TUYỂN TẬP CÁC BÀI TOÁN IMO

IMO là cuộc thi toán danh dự cho nhiều học sinh có đam mê với toán, nhằm cung cấp các dạng bài, hình thức đề thi của IMO trong nhiều năm, đây là cuốn EBOOK hữu ích In this book, all manuscripts have been collected into a single compendium of mathematics problems of the kind that usually appear on the IMOs. Therefore, we believe that this book will be the definitive and authoritative source for highschool students preparing for the IMO, and we suspect that it will be of particular benefit in countries lacking adequate preparation literature. A highschool student could spend an enjoyable year going through the numerous problems and novel ideas presented in the solutions and emerge ready to tackle even the most difficult problems on an IMO. In addition, the skill acquired in the process of successfully attacking difficult mathematics problems will prove to be invaluable in a serious and prosperous career in mathematics

Problem Books in Mathematics Edited by P Winkler Dusˇan Djukic´ Vladimir Jankovic´ Ivan Matic´ Nikola Petrovic´ The IMO Compendium A Collection of Problems Suggested for the International Mathematical Olympiads: 1959–2004 With 200 Figures Dusˇan Djukic´ Department of Mathematics University of Toronto Toronto ON, M5S3G3 Canada djdusan@EUnet.yu Vladimir Jankovic´ Department of Mathematics University of Belgrade 11000 Belgrade Serbia and Montenegro vjankovic@matf.bg.ac.yu Ivan Matic´ Department of Mathematics Berkeley, CA USA matic@matf.bf.ac.yu Nikola Petrovic´ Institute of Physics 11000 Belgrade Serbia and Montenegro nzpetr@eunet.yu Series Editor: Peter Winkler Department of Mathematics Dartmouth College Hanover, NH 03755-3551 USA Peter.winkler@dartmouth.edu Mathematics Subject Classification (2000): 00A07 Library of Congress Control Number: 2005934915 ISBN-10: 0-387-24299-6 ISBN-13: 978-0387-24299-6 © 2006 Springer Science+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 publisher (Springer Science+Business Media, Inc., 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly 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 terms, 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 Printed in the United States of America springer.com (MVY) Preface The International Mathematical Olympiad (IMO) is nearing its fiftieth anniversary and has already created a very rich legacy and firmly established itself as the most prestigious mathematical competition in which a high-school student could aspire to participate Apart from the opportunity to tackle interesting and very challenging mathematical problems, the IMO represents a great opportunity for high-school students to see how they measure up against students from the rest of the world Perhaps even more importantly, it is an opportunity to make friends and socialize with students who have similar interests, possibly even to become acquainted with their future colleagues on this first leg of their journey into the world of professional and scientific mathematics Above all, however pleasing or disappointing the final score may be, preparing for an IMO and participating in one is an adventure that will undoubtedly linger in one’s memory for the rest of one’s life It is to the high-school-aged aspiring mathematician and IMO participant that we devote this entire book The goal of this book is to include all problems ever shortlisted for the IMOs in a single volume Up to this point, only scattered manuscripts traded among different teams have been available, and a number of manuscripts were lost for many years or unavailable to many In this book, all manuscripts have been collected into a single compendium of mathematics problems of the kind that usually appear on the IMOs Therefore, we believe that this book will be the definitive and authoritative source for high-school students preparing for the IMO, and we suspect that it will be of particular benefit in countries lacking adequate preparation literature A high-school student could spend an enjoyable year going through the numerous problems and novel ideas presented in the solutions and emerge ready to tackle even the most difficult problems on an IMO In addition, the skill acquired in the process of successfully attacking difficult mathematics problems will prove to be invaluable in a serious and prosperous career in mathematics However, we must caution our aspiring IMO participant on the use of this book Any book of problems, no matter how large, quickly depletes itself if