  )) 2:: 0, and E (a, y, b) = a (y  a) 2 + (3 (y  b) 2 + 'Y (b  ) E ( b, y, a) = a (y  b) 2 + f3(y  a) 2 + 'Y (b  a) . We need to find the maximal values of the functions It ( y ) a(y  a) 2 + f3 (y  b) 2 and h (y) = a(y  b) 2 + (3(y  a)2 on the interval [a, b] . Since II (a) = f3 (b  a) 2 2:: a(b  a) II (b ) and the coefficient of y 2 in II is a + f3 2:: 0, it follows that the maximum value of II is obtained for y = a. Likewise, h(b) 2:: h(a) and the maximum value of h is obtained for y b. Therefore max E(a, y, b) = E ( a , a, b) = (f3 + 'Y)(b  a) , and max E( b , y , a) = E (b , b, a) ({3 + 'Y ) (b  a) . It follows that the maximum value of E is (f3 + 'Y)(b  a) 2 and is obtained for x a, y = a, z = b or x = b, y = b, z = a. (Dorin Andrica and loan Ra§a, Revista Matematica Timi§oara (RMT) , No. 1(1983), pp. 66, Problem C5:2) 1 1 ({b} ) , 9. For a function I : { 1, 2, . . . , n} + {a,b, e} let = 1 1 ({a}) , 1 respec = 1 ({e}) and let p, q, r be the number of elements of sets tively. Obviously p + q + r = n and without loss of generality we may assume that p 2:: q 2:: r . A term I /(i)  1(j ) 1 is different from 0 if the pair (i, j ) is in one of the sets , , , or Hence the number of nonzero terms in the sum L I / ( i )  l(j) 1 is 2 (pq + qr + rp). SO LUTIONS
CO MPREHENSIVE PROBLEMS
ca  a +
=
In
we
z
x
Z.
x
a
2
2
2
=
2
=
=
2
yE[a , b]
2
=
yE[ a , b]
=
Me
Ma
Mb
Ma
x
Mb Mb
X
Ma Ma
x
Me, Me n
i ,j=1
X
Ma Mb
X
Me
=
Ma, Mb , Me,
Me
X
Mb .
240
6. COMPREHENSIVE PROBLEMS
2 (pq+qT+TP) (Po , qo , TO) 2 (pq+qT+Tp) 1. Po  TO 2 = Po  1, ql qo , = TO + 1. Then + ql + = n and
p+q+T
The problem reduces to finding the maximal value of when == n and 2:: 0 are integers. Note that if then the absolute is a triplet that maximizes value of any difference of two numbers from this triplet is at most Indeed, assume that 2:: and define
p, q, T
=
PI
PI
,j
241
6.2. SOLUTIONS
{
� ... ( + . . . + A)] � �2 [( � A + A + + yo.;; ) + .Jb;+ 0. an
We have
Tl
Tl
= Poqo + POTo + qoro + Po  ro  1 > Poqo + POTO + qoTo , which contradicts the maximality of 2(Poqo + qoro + ropo) · We have the following cases. 1) n = 3k . Then Po + qo + TO 3k, Po 2:: qo 2:: ro 2:: Po  1, hence Po = k and then qo = ro k. In this case the maximal value is 2(k2 + k2 + k2 ) = 6k2 2n3 2) n = 3k + 1. Then Po + qo + ro = 3k + 1, Po 2:: qo 2:: TO 2:: Po 1, so 3po 2:: 3k + 1 2:: 3po  2. Hence Po k + 1 and then qo = TO = k. In this case the maximal value is 2((k + l)k + (k + l)k + k2 ) = 2(3k2 + 2k) = � (n  1). 3) n = 3k + 2. Then Po + qo + ro 3k + 2, 2:: qo 2:: TO � Po 1, so 3Po 2:: 3k + 1 2:: 3po  2. It follows that Po = k + 1 and qo + ro = 2k + 1. Because k + 1 2:: qo 2:: TO � k, qo = k + 1 and ro = k. The maximal value is in this case n2 1 2[(k + l)(k 1) + (k + l)k + (k + l)k] 2(k + 1)(3k + 1) = 2 · 3 2 2 n 1 n Therefore the requested number is 2 "3 if 3 divides n and 2  otherwise. Remark. The problem can be reformulated as follows: Suppose3 that n points in =
=
2
=
.

=
2
Po
=
+

(Dorin Andrica Pal Dalyai,
10. From the inequality � ( .fiil %I _ 4 fb1 ) + ( f02 2
Y Ul
+ . . . + ( A1 yO;1 ) (b2 + bJ) + yO;1 b .Jb; + 0. + . . . + A, + + . .. + va;; � .Jb; + 0. + . . . + A, __ __ _
hence
_
83,
4917)
'b) ) + . . . + (va; �
4 fb U22 Y
2
_
4 Un Y
2
>  0,
va;
as claimed .
_
1
=
.Ja2
(Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1977), pp. 63,
Problem 3046 )
1 1 . By multiplying the numbers aI , a2 , . . . , an with a suitable factor
M = '"'1 2p
n
� ai
we may reduce the problem to the case when Assume without loss of generality that
=
space are colored by three different colors. Find the maximum number of segments AB such that A and B are different colors. and Romanian IMO Selection Test, 1982; Revista Matematica Timi§oara ( RMT ) , No. 1 ( 1982), pp. Problem
it follows that
1
a
Let a
n
L a;P = 1. i=1
= � � . .. � a2
al
an ·
=� S (n, p) and suppose by way of contradiction that 2
Then hence
L a;P > L[a + a(i  1)] 2P. n
n
i= 1
i= 1
Consider the function <p : lR + ( 0, 00 ) , <p ( x ) =
L[x + a(i n
i=1

