Duˇsan Djuki´c Vladimir Jankovi´c Ivan Mati´c Nikola Petrovi´c IMO Shortlist 2007 From the book “The IMO Compendium” Springer c 2008 Springer Scien ce+Business Media, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publ isher (Springer Science+Business Media, Inc. 233, Spring Street, New York, NY 10013, USA), except for brief excerpts in c onnection with reviews or scholary analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar items, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. 1 Problems 1.1 The Forty-Eighth IMO Hanoi, Vietnam, July 19–31, 2007 1.1.1 Contest Problems First Day (July 25) 1. Real numbers a 1 , a 2 , . . ., a n are given. For each i (1 ≤ i ≤ n) define d i = max{a j | 1 ≤ j ≤ i}− min{a j | i ≤ j ≤ n} and let d = max{d i | 1 ≤ i ≤ n}. (a) Prove that, for any real numbers x 1 ≤ x 2 ≤ ··· ≤ x n , max{|x i − a i | | 1 ≤ i ≤ n} ≥ d 2 . (∗) (b) Show that there are real numbers x 1 ≤ x 2 ≤ ··· ≤ x n such that equality holds in (∗). 2. Consider five points A, B, C, D and E such that ABCD is a parallelogram and BCED is a cyclic quadrilater al. Le t ℓ be a line passing through A. Suppose that ℓ intersects the interior of the segment DC at F and intersec ts line BC at G. Suppose also that EF = EG = EC. Prove that ℓ is the bisector of angle DAB. 3. In a mathematical competition some competitors are friends. Friendship is always mutual. Ca ll a gro up of competitors a clique if each two of them are friends. (In particular, any group of fewer than two competitors is a clique.) The number of members of a clique is called its size. Given that, in this competition, the largest size of a clique is even, prove that the competitors can be arranged in two ro oms such that the largest size of a clique contained in one room is the same as the large st size of a clique contained in the other room. 2 1 Problems Second Day (July 26) 4. In triangle ABC the bisector of angle BCA intersects the circumcircle again at R, the perpendicular bisector of BC at P , and the per pendicular bisector of AC at Q. The midpoint of BC is K and the midpoint of AC is L. Prove that the triangles RP K and RQL have the same area. 5. Let a and b be positive integers. Show that if 4ab − 1 divides (4a 2 − 1) 2 , then a = b. 6. Let n be a positive integer. Consider S =  (x, y, z) | x, y, z ∈ {0,1, . . . , n}, x + y + z > 0  as a set of (n + 1) 3 − 1 points in three-dimensional space. Determine the smallest possible number of planes, the union of which contains S but does not include (0, 0,0). 1.1.2 Shortlisted Problems 1. A1 (NZL) IMO1 Given a sequence a 1 , a 2 , . . . , a n of real numbers, for each i (1 ≤ i ≤ n) define d i = max{a j : 1 ≤ j ≤ i} − min{a j : i ≤ j ≤ n} and let d = max{d i : 1 ≤ i ≤ n}. (a) Prove that for arbitrary r e al numbers x 1 ≤ x 2 ≤ ··· ≤ x n , max{|x i − a i | : 1 ≤ i ≤ n} ≥ d 2 . (1) (b) Show that there exis ts a sequence x 1 ≤ x 2 ≤ ··· ≤ x n of real numbers such that we have equality in (1). 2. A2 (BUL) Consider those functions f : N → N w hich satisfy the condi- tion f(m + n) ≥ f (m) + f (f(n)) − 1, for all m, n ∈ N. Find all possible values of f(20 07). 3. A3 (EST) L et n be a positive integer, and let x and y be positive real numbers such that x n + y n = 1. Prove that  n  k=1 1 + x 2k 1 + x 4k  n  k=1 1 + y 2k 1 + y 4k  < 1 (1 − x)(1 −y) . 4. A4 (THA) Find all functions f : R + → R + such that f(x + f(y)) = f(x + y) + f(y) for all x, y ∈ R + . 1.1 Copyright c : The Authors and Springer 3 5. A5 (CRO) Let c > 2, and let a(1), a(2), . . . be a sequence o f nonnegative real numbers such that a(m + n) ≤ 2a(m) + 2a(n) for all m, n ≥ 1, and a(2 k ) ≤ 1 (k + 1) c for all k ≥ 0. Prove that the sequence a(n) is bounded. 6. A6 (POL) Let a 1 , a 2 , . . . , a 100 be nonnegative real numbers such that a 2 1 + a 2 2 + ··· + a 2 100 = 1. Prove that a 2 1 a 2 + a 2 2 a 3 + ··· + a 2 100 a 1 < 12 25 . 