VI Preface the reader merely glances at a problem and then five minutes later, having determined that the problem seems unsolvable, glances at the solution The authors therefore propose the following plan for working through the book Each problem is to be attempted at least half an hour before the reader looks at the solution The reader is strongly encouraged to keep trying to solve the problem without looking at the solution as long as he or she is coming up with fresh ideas and possibilities for solving the problem Only after all venues seem to have been exhausted is the reader to look at the solution, and then only in order to study it in close detail, carefully noting any previously unseen ideas or methods used To condense the subject matter of this already very large book, most solutions have been streamlined, omitting obvious derivations and algebraic manipulations Thus, reading the solutions requires a certain mathematical maturity, and in any case, the solutions, especially in geometry, are intended to be followed through with pencil and paper, the reader filling in all the omitted details We highly recommend that the reader mark such unsolved problems and return to them in a few months to see whether they can be solved this time without looking at the solutions We believe this to be the most efficient and systematic way (as with any book of problems) to raise one’s level of skill and mathematical maturity We now leave our reader with final words of encouragement to persist in this journey even when the difficulties seem insurmountable and a sincere wish to the reader for all mathematical success one can hope to aspire to Belgrade, October 2004 Duˇsan Djuki´c Vladimir Jankovi´c Ivan Mati´c Nikola Petrovi´c For the most current information regarding The IMO Compendium you are invited to go to our website: www.imo.org.yu At this site you can also find, for several of the years, scanned versions of available original shortlist and longlist problems, which should give an illustration of the original state the IMO materials we used were in We are aware that this book may still contain errors If you find any, please notify us at imo@matf.bg.ac.yu A full list of discovered errors can be found at our website If you have any questions, comments, or suggestions regarding both our book and our website, please not hesitate to write to us at the above email address We would be more than happy to hear from you Preface VII Acknowledgements The making of this book would have never been possible without the help of numerous individuals, whom we wish to thank First and foremost, obtaining manuscripts containing suggestions for IMOs was vital in order for us to provide the most complete listing of problems possible We obtained manuscripts for many of the years from the former and current IMO team leaders of Yugoslavia / Serbia and Montenegro, who carefully preserved these valuable papers throughout the years Special thanks are due to Prof Vladimir Mi´ci´c, for some of the oldest manuscripts, and to Prof Zoran Kadelburg We also thank Prof Djordje Dugoˇsija and Prof Pavle Mladenovi´c In collecting shortlisted and longlisted problems we were also assisted by Prof Ioan Tomescu from Romania and Hà Duy Hưng from Vietnam A lot of work was invested in cleaning up our giant manuscript of errors Special thanks in this respect go to David Kramer, our copy-editor, and to Prof Titu Andreescu and his group for checking, in great detail, the validity of the solutions in this manuscript, and for their proposed corrections and alternative solutions to several problems We also thank Prof Abderrahim Ouardini from France for sending us the list of countries of origin for the shortlisted problems of 1998, Prof Dorin Andrica for helping us compile the ˇ c for proofreading part of list of books for reference, and Prof Ljubomir Cuki´ the manuscript and helping us correct several errors We would also like to express our thanks to all anonymous authors of the IMO problems It is a pity that authors’ names are not registered together with their proposed problems Without them, the IMO would obviously not be what it is today In many cases, the original solutions of the authors were used, and we duly acknowledge this immense contribution to our book, though once again, we regret that we cannot this individually In the same vein, we also thank all the students participating in the IMOs, since we have also included some of their original solutions in this book