1)] 2p .
i= l
·" .. .... . .... .
Then and
<p' ( x) = 2p
n
= 2p (2p
that
+ a (i
Dorin Andrica, Romanian Winter Camp, 1984; Revista Matematidi Timi§oara 1(1985), pp. 72, Problem 1)
( ( RMT ) , No.
1 )] 2p 1
I)i=l x  1) I)i=nl x _ 1) 2p2 > <p" (X) <p' <p" (x) > Xo (1  n ) Xo <p , n n [1 n 2P 2p ]  1) :, S (n, 8 a P > <p(Xo) 8 ; + a(i
13. a) Clearly,
O.
]
Because ( x ) is a polynomial of odd degree and <p' has a unique real zero:
1 1 au a2
0 for all real x, it follows
;
2 is also a minimum point of the function
=
ex
+
ex (i
=
a contradiction. Hence
=
1,
al,n a2,n
=0
an  2,n1 an  2,n
Expanding the determinant after the first row yields
hence
as desired.
Remark. For p = 1 we obtain the MitrinoviC's inequality Ipl. J? n (ai  aj )2 � (n212 1 � a2 . l :Sl<J:S ( Dorin Andrica)
n  ) i=l i (1 x) n > 1 (�) xk 1> (vn)n > 1 (�) (vn _1)k nl > (�) (vnl)k . n � 1 > (vn _ 1) k , () WJ > vnl,
as desired. b) Consider the determinants
L.t
1 2. Note that
1 1
a l,n  l a2,n 1
al2 a22
an 2, 1 an 2,2
so
p)
1 1
1 1
1
=  a
The number
243
6.2. SOLUTIONS
6. COMPREHENSIVE PROBLEMS
2 42
+
Ak
=
1
Xkl1 Xk1+l Xnkl2 Xnk 2l
1
Xl X nl 2
Xn X �2
+
+
for all positive real numbers x. Setting x = yfii 0, implies
+
then It follows that
and then
as desired.
From equality a) we obtain
n
lPkq)kk ( k, n=, l
L  = O.
k
k2 , k 1, 2, . .
c ) Set X = = manipulations we obtain
in the previous equality. After some algebraic
.
n
(  1) k k2
� (n k)!(n k)! = 0, _
as desired.
+
n
� 3,
( Dorin Andrica, Romanian Mathematical Regional Contest " Grigore Moisil" , 1995)
244
6.2. SOLUTIONS
6. CO MPREHENSIVE PROBLEMS
14. Let X l , X2, . . . , X n > zeros and coefficients yield
0 be the zeros of polynomial P. The relations between
L>X2 . . . Xm = (�)
and
L:x. X2 . . ' Xp =
(;)
(
Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 1(1978), pp.
Problem 3293)
.
16. We prove that
P� (x) = 2nPn  1 (x), n � O.
Then we have the equality case in the Generalized Mac Laurin 's Inequality: Note that
PI (x) 2 = 2 . 1 . Po (x) =
X, = X2 = . . . = Xn . From L: X1X2 . . ' Xm = i = 1, 2, . . . , n, hence P(x) = (x _ l ) n ,
hence
(�). it follows that Xi = 1,
as claimed.
(Titu Andreescu, Revista Matematidi Timi§oara (RMT) , No. 1(1977), pp. 24,
and assume that Then
Pn (x) = 2xPn  dx) + P�  l (x) = 2xPn  1 (x)  2(n  1)Pn 2 (x).
Differentiating we obtain
P� (x) =  2Pn  l (x)  2XP�_1 (x)  2(n  1)P�_2 (x) = = 2Pnl (x) + 4(n  1)xPn2 (x)  2(n  1)P�_2 (x) = = 2Pn l (x)  2(n  1)[2xPn  2(x) + P�_ 2 (X)] = =  2Pn  dx)  2(n  1)Pn  1 (x) = 2nPn 1 (x),
2300)
Problem
1 5 . It is obvious that degpo = 0 and deg pi = 1. Assuming that deg pk = k, from the given relation we obtain deg pk + l = k 1, hence by induction degpi = i for all
+
i = 0, 1, . . . , n .
Consider the function
f : 1R + lR, f(x) = xn ex . It is easy to prove that f Ck) (x) = Xnk Q k (x) eX ,
where Q k (X) is a polynomial with real coefficients of degree k. We prove that Q k (X) = Pk (X) for all k. Note that Qo(x) = 1  Po (x) and
xn  Ck +I ) Q k+ l (x) eX = f Ck+ 1 ) (x) = (f C k) (x)) ' = (Xnk Q k (x) eX )' = = Xn  Ck+1 ) [(n  k + X)Q k (x) + xQ � (x)],
hence
Q k+l (x) = (n  k + X)Q k (X) + xQ � (x). Since (Pkh= o ,n and (Q kh=o ,n satisfy the same recursive relation and Po = Qo it follows that Pk = Q k for all k . So
.
as needed. The initial relation becomes so Hence
Pn (O) =
d)!
if n is odd
if n is even
(e  X2 ) Cn) = Q n (x) e X2
for a polynomial
we obtain
k =O
It follows that
{�
1) "
Alternative solution. Note that
and, on the other hand,
n j Cn) (x) = (xn eX ) Cn) = L C� (xn ) Ck) e x .
P� l (x) = 2(n  1)Pn 2(x), n � 2.
Since
Q n . From ( e X2 ) Cn+ l ) = [(e  X2 ) Cn) y = (Q n (x) eX2 ), = = [Q � (x)  2xQ n (x)] eX2 = Q n+l (x) eX2 ,
Q n+ l (x) = 2xQ n (x)  Q�(x), n � O . Qo (x) = 1 = Po(x), we note that Q n (X) = Pn ( X) for all n � 0,
245 67,
6.
246
6.2.
CO MPREHENSIVE PROBLEMS
hence
with
2
2n + . . . x x2 + X4  x6 + . . . + n 3!
(1) (1) 1nl 11 21 If n is odd, then by differentiating the series ( 1) for an odd number of times, we deduce that P ( O ) = O. If n is even, set n = 2m. Differentiating (1) n times yields =
n
such that
Therefore
n
P (O )
=
{ � l) � (%)!
if n is odd if n is even
Ln ak xk be a polynomial with nonnegative real coefficients. By
=
=
It suffices to prove that P(I) < 1. Indeed
n 1 <1 L k=O n + l + l as needed. Therefore p2 (x) < P(x2 ) for all x, so the equation has no real roots. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 12(1989) , pp. 107, Problem C9:8) 18. If P
=
0, then for
t
=
0 and for any function
n kxk P(x) La k=O =
E
(n ; l) tbn+. ( +2 1) t2 bn+1 + ( ) tbn an l (� : �) tn+' bn+1 + . . . + G) tb. ao· = a,.
n
n
=
1

