IMO Shortlist 2004

Duˇsan Djuki´c Vladimir Jankovi´c Ivan Mati´c Nikola Petrovi´c IMO Shortlist 2004 From the book The IMO Compendium, www.imo.org.yu Springer Berlin Heidelberg NewYork Hong Kong London Milan Paris Tokyo c Copyright 2005 Springer-Verlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection 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. 1 Problems 1.1 The Forty-Fifth IMO Athens, Greece, July 7–19, 2004 1.1.1 Contest Problems First Day (July 12) 1. Let ABC be an acute-angled triangle with AB = AC. The circle with diameter BC intersects the sides AB and AC at M and N, respectively. Denote by O the midpoint of BC. The bisectors of the angles BAC and MON intersect at R. Prove that the circumcircles of the triangles BM R and CNR have a common point lying on the line segment BC. 2. Find all polynomials P (x) with real coefficients that satisfy the equality P (a −b) + P (b − c) + P (c −a) = 2P (a + b + c) for all triples a, b, c of real numbers such that ab + bc + ca = 0. 3. Determine all m × n rectangles that can be covered with hooks made up of 6 unit squares, as in the figure: Rotations and reflections of hooks are allowed. The rectangle must be covered without gaps and overlaps. No part of a hook may cover area outside the rectangle. Second Day (July 13) 2 1 Problems 4. Let n ≥ 3 be an integer and t 1 , t 2 , . . . , t n positive real numbers such that n 2 + 1 > (t 1 + t 2 + ···+ t n )  1 t 1 + 1 t 2 + ···+ 1 t n  . Show that t i , t j , t k are the side lengths of a triangle for all i, j, k with 1 ≤ i < j < k ≤ n. 5. In a convex quadrilateral ABCD the diagonal BD does not bisect the angles ABC and CDA. The point P lies inside ABCD and satisfies ∠P BC = ∠DBA and ∠P DC = ∠BDA. Prove that ABCD is a cyclic quadrilateral if and only if AP = CP . 6. We call a positive integer alternate if its decimal digits are alternately odd and even. Find all positive integers n such that n has an alternate multiple. 1.1.2 Shortlisted Problems 1. A1 (KOR) IMO4 Let n ≥ 3 be an integer and t 1 , t 2 , . . . , t n positive real numbers such that n 2 + 1 > (t 1 + t 2 + ···+ t n )  1 t 1 + 1 t 2 + ···+ 1 t n  . Show that t i , t j , t k are the side lengths of a triangle for all i, j, k with 1 ≤ i < j < k ≤ n. 2. A2 (ROM) An infinite sequence a 0 , a 1 , a 2 , . . . of real numbers satisfies the condition a n = |a n+1 − a n+2 | for every n ≥ 0 with a 0 and a 1 positive and distinct. Can this sequence be bounded? 3. A3 (CAN) Does there exist a function s: Q → {−1, 1} such that if x and y are distinct rational numbers satisfying xy = 1 or x + y ∈ {0, 1}, then s(x)s(y) = −1? Justify your answer. 4. A4 (KOR) IMO2 Find all polynomials P(x) with real coefficients that satisfy the equality P (a − b) + P (b − c) + P (c −a) = 2P (a + b + c) for all triples a, b, c of real numbers such that ab + bc + ca = 0. 5. A5 (THA) Let a, b, c > 0 and ab + bc + ca = 1. Prove the inequality 3  1 a + 6b + 3  1 b + 6c + 3  1 c + 6a ≤ 1 abc . 1.1 Copyright c : The Authors and Springer 3 6. A6 (RUS) Find all functions f : R → R satisfying the equation f  x 2 + y 2 + 2f (xy)  = (f(x + y)) 2 for all x, y ∈ R. 7. A7 (IRE) Let a 1 , a 2 , . . . , a n be positive real numbers, n > 1. Denote by g n their geometric mean, and by A 1 , A 2 , . . . , A n the sequence of arithmetic means defined by A k = a 1 +a 2 +···+a k k , k = 1, 2, . . . , n. Let G n be the geometric mean of A 1 , A 2 , . . . , A n . Prove the inequality n n  G n A n + g n G n ≤ n + 1 and establish the cases of equality. 8. C1 (PUR) There are 10001 students at a university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There are a total of k societies. Suppose that the following conditions hold: (i) Each pair of students are in exactly one club. (ii) For each student and each society, the student is in exactly one club of the society. (iii) Each club has an odd number of students. In addition, a club with 2m + 1 students (m is a positive integer) is in exactly m societies. Find all possible values of k. 9. C2 (GER) Let n and k be positive integers. There are given n circles in the plane. Every two of them intersect at two distinct points, and all points of intersection they determine are distinct. Each intersection point must be colored with one of n distinct colors so that each color is used at least once, and exactly k distinct colors occur on each circle. Find all values of n ≥ 2 and k for which such a coloring is possible. 