7. A7 (NET) IMO6 Let n > 1 be an integer. Cons ider the following subset of the space: S = {(x, y, z)|x, y, z ∈ {0, 1, . . . , n}, x + y + z > 0}. Find the smallest number of planes that jointly contain all (n + 1) 3 − 1 points of S but none of them passes through the origin. 8. C1 (SER) Let n be a n integer. Find all sequences a 1 , a 2 , . . . , a n 2 +n satisfying the following conditions: (i) a i ∈ {0, 1} for a ll 1 ≤ i ≤ n 2 + n; (ii) a i+1 + a i+2 + ··· + a i+n < a i+n+1 + a i+n+2 + ··· + a i+2n for all 0 ≤ i ≤ n 2 − n. 9. C2 (JAP) A unit square is dissected into n > 1 rectangles such that their sides are parallel to the sides of the square. Any line, parallel to a side of the square and intersecting its interior, also intersects the interior of some rectangle. Prove that in this diss e c tion, there exists a rectangle having no point on the boundary of the square. 10. C3 (NET) Find all positve integers n, for which the numbers in the set S = {1, 2, . . . , n} can be colored red and blue, with the following condition being satisfied: the set S × S × S contains exactly 20 07 ordered triples (x, y, z) such that (i) x, y, z are of the same color and (ii) x + y + z is divisible by n. 11. C4 (IRN) Let A 0 = {a 1 , . . . , a n } be a finite sequence of real numbers. For each k ≥ 0, from the sequence A k = (x 1 , . . . , x n ) we construct a new sequence A k+1 in the following way: (i) We choose a partition {1, . . . , n} = I ∪ J, where I and J are two discjoint sets, such that the expr e ssion        i∈I x i −  j∈J x j       4 1 Problems attains the smallest possible value. (We allow the sets I or J to be empty; in this case the corresponding sum is 0.) If there are several such partitions, one is chosen arbitrarily. (ii) We set A k+1 = (y 1 , . . . , y n ), where y i = x i + 1 if i ∈ I, and y i = x i −1 if i ∈ J. Prove that for some k, the sequence A k contains an element x such that |x| ≥ n/2. 12. C5 (ROM) In the Cartesian coor dinate pla ne define the strip S n = {(x, y) : n ≤ x < n + 1 } for every integer n. Assume that each strip S n is colored either red or blue, and let a and b be two distinct positive integers. Prove that there exists a re c tangle with side lengths a and b such that its vertices have the same color. 13. C6 (RUS) IMO3 In a mathematical competition some competitors are friends; friendship is always mutual. Call a group of competitors a clique if each two of them are friends. The number of members in a clique is called its size. It is known that the largest size of a clique is even. Prove that the com- petitors can be arranged in two rooms such that the lar gest size of a clique in one room is the same as the largest size of a clique in the other room. 14. C7 (AUT) Let α < 3− √ 5 2 be a positive real numb e r. Prove that there exist positive integers n and p such that p > α · 2 n and for which one can select 2p pairwise distinct subsets S 1 , . . . , S p , T 1 , . . . , T p of the set {1, 2, . . . , n} such that S i ∩ T j = ∅ for all 1 ≤ i, j ≤ p. 15. C8 (UKR) Given a convex n-g on P in the plane, for every three vertices of P , c onsider the tr iangle determined by them. Call such a triangle good if all its sides are of unit length. Prove that there are not more than 2n/3 good triangles. 