J:
JR
+
Note that the system has unique solution, hence there is a unique polynomial Qt such that Qt( x  Qt( x) = P ( ) , E JR Qt ( O ) = 0, and deg Qt = 1 deg P.
+ t)
k=O the CauchySchwarz Inequality we derive that 2 P (x) (� Vak. Vakx k ) � (�ak ) (�ak x 2 k ) P(l)P(x2 ).
Let
=
=
(Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1978), pp. 76,
Problem 3706)
=
+l x nk=l L bk X k Qt ( x + t)  Qt ( x ) P( x ) , x JR.
Identifying the coefficients from both sides yields
=
17. Let P (x)
We search for a polynomial
=
(e X' ) (2m) (_ l) m (:;) + x [( _ l)m+. (�:+;) ! + . . . ] ,
hence
E JR* .
Qt ( )
On the other hand,
e X
an f:. 0, and let t
247
SOLUTIONS
JR
the claim holds.
x x
+
g(x) x,
J
Set = J( x)  Qt (x) . Then satisfies the claim if and only if for all real Le. 9 has period Therefore
t.
J(x)
where
9
is a function of period
t.
=
Qt ( x )
g(x+t)g(x)
=
0
+ g(x),
(Dorin Andrica, Revista Matematidl Timi§oara (RMT) , No. 2(1984), pp. 103, Problem C6:10) 1 1 1 1 1 19. Because 2 a, 2 b, 2 c, 2 d, 2 e [ 1, 1] ' there are real numbers x, t, such that a 2 cos x, b 2 cos , 2 cos d 2 cos t, e 2 co U sing the identity 2 cos 5ex (2 cos ex) 5  5 (2 cos ex) 3 + 5 (2 cos ex) , we obtain 2 cos 5 x a5  5a3 + 5a and the analogous relations. Summing them up yields L 2 cos 5x L a5  5 L a3 + 5 L a 10 E
=
=
y
y,
C =
Z,
=
=
=
=
=
=
Z,
s u.
u
so
L cos 5x = 5. Hence
}
{
cos 5x = cos 5y = cos 5z = cos 5t = cos 5u = 1, V5  1 V5 + 1 a, b, c, d, e E 2,  '  2 . 2 From the relation a + b + c + d + e = 0, it follows that one of the numbers is 2,
and therefore
V5 + 1 V5  1 . easy to check that for these two of them  and the other two   . It 1S 2 2 numbers L a3 = O. Romanian Mathematical Olympiad  final round, 2002)
(Titu Andreescu,
N
2 0 . We will say that a subset of is pprogressionfree if it does not con tain an arithmetic progression of length p. Denote by b n the number obtained by writing n in base p  1 and reading it in base p. One can easily prove that an = b n for all n = 0, 1, 2, . . . by induction, using the following properties of the set B = {bo, b1 , . . . , b n , . . . } (whose proofs we postpone): 1 ° B is pprogressionfree; 2° If b n 1 < a < b n for some n � 1, then the set {bo, b 1 , . . . , b n 1 , a} is not pprogressionfree. Indeed, assume 1° and 2° hold. By the definition of ak and b k , we have a k = bk for k = 0, 1, . . . , p  2. Let al = b k for all k :::; n  1, where n � p  1. By 1°, the set is pprogressionfree, so an :::; b n  Also, the inequality an < b n is impossible, in view of 2° . Hence an = b n and we are done. So it suffices to prove 1° and 2°. Let us note first that B consists of all numbers whose base p representation does not contain the digit p  1 . Hence 1° follows from the fact that if a, a + d, . . . , a + (p  l)d is any arithmetic progression of length p, then all base p digits occur in the base p representation of its terms. To see this, represent d in the form d = pm k, where gcd ( k,p) = 1 . Then d ends in m zeros, and the digit o preceding them is nonzero. It is easy to see that if a is the ( m + l)st digit of a (from right to left) , then the corresponding digits of a, a + d, . . . , a + (p  l)d are the remainders of a, a + 0, . . . , a + (p  1)0 modulo p, respectively. It remains to note that a, a + 0, . . . , a + (p  1 ) 0 is a complete set of residues modulo p, because 0 is relatively prime to p. This finishes the proof of 1 ° . We start proving 2° by the remark that b n 1 < a < b n implies a (j. B. Since B consists precisely of the numbers whose base p representations do not contain the digit p  1, this very digit must occur in the base p representation of a. Let d be the
249
6.2. SO LUTIONS
6. CO MPREHENSIVE PROBLEMS
248
number obtained from a be replacing each of its digits by 0 if the digit is not p  1, and by 1 if it is p  1. Consider the progression
a  (p  l)d, a  (p  2)d, . . . , a  d, a. As the definition of d implies, the first p  1 terms do not contain p  1 in their base p representation. Hence, being less than a, they must belong to {bo, b 1 , . . . , b nI} . Therefore the set {bo, b 1 , . . . , bn1 , a} is not pprogressionfree, and the proof is fin ished.
0
(Titu Andreescu, USA Mathematical Olympiad, 1995)
2 1 . i) For a bounded region M of the plane we denote by n (M ) the number of points with nonnegative integral coordinates in M. Function f is increasing and bijective, hence continuous . Consider the sets
M1 = { (x, y) E ]�? I a :::; x :::; b, 0 :::; y :::; f (x)}, M2 = ((x, y) E ]�? I c :::; y :::; d, 0 :::; x :::; f  1 (y)}, M3 = { (x, y) E ]�? I 0 :::; x :::; b, 0 :::; y :::; d}, M4 = { (x, y) E ]I�? I 0 :::; x :::; a, 0 :::; y :::; c}. d
c a
o
Then
We have
b n (M2 ) =
L
c � k� d
(f  l (k)) ,
n (M3 ) = [b][dJ , n (M4 ) = a(a)a(c).
hence and the conclusion follows.
6. CO MPREHENSIVE PROBLEMS
250
ii) Consider the function
f : [1, n] + [1, n(\+ 1) ] , x (x + 1) I (x) = 2 .
Function I is increasing and bijective. Note that v'f+8X · £ormu Ia · ) we 0btam · . A pp Iymg
1 +
1
2
n and 1  1 (x)
n(n +l)
t, [ k(k; l) ] + t, [ 1 + �] _ n = n2 (n2+ 1) ,
hence
n(n+l)
t, [ 1 + �] = n2 (n2+ 1) + n  � t, k(k + 1) =
+ 1) = n(n2 + 2) = n2 (n2+ 1) + n  n(n 4+ 1)  n(n + 1)(2n 12 3 (Titu Andreescu and Dorin Andrica, " Asupra unor clase de identita�i" , Gazeta Matematica (GMB), No. 11(1978), pp. 472475) 22.
n( N1 ) n (N2 ), and n(N3 ) = ([b]  a(a))a(c), n(N4 ) = ([dJ  a (c))a (a) It follows that L [/ ( k)]  L [f  l (k)] n(N3 )  n( N4 ) c�k�d a�k�b [b]a(c)  [dJ a(a), desired. ii) Consider the function I : [1, n] + [1, n2], n2 I (x) = 2 x Note that I is decreasing and bijective and 1 Vxn · I  ( x) = =
n(Gf )
y
d
=
=
as
Using formula i), we obtain
t [�: ]  f [ vk . � ] = na(l)  n2 a(1) = 0,
hence
t [ �: ] = f [ vk .�] , n � 1, k= k= 1
as desired. pp.
k=l
1
(Dorin Andrica and Titu Andreescu, Gazeta Matematica (GMB), No. 6(1979),
254, Problem 0 . 48)
a > b � 0 such that Va  .,Jb > 1
([.,Jb] + 1) 2 . Jnl + . . . + nm+l  Jnl + . . . + nm > 1,
c
m
�
l.
This is equivalent to
a
o
Then
k=l
23. It is easy to prove that between numbers there is a perfect square  take for example It suffices to prove that
I (x)
=
=
i) Function I is decreasing and bijective, hence continuous. Consider the sets
Nl = {(x, y) E �2 1 a � x � b, c � y � I(x) }, N2 = ( (x, y) E �2 1 c � y � d, a � x � 1  1 (y) } , N3 = { (x, y) E �2 1 a � x � b, 0 � � c}, N4 = {(x, y) E �2 1 0 � x � a, c � y � d} .
251
6. 2 . SOLUTIONS
b
[/ (k)] = n(N1 ) + n(N3 ), L [f  l (k)] n(N2 ) + n(N4 ), c�k�d L
=
and then
nm+l > 1 + 2Jnl + n2 + . . . + nm, � l. We induct on For = 1 we have to prove that n2 > 1 + 2y'nl. Indeed, n2 � nl + 2 1 + (1 + nl) > 1 + 2y'nl. Assume that the claim holds for some � l.
Then
=
m.
m
m
m
,
6.
25 2
COMPREHENSIVE PROBLEMS
SOLUTIONS
(nm+l  1) 2 > 4(nl + . . . + nm) hence (nm+l + 1) 2 > 4(nl + nm+l) . This implies nm+l + 1 > 2.Jnl + . . . + nm+l, and since nm+ 2  nm+l 2:: 2, it follows
so
. . .
6. 2. 253 Lemma. There are gcd(k, n)  1 integers among 1 . n 2 · n (k  l)n k' k " ' " k Proof. Let gcd(k, n) = d, k = kld, n = nld and note that gcd( kl' nl) 1. The numbers are 1 · nl 2 · nl (k  l)nl kl r.;' r.;" ' " The number of multiples of kl in the set 1 , 2, . . . , k  1 is d  1, hence among the above numbers there are d  1 = gcd( k , n)  1 integers, as desired. The line OAk has the equation n Y = k · x.
+
that
as desired.
(Titu Andreescu, Gazeta Matematidi (GMB ) , No. 1(1980), pp. 41, Problem 0.113) 24. Substituting Xn +l = x�  2 in the relation
=
yields
U nX�  Vn Xn (2un + 1) = 0, n 2:: 1. (1) For a given n 2:: 1 the relation (1) is a quadratic equation with integral coefficient and with an integer root Xn . Hence the discriminant � = v� + 8u� + 4u n = t�, is a square, as desired. (Dorin Andrica) 2 5 . Let ( an ) n2: l be a sequence of real numbers. From the equality xn k 1  xn = xn + x2n + . . . + x n + . . . , I x i 1, n 2:: 1 we derive n  2: An Xn , 2: � n= l 1  xn n= l where An = 2: ad· din Using Gauss' formula 2: <p(d) = n, yields di n X . ""' <p ( n ) Xn = ""' nx n = n (1 X) 2 1 x n=l n= l 1 ' � <pn (n ) = 2 . Sett'mg X = 2 l. mp lles 2 1 (Dorin Andrica) n= l 

00
00
00
00
�
�
�
_
26. We start with a useful lemma.
D
o
<
From the lemma it follows that among numbers
1 . n 2 · n (k  l)n k' k" ' " k there are gcd(k, n)  1 integers. Hence the open segment OAk does not contain points with integral coordinates if and only if gcd( k, n) = 1 . There are <p(n) such numbers and we are done. Remark. An alternative version of this problem can be: " A hunter stays at the point in a forest where the trees are placed at points with integral coordinates. Deers stay at points Ao, Al, . . . ,A n l . How many chances of success has the hunter?" (Dorin Andrica) 0
27. We have a
n + bn {12 + cn W = (1 + 3"2 + W) n = ( {I2(l +( ij2{12) n+ {I4)) n = 2 i ( v2 + {14 + 2) n = 2  i (1 + (1 + v2 + W)) n Y L,
=
,
6.2. SOLUTIONS
6. CO MPREHENSIVE PROBL EMS
254
{
Relations (I) , (II), (III) imply hence
an + b n Y2 + Cn Y4
We study three cases. i) If n := 0 (mod 3 ) , then
2t t (�) a k = an , k=O
ii) If n := we obtain
2
(mod
3),
2 t E Q, 2 t
then
2
n
t (�) ak + k=l
t (�) b k bn , =
k=O
r i
t (�) Ck = Cn· k=O
Multiplying the relation
then 2
2 "
t (�) ak = bn Y2,
k=o
iii) if n := 1 (mod ij2, we obtain
3) ,
2  1}
then
t (�) bk Cn Y2, =
k=O
2 ;+ 1 E Q.
 n +2 3
(1) by 2� = .ij4,
2  1}
(;n (n) Ck k
t (�) Ck
k=O
By multiplying the relation
=
= 2
an ·
�.
n n .=.!!:.±1. 2 L ( k ) ak = 2cn , k=O 3
Hence
2
.=.!!..±l. 3
n
(; (n) bk k
=
an,
2
.=.!!..±l. 3
n
k
bn
� (�) Ck
Cn,
=
�'
n := O (mod n := O (mod
3) 3)
n := l (mod 3)
(Titu Andreescu and Dorin Andrica, Revista Matematica Timi§oara (RMT), No.
(II)
an = "21 X � , b n = Cn = 0 and dn = Y � .
+ 1. Then 2k+ (v0 + v'3) l = (v0 + v'3) (v0 + v'3) 2k = (X k + 3Yk ) v0 + (X k + 2Yk ) v'3 , for all k � 1, hence an = 0, bn = X n;:l + 3y n;:1 , Cn = X n;:l + 2y n;:1 , dn = 0 Let n
= 2k
(Dorin Andrica)
29. We have =
ij2 '
n := O (mod 3) n := O (mod 3) n := l (mod 3)
1 ( 1984), pp . 83, Problem C6:3) 28. Note that (v'2 + J3) n = an + bnv'2 + cnv'3 + dn v'6, n � 1. Let n = 2 k and 1 k k Xk = "2 [(5 + 2 v6) + (5  2 v6) ], Yk = "21 v6[(5 + 2 v6) k  (5  2 v6) k J . hence
( 1) by 2 t =
(; (n) Ck
1 { f'
n := O (mod 3) n := 2 (mod 3) n := l (mod 3)
Then
(3) then
r i
(I)
(2)
Hence
�
and
hence
2 / E Q. 
2 t
=
an, n = a � bn ?'2, ) ( k L k=O cn .ij4, bn, 2 i (�) bk = c;{J2,
2 �
255
=
1 (P  2 ,
q) = I (p  1, q) p  1 p (p  1) q) (P 2) + (p  1) 1 (1 , q) + 2 p 1) (p 1 (1 , q  1) + (q  1) + 1 ( 1 , 1) q(q ; 1) p (p; 1) 00 1 1(P ,
+
+