10. C3 (AUS) The following operation is allowed on a finite graph: Choose an arbitrary cycle of length 4 (if there is any), choose an arbitrary edge in that cycle, and delete it from the graph. For a fixed integer n ≥ 4, find the least number of edges of a graph that can be obtained by repeated ap- plications of this operation from the complete graph on n vertices (where each pair of vertices are joined by an edge). 11. C4 (POL) Consider a matrix of size n×n whose entries are real numbers of absolute value not exceeding 1, and the sum of all entries is 0. Let n be an even positive integer. Determine the least number C such that every such matrix necessarily has a row or a column with the sum of its entries not exceeding C in absolute value. 12. C5 (NZL) Let N be a positive integer. Two players A and B, taking turns, write numbers from the set {1, . . . , N} on a blackboard. A begins the game by writing 1 on his first move. Then, if a player has written n on 4 1 Problems a certain move, his adversary is allowed to write n+ 1 or 2n (provided the number he writes does not exceed N). The player who writes N wins. We say that N is of type A or of type B according as A or B has a winning strategy. (a) Determine whether N = 2004 is of type A or of type B. (b) Find the least N > 2004 whose type is different from that of 2004. 13. C6 (IRN) For an n × n matrix A, let X i be the set of entries in row i, and Y j the set of entries in column j, 1 ≤ i, j ≤ n. We say that A is golden if X 1 , . . . , X n , Y 1 , . . . , Y n are distinct sets. Find the least integer n such that there exists a 2004 ×2004 golden matrix with entries in the set {1, 2, . . . , n}. 14. C7 (EST) IMO3 Determine all m ×n rectangles that can be covered with hooks made up of 6 unit squares, as in the figure: Rotations and reflections of hooks are allowed. The rectangle must be covered without gaps and overlaps. No part of a hook may cover area outside the rectangle. 15. C8 (POL) For a finite graph G, let f (G) be the number of triangles and g(G) the number of tetrahedra formed by edges of G. Find the least constant c such that g(G) 3 ≤ c · f(G) 4 for every graph G. 16. G1 (ROM) IMO1 Let ABC be an acute-angled triangle with AB = AC. The circle with diameter BC intersects the sides AB and AC at M and N, respectively. Denote by O the midpoint of BC. The bisectors of the angles BAC and MON intersect at R. Prove that the circumcircles of the triangles BMR and CNR have a common point lying on the line segment BC. 17. G2 (KAZ) The circle Γ and the line  do not intersect. Let AB be the diameter of Γ perpendicular to , with B closer to  than A. An arbitrary point C = A, B is chosen on Γ . The line AC intersects  at D. The line DE is tangent to Γ at E, with B and E on the same side of AC. Let BE intersect  at F , and let AF intersect Γ at G = A. Prove that the reflection of G in AB lies on the line CF . 18. G3 (KOR) Let O be the circumcenter of an acute-angled triangle ABC with ∠B < ∠C. The line AO meets the side BC at D. The circumcenters of the triangles ABD and ACD are E and F , respectively. Extend the sides BA and CA beyond A, and choose on the respective extension points G and H such that AG = AC and AH = AB. Prove that the quadrilateral EF GH is a rectangle if and only if ∠ACB −∠ABC = 60 ◦ . 1.1 Copyright c : The Authors and Springer 5 19. G4 (POL) IMO5 In a convex quadrilateral ABCD the diagonal BD does not bisect the angles ABC and CDA. The point P lies inside ABCD and satisfies ∠P BC = ∠DBA and ∠P DC = ∠BDA. Prove that ABCD is a cyclic quadrilateral if and only if AP = CP . 20. G5 (Duˇsan Djuki´c, SMN) Let A 1 A 2 . . . A n be a regular n-gon. The points B 1 , . . . , B n−1 are defined as follows: (i) If i = 1 or i = n − 1, then B i is the midpoint of the side A i A i+1 . (ii) If i = 1, i = n − 1, and S is the intersection point of A 1 A i+1 and A n A i , then B i is the intersection point of the bisector of the angle A i SA i+1 with A i A i+1 . Prove the equality ∠A 1 B 1 A n + ∠A 1 B 2 A n + ···+ ∠A 1 B n−1 A n = 180 ◦ . 