Volume 1, Number1
January-February, 1995
Olympiad Corner The 35th International Mathematical Olympiad was held in Hong Kong last summer. The following are the six problems given to the contestants. How many can you solve? (The country names inside the parentheses are the problem proposers.) -Editors
Pigeonhole Principle Kin-Yin Li
What in the world is the pigeonhole principle? Well. this famous principle states that if n+ 1 objects (pigeons) are Problem 1. (France) Let m and n be positive integers. Let aI, a2, taken from n boxes (pigeonholes), then at least two of the objects will be from the ..., ambe distinct elements of {I, 2, ..., n} samebox. This is clear enough that it does suchthat whenever at + aJ ~ n for some i,j, 1 ~ i ~ j ~ m, there exists k, 1 ~ k ~ m, with not require much explanation. A problem solver who takes advantage of this at + aJ = at. Prove that principle can tackle certain combinatorial al+~+'..+am n+l problemsin a manner that is more elegant m 2 and systematicthancase-by-case.To show how to apply this principle, we give a few Problem 2. (Armenia/Australia) examples below. ABC is an isoscelestriangle with AB = AC. Suppose that Example 1. Suppose51 numbersare (i) M is the midpoint of BC and 0 is the chosenfrom 1, 2, 3, ...,99, 100. Show point on the line AM such that OB is thattherearetwo whichdo nothaveany perpendicular to AB; commonprimedivisor. (ii) Q is an arbitrary pOint on the segment BC different from Band C; Solution. Let us consider the 50 pairs of (iii) E lies on the line AB and F lies on the consecutive numbers (1,2), (3,4), ..., line AC such that E, Q and F are (99,100). Since 51 numbers are chosen, distinct and collinear. the pigeonhole principle tells us that there Prove thatOQ is perpendicular to EF if and will be a pair (k, k+l) among them. Now only if QE = QF. if a prime number p divides k+ 1 and k, thenp will divide (k+ 1) -k = 1, which is a (continued on page 4) contradiction. So, k and k+ 1 have no common prime divisor.
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Editors:
Artist:
Li, Kin- Yin, Math Dept, HKUST Ko,Tsz-Mei, EEE Dept, HKUST Ng, Keng Po Roger, lTC, HKP Yeung, Sau-Ying Camille, Fine Arts Dept, CU
Acknowledgment: Thanks to Martha A. Dahlen, Technical Writer, HKUST, for her comments. The editors welcome contributions from all students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in TeX, MS Word and WordPerfect, are encouraged. The deadline for receiving material for the next issue is January 31, 1995. Send all correspondence to: Dr. Li, Kin-Yin Department of Mathematics Hong Kong University of Science and Technology Clear Water Bay, Kowloon Hong Kong Fax: 2358-1643 Email: makyli@uxmail.ust.hk
Note that the two examples look alike, however the boxes fornled are quite different. By now, the readers must have observed that forming the right boxes is the key to success.Often a certain amount of experience as well as clever thinking are required to solve such problems. The additional examples below will help beginnersbecomefarniliar with this useful principle. Example 3. Show that aInQng any nine distinct real numbers,there are two, say a and b, such that 0 < (a-b)! Solution.
The
b)/(I+ab)
reminds
tan(x-y). the
Solution. Considerthe 50 odd numbers 1, 3, 5, ..., 99. For each one, form a box containingthe number and all powers of 2 timesthe number.So the first box contains 1,2,4,8, 16, ...and the next box contains 3,6,12,24,48, ...and so on. Then among the 51 numbers chosen, the pigeonhole principle tells us that there are two that arecontained in the samebox. They must be of the form 2mkand 2nk with the same odd number k. So one will divide the other.
middle
interval
expression
(a-
us of the fonnula
So we proceed
as follow.
(-1&12,1&12] into
for
Divide
8 intervals
(-1&/2,-31&/8], (-31&/8,-1&14], ..., (1&14,31&18], (31&/8,1&12]. Let the numbers
be ap a2, ...,
agandletxj=arctanaj, i=I,2,...,9. By the pigeonhole principle, two of the x/s, say Xjand Xk with xi> the 8 subintervals.
Xk' must be in one of
Then we have 0 < Xj -Xk
< 1&/8, so 0 < tan(Xj-xJ < tan(1&/8) = Ii
Example 2. Suppose 51 numbers are chosen from 1, 2, 3, ..., 99, 100. Show thattherearetwo such that one divides the other.
(1 +ab) <.J2 -1.
= (aj-aJ/(1
+aJaJ
-1.
Example 4. Suppose a triangle can be placed inside a square of unit area in such a way that the center of the square is not inside the triangle. Show that one side of the triangle has length less than 1. (This examplecame from the XLI Mathematical Olympiad in Poland.)
Solution. Through the center C of the square, draw a line L) parallel to the closestside of the triangle and a second line ~ perpendicularto L) atC. The lines L) and ~ divide the square into four congruentquadrilaterals.Since C is not (continuedon page 2)