=
=
=
=
... =

2
_
+
=
= 2
... = .
=
6. COMPREHENSIVE PROBLEMS
256 Therefore
1) q (q 1) 2 _ 2 = 1999 (p  q) (p + q  1) = 2 · 1999. p(p

Thus

Note that 1999 is a prime number and that p  q < p + q  1 for p, q have the following two cases: (a) p  q = 1 and p + q  1 = 3998. Hence p = 2000 and q = 1999. (b) p  q = 2 and p + q  1 = 1999. Hence p = 1001 and q = 999. Therefore (p, q ) = (2000, 1999) or (1001, 999). Korean Mathematics Competition, 2001)
pp. E N* .
We
(Titu Andreescu,
30. The only solutions are f(x) = 0, f(x) = x, and f(x) = x. First, it is clear that these three are solutions. Next, setting x = y = z = 0, we find f(O) = 3(f(0)) 3 , the only integer solution of which is f(O) = O. Next, with y = x and z = 0, we obtain f(O) = (f(X))3 + (f( X))3 + (f(0)) 3. This yields f(x) = f(x), so f is an odd function. With (x, y, z) = (1, 0, 0), we obtain f(l) = (f(1))3 + 2(f(0))3 = f(1)3; thus f(l) E {1, 0, 1 } . Continuing with (x, y , z) = (1, 1, 0) and (x, y , z) = (1, 1, 1) yields f(2) = 2(f(1))3 = 2f(1) and f(3) = 3(f(1))3 = 3f(1). To continue, we need a lemma.
x is an integer greater than 3, then x3 can be written as the sum of five cubes that are smaller in magnitude than x3 . Proof. We have 43 = 33 + 33 + 23 + 13 + 13, 53 = 43 + 43 + (_1) 3 + (_1) 3 + (_1) 3 , 63 = 53 + 43 + 33 + 03 + 03, and 73 = 63 + 53 + 13 + 13 + 03. If x = 2k + 1 with k > 3, Lemma. If
then
x3 = (2k + 1) 3 = (2k _ 1) 3 + (k + 4) 3 + (4  k) 3 + (_5) 3 + (_1) 3 , and all of {2k  1, k + 3, 1 4  k l , 5, I } are less than 2k + 1. If x > 3 is an arbitrary integer, then write x = my, where y is 4 or 6 or an odd number greater than 3 and m is a natural number. Express y 3 as yr + y� + y� + y� + y� . The number x 3 can then be expressed as (mY l )3 + (mY2 )3 + (mY3 )3 + (mY4 )3 + (mY5 )3 . 0 Since f is an odd function and f(l) E {I, 0, 1}, it suffices to prove that f(x) = xf(l) for every integer x. We have proved this for Ixl ::; 3. For x � 4, suppose that the claim is true for all values with magnitude smaller than x. By the lemma, x3 = xr +x�+x�+x�+x�, where IXil < x for all i. After writing x3+( X4 )3+( X5 )3 = xr + x� + x� , we apply f to both sides. By the stated condition of f and the oddness of f, we have (f(X)) 3  (f(X4 )) 3  (f(X5 )) 3 = (f(X r )) 3 + (f(X2 )) 3 + (f(X3 )) 3 . Therefore, the inductive hypothesis yields
5 5 (f(X)) 3 = 2: (xi f(1)) 3 = (f(1)) 3 2: xr = (f(1)) 3 x3 . i=l i= l
257
6.2. SOLUTIONS f(x) = xf(l), and the result follows by induction.
(Titu Andreescu, The American Mathematical Monthly, Volume 108, No. 4(2001),
372, Problem 10728)
3 1 . Assume by way of contradiction that the distance between any two points is greater than or equal to 1. Then the spheres of radius 1/2 with centers at these 1981 points have disjoint interiors and are included in the cube of side 10 determined by the six parallel planes to the given cube's faces and situated in the exterior at a distance of 1/2. It follows that the sum of the volumes of the 1981 spheres is less than the volume of the cube of side 10, hence
1981 ·
4rr o N 2 . 3 = 1981 � > 1000,
a contradiction. The proof is complete. The pigeonhole principle does not help us here. Indeed, dividing each side of the cube in [{f1981] = 12 congruent segments we obtain 123 = 1728 small cubes of side � = � . In such a cube there will be two points from the initial 1981 points.
Remark.
12 4
3
�
The distance between them is less than J3 which is not enough, since 4 J3 > 1. Revista Matematica Timi§oara (RMT) , No. 2(1981), pp. 68, Problem 4627)
(Titu Andreescu, 32. Note that
63 6 2: In k  k l = 2:3 ck (n k  k), where C k E {I, I } , k=l k=l hence 6 S = 2:3 I n k  k l = ±63 ± 63 ± 62 ± 62 ± . . . ± 2 ± 2 ± 1 ± 1, k=l with 63 signs of + and 63 signs of  . Then S ::; (63 + 63 + 62 + 62 + . . . + 33 + 33 + 32)  (32 + 31 + 31 + . . . + 1 + 1) = 1984, as desired.
We prove that, for some labeling, = 1984. It is known that a knight can pass through all the 64 squares of the board only once and then come back to the initial square. Now label the squares from 1 to 64 in the order given by these knight moves. The free position can be made successively 64, 1, 2, . . . , 63, 64, . . . so we can reach the situation n l = 32, n 2 = 33, . . . , n3 2 = 63, n33 = 1, n 34 = 2, . . . , n6 3 = 31. For this diagram we have = 1984. Revista Matematica Timi§oara (RMT) , No. 2(1984), pp. 103, Problem C6:6)
Remark.
(Titu Andreescu,
S
S
258 33.
6. COMPREHENSIVE PROBLEMS
Note that
hence for all n 2: 2. Setting a
(1)
3 F2n , b =  F2n+2 , and e =  F2n 2 in the algebraic identity a3 + b3 + e3  3abe = (a + b + e)(a2 + b2 + e2  ab  be  ea) =
gives 27F�n  F�n+2  F�n  2  9F2n+2F2nF2n2 Applying (1) twice gives =
=
+
n  2(k  1), . . . , n
9F2n+2F2 nF2n2  9F�n = 9 F2n ( F2n+2F2 n 2  Fin ) = 9 F2n · (Titu Andreescu, Korean Mathematics Competition, 2000) 34. Let r be the ratio of an arithmetic progression. We have 1 + r(k 1) ::; n, so r :S � = � . It follows that the maximum ratio is q = 1 � = � I· Let r E {I, 2, . . . , q} and let ar E {I, 2, . . . , n} be the greatest first term of a 
so ar is equal to the number of arithmetic progressions of ratio r from the set {I, 2, . . . , n}. Hence N(n, k) al a2 aq • Because al n  k 1, using (1) gives =
+... +
+
N ::; al + n  k + n  2k + . . . + n  (q  l)k = (q  l)qk + 1 = = nq  k + 1  k(1 + 2 + . . . + (q  1)) = nq  k 2 =
so a total of n  2(k  1) progressions. It follows that q
Nk = I:[n  d(k  1)], d=l
where q is the greatest ratio of an arithmetic progression with k terms from the set {I, 2, . . . , n}. We have proved that q [ � = � 1 hence =
 � k + (n + � ) q + 1  k.
It suffices to prove that � k + (n + � ) q + 1  k :S  � q2 (n + � ) q +
'
 1) . 1)) = nq q(q + l)(k Nk = � L..J (n  d(k 2 d=1 _
It suffices to show that
(1)
ar + (r  1) k ::; n, +
=
1, 3, . . . , 2k  1 2, 4, . . . , 2k
The desired result follows from
=
n  (k  l), . . . , n
so there are n  k 1 such progressions. The arithmetic progressions with k terms and ratio r 2 are
O.
F2n (3 F2n2  F2n )  F:jn2 = F2nF2n4  D �n 2 = = . . . = F6 F2  F; = 1.
progression with ratio r. Then
259 6.2. SOLUTIONS k  1 k  1 q2 . ThIS" mequa1"Ity IS C1ear1y true, so we are whlC· h IS. eqUlva . 1ent to  q ::; 2 2done. Alternative proof. Note that n 2: k 2: 2. The arithmetic progressions with k terms and ratio r = 1 are 1, 2, . . . , k 2, 3, . . . , k + 1
(k  2)q2 + kq + 2  2k 2: O.
(1)
The roots of the quadratic polynomial from the lefthand side of the inequality  1) ql 2(k k 2 and q2 1. Nate that q, 0 for k > 2. Since q [ � = n 2: 1, the inequality (1) is true for all k > 2. If k 2, then is easy to check that the claim holds as well. (Titu Andreescu and Dorin Andrica, Revista Matematica Timi§oara (RMT), No. 1(1982), pp. 104, Problem C4:2) 35. Denote by Pn (x) the (unique) polynomial of degree n such that Pn (k) Fk for k n + 2, n + 3, . . . , 2n 2. (1)
are
=
=
_
=
<
=
=
=
+
6. COMPREHENSIVE PROBLEMS We are going to show that Pn (2n + 3) = F2n+3  1 for all n � O. Clearly, Po (x) = 1 and the claim is true for n = O. Suppose it holds for Pn 1 (x), and consider Pn (x). The polynomial 260
has degree at most n  1 . In view of (1), for each k = n+1, n+2, . . . , 2n . Therefore Q(x) and Pn 1 (x) agree at n distinct points, and hence Q(x) = Pn 1 (x) for all x. In other words, Pn (x +2) = Pn (x + 1) + Pn1 (x) for all x. Combined with the inductive hypothesis P2n1 (2n + 1) = F2n+1  1, this implies Pn (2n + 3) = F2n+2 + F2n+1  1 = F2n+ 3  1. (Titu Andreescu, IMO 1983 Shortlist) 36� For an arbitrary integer n we have Xn+l = aX n + (3Xn l, (1) Xn = aXn l + (3Xn 2 , n � 2 . This is a system of linear equations with unknowns a and (3. The solutions are 6.a = XnXn l  Xn+ l Xn 2 a= � X�_ l  XnXn 2
and
_
Since a and (3 are constant, the conclusion follows. (Dorin Andrica) 37. Let 6.n = a n  a n l for n � 2 . Because an+l = k1 (b + an + an2 1 + . . . + anklk+2 )
and we have where
\
so since b 1. Therefore the sequence is upper bounded, so is convergent. Let x = nlim+oo an. Then (2) Xk1 + X k2 + . . . + x2  (k  l)x + b = 0 If b = 0, then an = 0 for all n � 1, hence x = O. If b (0, 1), we prove that the equation (2) has a unique solution in the interval (0, 1). Let I [0 , 1] I (x) = X k1 + X k2 + . . . + x2  (k  l)x + b. Then 1(0) = b, 1(1) = b  1, 1(0) / (1) = b(b 1) 0 hence the equation (2) has an odd number of solutions in the interval (0, 1). The function I is twice differentiable and since I" (x)  (k  l)(k  2)Xk 3 + 2 > 0, x (0, 1), I is concave up on (0, 1). It follows that equation (2) has at most two solution in (0, 1) therefore the conclusion follows. Remark. The claim that equation (2) has a unique positive solution it follows from 1(0) / (1) = b(b  1) 0 and from the fact that I (x) = Xk1 + Xk2 + . . . + X2  (k  l)x + b has a unique variation of sign (Descartes). (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 12(1979), pp. 56, Problem 3866) 38. From the relation iii) we obtain an+l  an + bn+1  bn  0 , n � 1, an bn <
E
:
+
JR,
<
<
(1) Al = anl + an2 A2 = a�_ 2 + an2 an  3 + a�_ 3 Akl = ank2k+2 + . . . + an k+2 ank_3k+l + ank2k+1 ·
.
E
an = k1 (b + an l + an2 2 + . . . + anklk+1 ) ,
261
> 0 for n � k, so Note that al = a2 = . . . Al , A2 , . . . , Akl � O. On the other hand, 6. 1 = 6.2 = . . , 6.k1 = 0 and from relation (1) it follows that 6.n > 0 for all n � k . Hence the sequence (an ) n�k is increasing. We prove that an < 1 for all n � 1 . Assume that