16. G1 (CZE) IMO4 In a triangle ABC the bisector of angle BCA intersects the circumcircle again at R, the perpendicular bisector of BC at P , and the perpendicular bisector of AC at Q. The midpoint of BC is K and the midpoint of AC is L. Prove that the triangles RP K and RQL have the same area. 17. G2 (CAN) Given an isosceles triangle ABC, assume that AB = AC. The midp oint of the side BC is denoted by M . Let X be a variable point on the shorter ar c MA of the circumcircle of triangle ABM. Let T be the point in the angle domain BM A for which ∠T MX = 90 ◦ and T X = BX. Prove that ∠MT B −∠CT M doe s not depend on X. 18. G3 (UKR) The diagonals of a trapezoid ABCD intersect at point P . Point Q lies between the parallel lines BC and AD such that ∠AQD = ∠CQB, and the line CD separates the points P and Q. Prove that ∠BQP = ∠DAQ. 1.1 Copyright c : The Authors and Springer 5 19. G4 (LUX) IMO2 Consider five points A, B, C, D and E such that ABCD is a parallelogram and BCED is a cyclic quadrilateral. Let ℓ be a line passing through A. Suppose that ℓ intersects the interior of the segment DC at F and intersects line BC at G. Suppose also that EF = EG = EC. Prove that ℓ is the bisector of a ngle DAB. 20. G5 (GBR) Let ABC be a fixed triangle, and let A 1 , B 1 , C 1 be the modpoints of sides BC, CA, AB respe c tively. Let P be a variable point on the circumcirc le . Let lines P A 1 , P B 1 , P C 1 meet the circumcircle ag ain at A ′ , B ′ , C ′ respectively. Assume that the points A, B, C, A ′ , B ′ , C ′ are distinct, and lines AA ′ , BB ′ , CC ′ form a triangle. Prove that the area of this triangle does not depend on P . 21. G6 (USA) Let ABCD be a convex quadrilateral, and let points A 1 , B 1 , C 1 , and D 1 lie on sides AB, BC, CD, and DA respectively. Consider the areas of triangles AA 1 D 1 , BB 1 A 1 , CC 1 B 1 , and DD 1 C 1 ; let S be the sum of the two smallest ones, and let S 1 be the area of the quadrilateral A 1 B 1 C 1 D 1 . Find the smallest positive real number k such that kS 1 ≥ S holds for every convex quadrilateral ABCD. 22. G7 (IRN) Given an acute triangle ABC with angles α, β, and γ at vertices A, B, and C respectively such that β > γ, let I be its incenter, and R the circumradius. Point D is the foot of the altitude from vertex A. Point K lies on line AD such that AK = 2R, and D separates A and K. Finally lines DI and KI meet sides AC and BC at E and F resp e c tively. Prove that if IE = IF then β ≤ 3γ. 23. G8 (POL) A point P lies on the side AB o f a convex quadrilateral ABCD. Let ω b e the incircle of the triangle CP D, and let I be its incenter. Suppose that ω is tangent to the incircles of triangles AP D and BP C at points K and L, respectively. Let the lines AC and BD meet at E, and let the lines AK and BL meet at F . Prove that the points E, I, and F are colinear. 24. N1 (AUT) Find all pairs (k, n) of positive integers for which 7 k − 3 n divides k 4 + n 2 . 25. N2 (CAN) Let b, n > 1 be integers. Suppose that for each k > 1 there exists an integer a k such that b − a n k is divisible by k. Prove that b = A n for some intege r A. 26. N3 (NET) Let X be a set of 10000 integers, none of which is divisible by 47. Prove that there exists a 2007-element subset Y of X such that a − b + c − d + e is not divisible by 47 for any a, b, c, d, e ∈ Y . 27. N4 (POL) For every integer k ≥ 2, prove that 2 3k divides the number  2 k+1 2 k  −  2 k 2 k−1  6 1 Problems but 2 3k+1 does not. 28. N5 (IRN) Find all sur jective functions f : N → N such that for every m, n ∈ N and every prime p, the number f (m + n) is divisible by p if and only if f(m) + f(n) is divisible by p. 29. N6 (GBR) IMO5 Let k be a positive integer. Pr ove that the number (4k 2 − 1) 2 has a po sitive divisor of the form 8kn − 1 if and only if k is even. 30. N7 (IND) For a prime p and a positive integer n, denote by ν p (n) the exp onent o f p in the prime factorization of n!. Given a positive integer d and a finite set {p 1 , . . . , p k } of pr imes , show that there are infinitely many positive integers n such that d|ν p i (n) for all 1 ≤ i ≤ k. 2 Solutions 8 2 Solutions 2.1 Solutions to the Shortlisted Problems of IMO 2007 1. (a) Assume that d = d m for some index m, and let k and l (k ≤ m ≤ l) be the indeces such that d m = a k − a l . Then d m = a k − a l ≤ (a k − x k ) + (x l −a l ) hence a k −x k ≥ d/2 or x l −a l ≥ d/2. The claim follows immediately. (b) Let M i = max{a j : 1 ≤ j ≤ i} and m i = min{a j : i ≤ j ≤ n}. Set x i = m i +M i 2 . Clearly, m i ≤ a i ≤ M i and both (m i ) and (M i ) are non- decreasing. Furthermore , − d i 2 = m i −M i 2 = x i −M i ≤ x i −a i . Similarly x i −a i ≤ d i 2 , hence max{|x i −a i | : 1 ≤ i ≤ n} ≤ max  d i 2 , 1 ≤ i ≤ n  . Thus, the equality ho lds in (1) for the sequence {x i }. 