21. G6 (GBR) Let P be a convex polygon. Prove that there is a convex hexagon that is contained in P and that occupies at least 75 percent of the area of P. 22. G7 (RUS) For a given triangle ABC, let X be a variable point on the line BC such that C lies between B and X and the incircles of the triangles ABX and ACX intersect at two distinct points P and Q. Prove that the line P Q passes through a point independent of X. 23. G8 (Duˇsan Djuki´c, SMN) A cyclic quadrilateral ABCD is given. The lines AD and BC intersect at E, with C between B and E; the diagonals AC and BD intersect at F . Let M be the midpoint of the side CD, and let N = M be a point on the circumcircle of the triangle ABM such that AN/BN = AM/BM. Prove that the points E, F , and N are collinear. 24. N1 (BLR) Let τ (n) denote the number of positive divisors of the positive integer n. Prove that there exist infinitely many positive integers a such that the equation τ(an) = n does not have a positive integer solution n. 25. N2 (RUS) The function ψ from the set N of positive integers into itself is defined by the equality ψ(n) = n  k=1 (k, n), n ∈ N, where (k, n) denotes the greatest common divisor of k and n. (a) Prove that ψ(mn) = ψ(m)ψ(n) for every two relatively prime m, n ∈ N. (b) Prove that for each a ∈ N the equation ψ(x) = ax has a solution. 6 1 Problems (c) Find all a ∈ N such that the equation ψ(x) = ax has a unique solution. 26. N3 (IRN) A function f from the set of positive integers N into itself is such that for all m, n ∈ N the number (m 2 + n) 2 is divisible by f 2 (m) + f(n). Prove that f(n) = n for each n ∈ N. 27. N4 (POL) Let k be a fixed integer greater than 1, and let m = 4k 2 −5. Show that there exist positive integers a and b such that the sequence (x n ) defined by x 0 = a, x 1 = b, x n+2 = x n+1 + x n for n = 0, 1, 2, . . . has all of its terms relatively prime to m. 28. N5 (IRN) IMO6 We call a positive integer alternate if its decimal digits are alternately odd and even. Find all positive integers n such that n has an alternate multiple. 29. N6 (IRE) Given an integer n > 1, denote by P n the product of all positive integers x less than n and such that n divides x 2 − 1. For each n > 1, find the remainder of P n on division by n. 30. N7 (BUL) Let p be an odd prime and n a positive integer. In the coordinate plane, eight distinct points with integer coordinates lie on a circle with diameter of length p n . Prove that there exists a triangle with vertices at three of the given points such that the squares of its side lengths are integers divisible by p n+1 . 2 Solutions 2.1 Solutions to the Shortlisted Problems of IMO 2004 1. By symmetry, it is enough to prove that t 1 + t 2 > t 3 . We have  n  i=1 t i  n  i=1 1 t i  = n 2 +  i<j  t i t j + t j t i − 2  . (1) All the summands on the RHS are positive, and therefore the RHS is not smaller than n 2 + T , where T = (t 1 /t 3 + t 3 /t 1 − 2) + (t 2 /t 3 + t 3 /t 2 − 2). We note that T is increasing as a function in t 3 for t 3 ≥ max{t 1 , t 2 }. If t 1 +t 2 = t 3 , then T = (t 1 +t 2 )(1/t 1 +1/t 2 )−1 ≥ 3 by the Cauchy–Schwarz inequality. Hence, if t 1 + t 2 ≤ t 3 , we have T ≥ 1, and consequently the RHS in (1) is greater than or equal to n 2 + 1, a contradiction. Remark. In can be proved, for example using Lagrange multipliers, that if n 2 + 1 in the problem is replaced by (n + √ 10 −3) 2 , then the statement remains true. This estimate is the best possible. 2. We claim that the sequence {a n } must be unbounded. The condition of the sequence is equivalent to a n > 0 and a n+1 = a n +a n−1 or a n − a n−1 . In particular, if a n < a n−1 , then a n+1 > max{a n , a n−1 }. Let us remove all a n such that a n < a n−1 . The obtained sequence (b m ) m∈N is strictly increasing. Thus the statement of the problem will follow if we prove that b m+1 − b m ≥ b m − b m−1 for all m ≥ 2. Let b m+1 = a n+2 for some n. Then a n+2 > a n+1 . We distinguish two cases: (i) If a n+1 > a n , we have b m = a n+1 and b m−1 ≥ a n−1 (since b m−1 is either a n−1 or a n ). Then b m+1 − b m = a n+2 − a n+1 = a n = a n+1 − a n−1 = b m − a n−1 ≥ b m − b m−1 . (ii) If a n+1 < a n , we have b m = a n and b m−1 ≥ a n−1 . Consequently, b m+1 −b m = a n+2 −a n = a n+1 = a n −a n−1 = b m −a n−1 ≥ b m −b m−1 . 