x �  Xn l Xn+l (3  6.(3 6.  X�_ l  XnXn 2 _
6.2. SOLUTIONS akl = 0 implies an

2 62
then Using ii) yields
(1)
=
(2)
It is easy to see that relations (1) and (2) imply an 0, Im n+oo bn as desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1978) , pp. 69, Problem 3304) 39. Consider the determinant 1·
=
X l X2 X l 0 X2 X l X2 0
Xn Xn Xn
Xl X2 X3
0
0
� (Xl , X2 , . . . , X n ) =
Note that
�(O, X 2 , . . . , xn ) = �(XI ' 0, X3 , ' . . , xn ) = . . . = �(XI ' X2 , . · . , Xn l , 0) = O . Moreover, if Xl + X2 + . . . + Xn = 0, then �(XI ' X2 , . . . , xn ) = 0, therefore �(XI' X2 , · · . , xn) = aXIX2 . . . Xn(XI + X2 + . . . + x n), for some real number a. By identifying the coefficients of XI X2 . . . Xn from both sides we obtain a = 1 .
then Since U2
UI + U2 + . . . + Uk+l . . . Uk = 1 + UI +U.k+l . . + Uk UI + U2 + . . . + Uk UI + U2 + . . . + Un� +l II (1 + UI . . . U k ) = + U UI 2 k=2 I we obtain n 1 UI + . . . + Un+l  II (1 + UI . . . U k ) = nU l ( 1  UI) n k=2
1 + UI
n
=
u ,
Let U nlim+oo Un and note that UI + . . . + Un+l nlim +oo n It follows that =
_ an +oo anbn+l+ l  anbn nlim +oo ( bn ) = nlim On the other hand, by the StolzCesaro theorem we have lim anbn+l+l  anbn nlim+oo anbn n+oo
Hence We have
26 3
6.2. SOLUTIONS
6. CO MPREHENSIVE PROBLEMS

=
U.
(Titu Andreescu, Revista Matematica Timi§oara (RMT), No. 2(1978), pp. 52, Problem 3533) an 40. Let a = lim n+oo bn and let c > O. There is an integer n (c ) > 0 such that an < a + c k + 1 for n 2: n (c ) a  c kk + 11 < k1 bn Since bn > 0 it follows that abn  c kk +_ 11 bn < an < abn + c k _+ 11 bn for all n 2: n (c ) . Then k+1 

k
a(bn+l  bn)  c k < a(bn+l
_
1
(b n+l + bn ) < an+l  an <
 bn) + c kk + 11 (bn+ 1 + bn), _
and, dividing by bn+1  bn > 0, we obtain k+1 k1
a  c  '
bn+l + bn < an+l  an bn+ I  bn bn+ I  bn
for all n 2: n (c) . From relation ii) we deduce
< a+c
k + 1 b n+l + bn ' , k + 1 b n+ I  bn
k1 bn+�l + bn < bn+l  b n  k + 1 , n 2: 1, +l  an < a + c for all n 2: n (c ) a  c < anbn+l  bn 

hence Therefore as desired. (Dorin Andrica, reciproca a teoremei StolzCesaro §i aplicatii ale acesteia", Revista Matematica Timi§oara (RMT), No. 2(1978) , pp. 612) 41. Let Xl k + Jk2 + 1 and X 2 k  Jk2 + 1. We have "0
=
=
IX21 = Xl1

<
1 1 < 2k  2 '