2. Placing n = 1 we get f(m + 1) ≥ f (m) + f(f(1)) − 1 ≥ f(m) hence the function is non-decreasing. Let n 0 be the smallest integer such that f(n 0 ) > 1. If f(n) = n + k for some k, n ≥ 1 then placing m = 1 gives that f(f(n)) = f(n + k) ≥ f (k) + f(f(n)) − 1 which implies f(k) = 1. We immediately get k < n 0 . Choose maximal k 0 such that there exists n ∈ N fo r which f(n) = n + k 0 . Then we have 2n + k 0 ≥ f(2n) ≥ f(n)+f(f(n))−1 = n+k 0 +f(n+k 0 )−1 ≥ n+k 0 +f(n)−1 = 2n+(2k 0 −1) hence 2k 0 −1 ≤ k 0 , or k 0 ≤ 1. Therefore f(n) ≤ n+1 and f(2007) ≤ 2008. Now we will prove that f(2007) can be any of the numbers 1, 2, . . . , 2008. Define the functions f j (n) =  1, n ≤ 2007 − j, n + j −2007, otherwise. , j ≤ 2007, and f 2008 (n) =  n, 2007 ∤ n, n + 1, 2007 | n. It is easy to verify that f j satisfy the conditions of the problem for j = 1, 2, . . . , 2008. 3. The inequality 1+t 2 1+t 4 < 1 t holds for all t ∈ (0, 1) because it is equivalent to 0 < t 4 − t 3 − t + 1 = (1 −t)(1 − t 3 ). Applying it to t = x k and summing over k = 1, . . . , n we get  n k=1 1+x 2k 1+x 4k <  n k=1 1 x k = x n −1 x n (x−1) = y n x n (1−x) . Writing the same relation for y and multiplying by this one gives the desired inequa lity. 4. Notice that f(x) > x for all x. Indeed, f(x + f(y)) = f(x + y) and if f(y) < y for some y, setting x = y −f(y) yields to a contradictio n. Now we will prove that f(x)−x is injective. If we assume that f(x) −x = f(y) − y for some x = y we would have x + f(y) = y + f(x) hence f(x+y)+f (y) = f(x+y)+f(x) implying f(x) = f (y), which is impossible. From the functional equation we conclude that f(f(x) + f(y)) −(f(x) + f(y)) = f(x+ y), hence f(x)+f(y) = f(x ′ )+f (y ′ ) whenever x+y = x ′ +y ′ . In particular, we have f (x) + f(y) = 2f( x+y 2 ). Now we will prove that f is injective. If f(x) = f(x + h) for some h > 0 then f(x) + f(x + 2h) = 2f(x + h) = 2f(x) hence f(x) = f (x + 2h), and [...]... s) = F (s, p, q) = F (q, s, p) = (p, q) Hence |T \ M | = 3|R × B| = 3r(n − r) and |T | = n2 − 3r(n − r) = n2 − 3rn + 3r2 It remains to solve n2 − 3nr + 3r2 = 2007 in the set N × N First of all, n = 3k for some k ∈ N Therefore 9k 2 − 9kr + 3r2 = 2007 and we see that 3|r Let r = 3s The equation becomes k 2 − 3kr + 3r2 = 223 From our assumption r ≥ n/2 we get 223 = k 2 −3kr+3r2 = (k−r)(k−2r)+r2 ≤ r2 Furthermore... ka3 − ka4 + ka5 There are elements bi ∈ G for which kai ≡ bi (mod 47) which is impossible Each element of x is contained in exactly 10 of the sets Gk hence 10|X| = 46 i=1 |Ak | therefore |Ak | > 2173 > 2007 for at least one k 27 The difference of the two binomial coefficients can be written as D= 2k+1 2k 2k − 2k−1 (2k+1 )! 1 = k − · (2 )! · (2k )! (2k )! (2k )! (2k−1 )! k−1 k = 2 22 ·2 22 · (2k+1 − 1)!! . coordinates of M and N is τ = √ a 2 1 −4 a 1 b ∈ Q. Let s be one of longest single-colored (say red) segments on the x-axis. The translation s ′ of s to the left by τ has non-integer e nd- points,. that maps a to k; K is the center of neg- ative homothethy that maps a to ω; and denote by ˆ F the center o f neg- ative homothethy that maps ω to k. By the consequence of the De- sargue’s theorem we. dissected into n > 1 rectangles such that their sides are parallel to the sides of the square. Any line, parallel to a side of the square and intersecting its interior, also intersects the interior of