8 2 Solutions 3. The answer is yes. Every rational number x > 0 can be uniquely expressed as a continued fraction of the form a 0 + 1/(a 1 + 1/(a 2 + 1/(··· + 1/a n ))) (where a 0 ∈ N 0 , a 1 , . . . , a n ∈ N). Then we write x = [a 0 ; a 1 , a 2 , . . . , a n ]. Since n depends only on x, the function s(x) = (−1) n is well-defined. For x < 0 we define s(x) = −s(−x), and set s(0) = 1. We claim that this s(x) satisfies the requirements of the problem. The equality s(x)s(y) = −1 trivially holds if x + y = 0. Suppose that xy = 1. We may assume w.l.o.g. that x > y > 0. Then x > 1, so if x = [a 0 ; a 1 , a 2 , . . . , a n ], then a 0 ≥ 1 and y = 0 + 1/x = [0; a 0 , a 1 , a 2 , . . . , a n ]. It follows that s(x) = (−1) n , s(y) = (−1) n+1 , and hence s(x)s(y) = −1. Finally, suppose that x + y = 1. We consider two cases: (i) Let x, y > 0. We may assume w.l.o.g. that x > 1/2. Then there exist natural numbers a 2 , . . . , a n such that x = [0; 1, a 2 , . . . , a n ] = 1/(1 + 1/t), where t = [a 2 , . . . , a n ]. Since y = 1 − x = 1/(1 + t) = [0; 1 + a 2 , a 3 , . . . , a n ], we have s(x) = (−1) n and s(y) = (−1) n−1 , giving us s(x)s(y) = −1. (ii) Let x > 0 > y. If a 0 , . . . , a n ∈ N are such that −y = [a 0 ; a 1 , . . . , a n ], then x = [1 + a 0 ; a 1 , . . . , a n ]. Thus s(y) = −s(−y) = −(−1) n and s(x) = (−1) n , so again s(x)s(y) = −1. 4. Let P (x) = a 0 + a 1 x + ··· + a n x n . For every x ∈ R the triple (a, b, c) = (6x, 3x, −2x) satisfies the condition ab + bc + ca = 0. Then the condition on P gives us P(3x) + P (5x) + P (−8x) = 2P (7x) for all x, implying that for all i = 0, 1, 2, . . . , n the following equality holds:  3 i + 5 i + (−8) i − 2 · 7 i  a i = 0. Suppose that a i = 0. Then K(i) = 3 i + 5 i + (−8) i −2 ·7 i = 0. But K(i) is negative for i odd and positive for i = 0 or i ≥ 6 even. Only for i = 2 and i = 4 do we have K(i) = 0. It follows that P (x) = a 2 x 2 + a 4 x 4 for some real numbers a 2 , a 4 . It is easily verified that all such P (x) satisfy the required condition. 5. By the general mean inequality (M 1 ≤ M 3 ), the LHS of the inequality to be proved does not exceed E = 3 3 √ 3 3  1 a + 1 b + 1 c + 6(a + b + c). From ab + bc + ca = 1 we obtain that 3abc(a + b + c) = 3(ab · ac + ab · bc + ac · bc) ≤ (ab + ac + bc) 2 = 1; hence 6(a + b + c) ≤ 2 abc . Since 1 a + 1 b + 1 c = ab+bc+ca abc = 1 abc , it follows that E ≤ 3 3 √ 3 3  3 abc ≤ 1 abc , [...]... 2004 = 11111010100 in the binary system, 2004 is of type A The least N > 2004 that is of type B is 100000000000 = 211 = 2048 Thus the answer to part (b) is 2048 2.1 Copyright c : The Authors and Springer 13 13 Since Xi , Yi , i = 1, , 2004, are 4008 distinct subsets of the set Sn = {1, 2, , n}, it follows that 2n ≥ 4008, i.e n ≥ 12 Suppose n = 12 Let X = {X1 , , X2004 }, Y = {Y1 , , Y2004... can be easily proved by induction that each of the matrices Am is golden Moreover, every upper-left square submatrix of Am of size greater than 2m−1 is also golden Since 210 < 2004 < 211 , we thus obtain a golden matrix of size 2004 with entries in S13 14 Suppose that an m×n rectangle can be covered by “hooks” For any hook H there is a unique hook K that covers its “ inside” square Then also H covers . Duˇsan Djuki´c Vladimir Jankovi´c Ivan Mati´c Nikola Petrovi´c IMO Shortlist 2004 From the book The IMO Compendium, www .imo. org.yu Springer Berlin Heidelberg NewYork Hong Kong London Milan. a winning strategy. (a) Determine whether N = 2004 is of type A or of type B. (b) Find the least N > 2004 whose type is different from that of 2004. 13. C6 (IRN) For an n × n matrix A, let. sets. Find the least integer n such that there exists a 2004 2004 golden matrix with entries in the set {1, 2, . . . , n}. 14. C7 (EST) IMO3 Determine all m ×n rectangles that can be covered