264
6.2. SOLUTIONS 265 It follows that a� = ai for some k =I l i.e. lak I = l ad , which is a contradiction.
6. COMPREHENSIVE PROBLEMS
so
Hence
n x� + x�  1 < x� + G )  1 < !In < x�  G r + 1 <: x� + x� + 1 for all n � 1. The identities X�+l + X�+l = (X l + X2 )(Xr + x�)  XIX2 (X � 1 + X�  l ) = = 2k (xr + x�) + (X � l + X� l ), n � 1 show that xr + x� is an integer for all n, and since an is an integer, it follows that an = xr + x� for all n � 0, and that an+l = 2kan + an l for all n � 1.
Then
The solution is complete . (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 12(1980), pp. 69, Problem 4148) 43. Assume by way of contradiction that g g(x) = sin f(x) , is periodical. We have g' (x) = f'(x) cos f(x), and since f and f' are continuous then g' is also continuous. Note that if g is periodical, then g' is periodical. Moreover, g' is continuous, so it is bounded. Consider the sequence Yn = (4n + 1) �, n � 1. Function f is continuous and lim n+oo Xn = ' hence f(x n ) = Yn for n sufficiently large. Then g (xn ) = f' (xn ) so nlim + oo f'(x n ) = +oo g'(x n ) = nlim which is a contradiction, since g' is bounded. This concludes the proof. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 54, Problem 3544) 44. i) Let f be injective and assume by way of contradiction that a is rational. Hence there are integers p, q with a = P.q and gcd(p, q) = 1. Then f(q) = f(2q) = 0 a contradiction. Therefore a is irrational. Conversely, let a be irrational and assume by way of contradiction that f(m) = f(n) for some integers m =I n � O . Then {am } = {an} so am  [am] = an  [an]. We obtain a = [am]m  n[an] Q, which is a contradiction, and the conclusion follows. ii) Let p, q be relatively prime integers such that a = P.q . We have f (n) = {an} = { r; } . By the division algorithm, there are integers q and r such that n = tq + r, r {O, 1, 2, . . . , q  I } . Then f(n) = { P (tqq+ r) } = { pt + r; } = { r; } = f(r) . We prove that f(O), f(l), . . . , f(q  1) are all distinct. Indeed, if f(i) f(j), then {�} = { j;} :
JR
+
JR,
00,
00
and Therefore (Titu Andreescu) 42 . Assume by way of contradiction that f (xo) =I 0 for all X Clearly, f(p) (x) = 0 for all integers p > O . For p = 0, 2, 4, . . . , 2 (n  1), we obtain sin al X + sina2 x + . . . + sin anx = 0 ai sin al X + a� sina2 x + . . . + a� sin an x = 0 (1) E JR.
for all x Consider a number x such that at least one of sin al x, sin a2X, . . . , sin anx is not zero. Then the homogeneous system of linear equation (1) has a nontrivial solution, hence the determinant is zero: eR
1
.6.8 =
1
a21 (n  l ) a22 ( n  l )
an2 ( nl)
= II (a;  aJ) = 0 i , i=l i >i
.
E
E
=
It follows that (i qj )p is an integer. Note that (p, q) = 1 and Ii  j l < q, hence i  j = 0, so i = j . Therefore M {f(O), f(I), . . . ,f(q  I)} = { 0, qI , q2 , . . . , q q l } since M has q elements. (Dorin Andrica, Romanian Winter Camp, 1984; Revista Matematidi Timi§oara (RMT), No. 1(1985), pp. 67, Problem 3) 45. Assume by way of contradiction that there is a number t > 0 such that g(x + t) = g(x), x Then f(x + t) + f(xB + tB) f(x) + f(xB), x JR, hence f(t) + f(tB) = 2f(0) = 2M. From the relation iii) it follows that f(t) = f(tB) = M and then t = kI T and tB k2 T, for some integers kl ' k2 :f. O. This gives 
=


E R
E
=
=
B = kki2
E
Q,
a contradiction. (Dorin Andrica) 46. We start with an useful lemma. Lemma. If B is an irrational number, then the set M = {mB + nl m,n integers} is dense in JR. Proof. We prove that in any open bounded interval J � JR\ {O} there is an element of M, i.e . M :f. 0. Let J be such an interval and without loss of generality consider J (0, 00). There is an integer n (J) such that 1 n(J) J (0, 1). We consider two cases: C
In
C
267
6.2. SO LUTIONS
6. COMPREHENSIVE PROBLEMS
266
JI = n(J)1 J = (O , ), with 0 < < 1 . Let N be an integer such that � < and consider the numbers {B}, {2B}, . . . , {NB} . There are p, q {I, 2, . . . , N, N + I} such that o < {pB}  {qB} :::; N1 On the other hand, {pB}  {qB} = [qB]  [PB] + (p  q)B M, hence J1 M 1:f. 0. It follows that n(J)( {pB}  {qB}) J M, as desired. 2. JI = J = (a, b) with 0 < a < b < 1. Then 0 <n(J)b  a < 1 and by case 1), there is M such that 0 < < b  a. Let no = [�] + 1. Then a < noc < b and no M JI . Likewise, J M :f. 0, as desired. The lemma is now proved. a) Let A [min f(x), max f(X)] . Hence there is Xo JR such that f(xo) = A. From the lemma we deduce that there are sequences (xn ) n2:I and (Yn ) n2:I such that nlim +oo (xn + Yn T) = Xo. The function f is continuous, so nlim +oo f(xn + Yn T) = f(xo) = A. Note that f(xn + Yn T) = f(xn ), therefore nlim +oo f(x n ) = A, as desired. b) Let B beT an irrational number and consider the function g(x) = f(xB), x The number 7f is irrational and a period for the function g. Using the result from a), there is a sequence (xn ) n2:I of integers such that 1.
e
e
e
E
E
n
E
c
c
E
x E IR
E
n
n
E
c
n
E
x EIR
E R
as desired. (Dorin Andrica, "Asupra unor §iruri care au multimea punctelor limita intervale", Gazeta Matematica (GMB), No. 11 (1979) , pp. 404406) 47. i) Let u = x + Y = + v and assume that x < Y and < v. Then we have x < "2u ' < u"2 ' Y = u  x and v = u z
z
z.
z
6. CO MPREHENSIVE PROBL EMS
268
Suppose, by way of contradiction, that there is a number A > 1 such that xA + yA = Z A + VA . Consider the function f : (0, ) (0, ) f(t) = tA + (  t)A, and note that f is differentiable. We have f' (t) A[tA1  (  t)A1 J , t (0, ) and since A > 1, it follows that f is increasing on (0, �) . Both x, z are in (0, �) , so f(x) f:. f(z), because x f:. z. This implies XA + yA f:. ZA + vA, which is a contradiction. ii) Because is a prime, by Little Fermat's Theorem we have aP  a == bP  b == cP  c == d == 0 (mod p) , hence (aP  a) + (bP  b)  ( cP c) + (dP  d) 0 (mod p) . From aP + bP = cP + dP, we deduce that (1) a  c + b  d == 0 (mod p) By i) we note that a + b f:. c + d, therefore u
+
(0 ,
u
=
E
u
u ,
p
at

==

la  c + b  dl � p,
and then la  c l + I b  d l � p, as desired. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 2(1978), pp. 55, Problem 3550) 48. i) Consider the function <p : (0, 00) <p(t) = t ln t  t. The Mean Value Theorem yields <p(x)  <p(y) = <pI (()) xy for some () (y, x), then x In x  y In y  x + Y = In () . xy It follows that +
JR,
6.2. SOLUTIONS
269
and so XXY = (e()) X  Y . y The inequalities y < () < x, imply (ey) X  Y ::; yXyX ::; (ex) X Y , as desired. ii) Setting X k 1 and y k yields ek (k + l) k k + 1 < kk < e, k > O. Multiplying these inequalities from k = 1 to k = n gives e n < 2 . 3 432 (n + l) n < en , n + 1 1 . 2 . 33 . . . n n and then en 1 en (n + 1) n+1 < n! < (n + l)n ' Hence (n +nl) n < n ., < (n 1)n n+ 1 e e ' as claimed. (Dorin Andrica, Gazeta Matematica (GMB), No. 8(1977), pp. 327, Problem 16820; Revista Matematica Timi§oara (RMT), No. 12(1980), pp. 70, Problem 4153) 49. i) Note that f is bijective and increasing" therefore f is continuous. In this diagram, we have S1 = [ f (t)dt , S 1d r1 (t)dt, hence S1 + S = bd  ac, as desired. y _
=
+
=
2
•
• • •
+
2 =
2
d
E
c o
a
b
x
270
6 . COMPREHENSIVE PROBLEMS
ii ) Again, f is continuous and the diagram shows that Sl  S2 (b  a)c  (d  c)a bc  ad, as desired. y
then eX <p (x) eX a  a + 1 etw(t)dt, by integrating on (0, x]. It follows that i.p(x) a  eaX + e x l0 etw(t)dt, and from i) we obtain lim i.p(x) a. Denote x
=
=
=
x
=
d
x+oo
c s
b x (Dorin Andrica, Revista Matematica Timi§oara (RMT ) , No. 1 (1981) , pp. 62, Problem 4363) 50. i ) Let c > O. There is > 0 such that I I1 (t)1 c for all t > and there is 62 > 0 such that e X < c ° 1 x>U . 1 1 et l l1(t) I dt 2 For x > 6 max(61 , 62 ), we have l e 1 etl' (t)dt l :::; e X 1°' et l l'(t)ldt + eX 1� et I I' (t) I dt (h
<
'
(h
�
=

x
x
<
c + ce  X (eX  e01 ) < 2c,
and the conclusion follows. ii) We start with the following lemma. Lemma. If i.p [0, 00) is a differentiable function with continuous derivative such that lim (i.p(x) + i.p1(X)) = a, then the limit lim i.p(x) exists and equals a. Proof. Without loss of generality we may assume that i.p(0) = O . Define w (0, 00) w(t) i.p(t) + i.pl (t)  a. Function w is continuous and lim w(t) = O. Note that <
:
x+oo
x+oo
:
+
JR,
=
t+oo
=
and note that fm (x) fm  1 (x) + f:n  1 (x), > O. Using the lemma, lim fn (x) A implies lim fn 1 (X) A. Applying the same argument, we finally obtain lim fo(x) lim f(x) A, as desired. (Dorin Andrica) 5 1 . Note that (f(x)  c)(f(x)  d) :::; 0 so f2 (X) + cd :::; (c + d)f(x), x E (a, b]. Then [ f2 (X)dx + (b  a)cd :::; (c + d) [ f(x)dx, and, since the lefthand side is 0, by hypothesis we obtain m
=
a
o
271
6.2. SOLUTIONS
=
x+oo
=
X+OO
=
x+oo
0 :::;
and then
o
On the other hand, 2 Then
(f(x)  C + d ) 2
=
x+oo
=
(c + d) [ f(x)dx,
:::; c +1 d lab f(x)dx.
d) 2 � 0, f2 (X)  (c + d)f(x) + (C +4
{b la f2 (X)dx +
(1)


x (a,b]
(b  a) (c +4 d) 2 � (c + d) l{a b f(x)dx, 
E
272
6. CO MPREHENSIVE PROBLEMS
so
Hence
We have
1 [ (b  a)cd + (b  a) 1 1b (c + d) 2 ] >  c + d f(x)dx. 4 (c + d)2 1 lb b  a (c  d) 2 � 4 c + d f(x)dx. c+d a
From (1) and (2) the conclusion follows. ( Titu Andreescu, Revista Matematidi Timi§oara (RMT), No. 2(1986), Problem 6004) 5 2 . By the Mean Value Theorem there are numbers Cx (x, b) and c� such that
(2)
a
E
pp.
E
76,
(a, x)
F (x) = (x  a)(b  x)f(cx ) + (x  b)(x  a)f(c� ) = = (x  a)(b  x)[J(cx )  f(c�)]. We have c� < Cx for all x E [a, b]. Since f is monotonic f(cx )  f(c�) has constant sign on [a, b]. Moreover, (x  a) (b  x) � 0 for all x E [a, b] and the conclusion follows. (Titu Andreescu, Revista Matematica Timi§oara (RMT) , No. 1(1985), pp. 63, Problem 5505)
a) The function h is continuous, hence it is a derivative. Let F be an anti derivative of h. By the Mean Value Theorem we have F (k + 1 )  F (k) = h( Ck ) for some Ck (k, k + 1) . Since h is nonincreasing, h(k + 1) ::; h( Ck ) ::; h(k) so h(k + 1) � F (k + 1)  F (k) � h(k). Summing these inequalities from k = 1 to k = n yields 53.
E
hence Because
an+1 h(l) � F (n + 1)  F(l) � an, ;:n+l g' (x)f(g(x))dx � an, n � l. an+1  h(l) � 
19(l n+1) f(x)dx =  ;:1 cot xdx = g( ) 1 ) . =  In(sinx) I 1 =  In (sm n + 1 + In ( sm 1) . Hence r g ( n+1 ) f(x)dx = lim [ In (sin _1_ ) + In (sin 1) ] = +00 lim n+oo n+oo }g ( l ) n+1
1
the conclusion follows. b) Setting f (O, �) f(x) =  cotx and g [1, ) obtain h(x) = ;x cot �,x which is decreasing on the interval [1, :
+
JR,
:
(0
+
)
(0 .
JR,
g(x) = �, we
•
Using the lefthand side inequality from a) it follows that n I l cot  = lim = lim an n +oo n +oo L k=l k k 2
00 .
(Dorin Andrica)
54.
Recall that
at  1 = ln a. lim o + t For c > 0 there is 8 > 0 such tthat
at  1 ln a + c, Itl < 8 (1) lna  c < t < Function f is integrable on [0, 1] hence is bounded. Let M > 0 such that f(x) ::; M for all x E [0, 1] . There is an integer no such that � f ( � ) < J for all n � no and k = 0, 1, . . . , n. The inequality (1) gives In a  c < a *f1 ( � )k 1 < In a + c, k = 0, 1, . . . , n nf ( n )
then
It follows that
g (l)
n�l
n�l
1
;: n+1 g' (x)f(g(x))dx = 19(n+1) f(x)dx,
273
6.2. SOLUTIONS
t (a *f ( � )  l ) In a  c < k=l n < In a + c .
(;, H �)
274
On the other hand
6. COMPREHENSIVE PROBLEMS
The function
n 1 ( k ) 11 lim L n+oo k=l n ! n = ! (x) dx,
fr,l (X ) =
0
therefore J��
(t,a;f(!c)  n) = ([ f(X)da;) In a.
(Dorin Andrica, Romanian Mathematical Regional Contest "Grigore Moisil",
1997)
55.
1) We have
nk7rt ) 2  ( 1m sin nk7rt ) 2 1m& fn (X)  tl+l1m' ( sinsm.k7rt t+l sin k7rt nk7r cos nk7rt ) 2 2 nkk 2 2 = (tlim +l k7r cos k7rt = n [( _1) ] = n . Hence In is continuous on JR* if and only if an,k = n2 for all k Z*. 2) It is known that hex : JR JR, a cos { �, hex (x ) = is a derivative. Recall the identity x ) 2 = n + 2 L cos2(k  l ) x, x JR \ {m7r,m Z}. ( Sinn sinx 19<k� n Then !n (x) = { n,an , xx =� OO + 2 19<k� n so In is a derivative function if and only if an = n and an,k = n2 for all k Z*. (Dorin Andrica, Romanian Mathematical Regional Contest "Grigore MoisH", 1.
_
x +
1'
E
+
0,
E
E
E
1995)
56.
We have
{
!o , o(x) = co1, ,o, so !o ,o is a derivative if Co,o = l.
Functions
{ sin x
x�0 0, x=0 are derivatives, so Cl,O = CO ,l = O. u
(x ) =
�,
and
if if
{ cos � , xx =� OO 0,
{ SIn. X1 cOS ,X1 x � = � { sin �, x � 0
0
x=0 2Cl,l , X = 0 Cl,l , = 21 ( 2x ) + 21 20,C , xx �= oo l,l 
{
u
is a derivative if Cl , l = O. For p, q > 1 consider the differentiable function G : JR JR, 2 1 1 G(x) = x smP x cosq , x
1
with G' (x) =
Function
{
.
0,
2x sinP x1 cosq x1 0,
x
t+
+
psmP.  l x1 cosq+l x1 + +q sinP+l �X cosql �X ,
{ 2x sinP x cosq x ' xx =� OO 0,
�
{
{
0,
x�o x=O
�
is continuous, hence a derivative. Therefore +l ql  l q+l g(x ) = p sinP �x cos �x + q sinP �x cos �x ,
I I
x�O x=O
is a derivative. Using the fact that sin2 t = 1  cos2 t and cos2 t = 1 sin2 t, we obtain +l 1 ql 1 q l g(x) = psinPl x cos x + (p + q) sinP x cos x' X � Hence
x�0 x=0
v(x ) =
275
6.2. SOLUTIONS
Therefore
0,
g(x ) = P!pl,ql (x ) + (p + q) !P+l,ql (x )+ x�0 0, + PC ( p + q , l, )C p+l,ql X � 0 p ql x�0 0, (p + q)Cpl,q+l  qCp l,ql , X = 0
{ ={
PCp l,ql = (p + q)Cp+l,ql qCpl,ql = (p + q)Cpl,q+l ,
0
x=0
276
6. COMPREHENSIVE PROBLEMS
and so
Cp+l , q l = p + q Cpl , ql q Cp l , q +l = p+ q Cp l , ql · P 
For k, l 2:: 1 we obtain C2k,2l = 2k2k  2l1 C2k  2,2l = 2k2k  2l1 . 2k2k+ 2l 3 2 C2k4,2l = . . . = = 2k2k + 2l1 . 2k2k+ 2l 3 2 . . . 2l 1+ 2 CO,2l = 2l  1 CO,2l 2 = . . . = = 2k2k + 2l1 . 2k 2k+ 2l 3 2 . . . 2l +1 2 . � 2l  1 . 2l  3 . . . "21 co,o = = 2k2k + 2l1 . 2k2k+ 2l 3 2 . . . 2l +1 2 . � 2l  2 1 · 3 . . . (2l k+l1) . 1 · 3 . . . (2k  1) 2 (k + l)! Note that C2k,2l+l = A · CO , l , C2k+l,2l = B . Cl ,O and C2k+l,2l+l = C · Cl , l , where A, B, C are rational numbers and CO ,l = Cl , O = Cl,l = 0, therefore C2k,2l+l = C2k+l,2l = C2k+l,2l+l = 0, for all k, l 2:: o . To conclude, (2kk+l'1)!!(2l 'I)!! , if p = 2k and q = 2l ''(k + l)! 2 Cp,q = 0, otherwise (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 1(1986), pp. 78, Problem 5773) 57. We start with a useful lemma. Lemma. For any real numbers X l , X 2 , . . . , xn , we have cos Xl . COS X2 · · · COS Xn = 21n � L..t COS(±X l ± X2 ± . . . ± Xn) where the sum is taken over all 2n possible choices of signs . The proof can be made by induction on n. On the other hand recall that function : x =I= O x=o is a derivative. We have two cases. 1 q qn l , x =1= O. From the lemma we Case If q 2:: 2, set Xl x , X2 x , . . . , Xn x obtain n l q = 1 � cos ( ± l ± q ± . . . ± qn l )1 cos x1 · cos xq . . . cos x 2n L..t x +
+
_
{
get
1.
=
=
JR
t
=
JR,
6.2. SO LUTIONS
hence
277
{
(x) + O!�2nn O!(q),n (q), xX =l== O0 fn (x) = 21n � L..t where O!n (q) is the number of choices of signs +,  such that ±1±q±q2 ± . . . ±qnl =2 0, becausen inl the sum are considered only the choices of signs such that ± 1 ± q ± q ± . . . ± q =1= O . Note that if ±1 ± q ± q2 ± . . . ± qn l 0, then q 1, which is false. Hence O!n (q) = 0 and therefore fn is derivative if and only if an (q) = O . Case 2. If q = 1 then for n odd we have O!n ( l) = 0 because ±1 ± 1 ± . . . ± 1 cannot be 0, having an odd number of terms. n is even, let n = 2 . There are (2 ) choices of signs  and signs + so g±l± q ± . . . ± q n  l
=
<
If
m
m
m
To conclude, we have
m
m
if q 2:: 2 O!n (q) = °i (n if q = 1 and n odd ) if q = 1 and n even 2n � , (Dorin A ndrica, Romanian Mathematical Regional Contest "Grigore MoisH", 1999) 58. 1) Function f is continuous, so it is a derivative. Let F be an antiderivative of f. For x =1= 0 we have
1
0,
therefore
f G) = 2xF G)  (X2 F G))' for all # O . Consider the function h : 2XF(1 /X), x =1= 0 h(x) = { 2M(f), x = o. and note that lim h(x) = lim 2xF ( �X ) = yH::O lim �Y 10(Y f(s)ds = 2M(f). �;g �;g Hence h is continuous so it is a derivative function. Let H be an antiderivative of h. We have /x)  (X2 F(l/x))', x i 0 g(X) = { 2xF(1 x=0 M(f), x
JR
t
JR,
278
The function u :
1R
is a deri vative since U :
6. COMPREHENSIVE PROBLEMS (X2 F (l/x)) " X 1= 0 h(x) = M( X = o. f), t
1R
{
lR,
We have
lR,
1 1t 1o a(t) f(x)dx I ::; 1t 10a(t) If(x) l dx ::; 1t 10T If(x)ldx t"4OO 0 hence lim 1t 10 a(t) f(x)dx = O. ttoo Moreover lim n(t)t = ttoo lim n( t)Tn(t)+ a(t) ttoo = ttoo lim 1a(t);: T1 ' T + n(t) therefore lim 1t 1o t f(x)dx = T1 1o T f(x)dx, M (f) = ttoo as claimed. The proof is similar for the case t 00 (Dorin Andrica) 59. The answer is negative. Indeed, consider the derivative function f : f (x) = cos � , x 1= 0 0, x=0 U sing the previous problem, the function cos .!. 1 x 1= 0 g(x) = I2 , x ' ; x=o is also a derivative. Therefore the function 0 ::;
{
(x2 F (l/x)) " X 1= 0 u(x) = M( x=0 f), t
279
6.2. SOLUTIONS
{
2 U(x) = x0, F (l/x), xx =1= O.0 is differentiable and U' = u. It follows that function G = H  U is an antiderivative of g, as desired.
2) We prove that
lim �t 1r0 t f(x)dx = M (f). ttoo For t > 0 there is an integer n = n(t) and a number a = a(t) [0, T] such that t = nT + a. Then l'o t f(x)dx = l0nT f(x)dx + inTt f(x)dx (1) On the other hand !.(k+l)T f(x)dx. lo nT f(x)dx = nl � (2) k=O kT Setting x = B + kT yields !.(k+l)T f(x)dx = rT f(B + kT)dB = rT f(B)dB 10 10 kT so relation (2) gives T T f f(x)dx = n fo f(x)dx Setting B = x  nT yields inTt f(x)dx = I0tnT f(B + nT)dB = l0a(t) f(B)dB The relation (1) becomes r10t f(x)dx = n 1r0T f(x)dx + 1r0 a (t) f(x)dx (3) and hence t n(t) I T f(x)dx + 1 1 a ( t) f(x)dx for all t > O. 1t 1o f(x)dx = t 0 t 0 E

t
.
{ {
x=o
1R
t
lR,
is not a derivative and we are done. (Dorin Andrica, Romanian Mathematical Regional Contest "Grigore Moisil", 1992) 60. The answer is negative. For example, consider the derivative functions il, h : lR, 1  cos � , x 1= 0 f( h (x) = cos ; , x 1= O
1R
t
){
1 X 
0,
X
x=0
,
{
0,
x=o
2 80
6. COMPREHENSIVE PROBLEMS
Then f(x) = ma:x ( h (x) , 12 (x)) =
{l
eos
�I ,
0, which is not a derivative function (see problem 59). Consider the derivative functions
Alternative proof.
{ { {
cos 0,
h (x) =
12 (x) = h (x) =
�,
x#0 x=0
x =J 0 x=0
�
cos , x =J 0 0, x=0 
COS2 � ' x =J 0 _1 '
x
x=O
2
Assume for the sake of contradiction that the statement is valid. Then 9
= m a:x ( h , 12 , h)
{

m a:x ( h , h )
is a derivative function, which is a contradiction, since g (x) =
0,
x =J O
2'
x=O
�
Therefore the answer is negative. Romanian Mathematical Regional Contest " Grigore MoisH" , 1997)
(Dorin A ndrica,
"I take great pleasure in recommending to all readers Romanians or from abroad  the book of professors Titu Andreescu and Dorin Andrica. " "All featu red problems are interesting, with an increased level of difficulty; some of them are real gems that will give great satisfaction to any math lover ... tďż˝ " One more time, I strongly express my belief that the 360 mathematics problems featured in this book will reveal the beauty of mathematics to all students and it will be a guide to their teachers and professors. "
Professor loan Tomescu Department of Mathematics and Computer Science University of Bucharest Associate member of the Romanian Academy
I.S.B.N. 97394 1 7 1 24