The illustrations of geometry problems were done in WinGCLC, a program created by Prof Predrag Janiˇci´c This program is specifically designed for creating geometric pictures of unparalleled complexity quickly and efficiently Even though it is still in its testing phase, its capabilities and utility are already remarkable and worthy of highest compliment Finally, we would like to thank our families for all their love and support during the making of this book Contents Preface v Introduction 1.1 The International Mathematical Olympiad 1.2 The IMO Compendium 1 2 Basic Concepts and Facts 2.1 Algebra 2.1.1 Polynomials 2.1.2 Recurrence Relations 2.1.3 Inequalities 2.1.4 Groups and Fields 2.2 Analysis 2.3 Geometry 2.3.1 Triangle Geometry 2.3.2 Vectors in Geometry 2.3.3 Barycenters 2.3.4 Quadrilaterals 2.3.5 Circle Geometry 2.3.6 Inversion 2.3.7 Geometric Inequalities 2.3.8 Trigonometry 2.3.9 Formulas in Geometry 2.4 Number Theory 2.4.1 Divisibility and Congruences 2.4.2 Exponential Congruences 2.4.3 Quadratic Diophantine Equations 2.4.4 Farey Sequences 2.5 Combinatorics 2.5.1 Counting of Objects 2.5.2 Graph Theory 5 10 12 12 13 14 14 15 16 16 17 18 19 19 20 21 22 22 22 23 X Contents Problems 3.1 IMO 1959 3.1.1 Contest Problems 3.2 IMO 1960 3.2.1 Contest Problems 3.3 IMO 1961 3.3.1 Contest Problems 3.4 IMO 1962 3.4.1 Contest Problems 3.5 IMO 1963 3.5.1 Contest Problems 3.6 IMO 1964 3.6.1 Contest Problems 3.7 IMO 1965 3.7.1 Contest Problems 3.8 IMO 1966 3.8.1 Contest Problems 3.8.2 Some Longlisted Problems 1959–1966 3.9 IMO 1967 3.9.1 Contest Problems 3.9.2 Longlisted Problems 3.10 IMO 1968 3.10.1 Contest Problems 3.10.2 Shortlisted Problems 3.11 IMO 1969 3.11.1 Contest Problems 3.11.2 Longlisted Problems 3.12 IMO 1970 3.12.1 Contest Problems 3.12.2 Longlisted Problems 3.12.3 Shortlisted Problems 3.13 IMO 1971 3.13.1 Contest Problems 3.13.2 Longlisted Problems 3.13.3 Shortlisted Problems 3.14 IMO 1972 3.14.1 Contest Problems 3.14.2 Longlisted Problems 3.14.3 Shortlisted Problems 3.15 IMO 1973 3.15.1 Contest Problems 3.15.2 Shortlisted Problems 3.16 IMO 1974 3.16.1 Contest Problems 3.16.2 Longlisted Problems 27 27 27 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 42 42 42 51 51 52 55 55 55 64 64 65 72 74 74 75 81 84 84 84 89 91 91 92 94 94 95 Contents XI 3.16.3 Shortlisted Problems 100 3.17 IMO 1975 103 3.17.1 Contest Problems 103 3.17.2 Shortlisted Problems 103 3.18 IMO 1976 106 3.18.1 Contest Problems 106 3.18.2 Longlisted Problems 106 3.18.3 Shortlisted Problems 112 3.19 IMO 1977 114 3.19.1 Contest Problems 114 3.19.2 Longlisted Problems 114 3.19.3 Shortlisted Problems 120 3.20 IMO 1978 123 3.20.1 Contest Problems 123 3.20.2 Longlisted Problems 123 3.20.3 Shortlisted Problems 128 3.21 IMO 1979 131 3.21.1 Contest Problems 131 3.21.2 Longlisted Problems 132 3.21.3 Shortlisted Problems 139 3.22 IMO 1981 143 3.22.1 Contest Problems 143 3.22.2 Shortlisted Problems 144 3.23 IMO 1982 147 3.23.1 Contest Problems 147 3.23.2 Longlisted Problems 148 3.23.3 Shortlisted Problems 153 3.24 IMO 1983 157 3.24.1 Contest Problems 157 3.24.2 Longlisted Problems 157 3.24.3 Shortlisted Problems 165 3.25 IMO 1984 169 3.25.1 Contest Problems 169 3.25.2 Longlisted Problems 169 3.25.3 Shortlisted Problems 176 3.26 IMO 1985 180 3.26.1 Contest Problems 180 3.26.2 Longlisted Problems 180 3.26.3 Shortlisted Problems 190 3.27 IMO 1986 193 3.27.1 Contest Problems 193 3.27.2 Longlisted Problems 194 3.27.3 Shortlisted Problems 201 3.28 IMO 1987 204 3.28.1 Contest Problems 204 724 Solutions Denote by Ci the midpoint of the side Ai Ai+1 , i = 1, , n − By definition C1 = B1 and Cn−1 = Bn−1 Since A1 Ai Ai+1 An is an isosceles trapezoid with A1 Ai Ai+1 An for i = 2, , n − 2, it follows from the lemma that ∠A1 Bi An = ∠A1 Ci An for all i A3 B3 A4 The sum in consideration thus equals ∠A1 C1 An +∠A1 C2 An +· · ·+ ∠A1 Cn−1 An Moreover, the trianC2 B2 gles A1 Ci An and An+2−i C1 An+1−i are congruent (a rotation about A2 the center of the n-gon carries the first one to the second), and conseBn−1 B1 quently ∠A1 Ci An = ∠An+2−i C1 An+1−i for i = 2, , n − A1 An Hence Σ = ∠A1 C1 An +∠An C1 An−1 +· · ·+∠A3 C1 A2 = ∠A1 C1 A2 = 180◦ 21 Let ABC be the triangle of maximum area S contained in P (it exists because of compactness of P) Draw parallels to BC, CA, AB through A, B, C, respectively, and denote the triangle thus obtained by A1 B1 C1 (A ∈ B1 C1 , etc.) Since each triangle with vertices in P has area at most S, the entire polygon P is contained in A1 B1 C1 Next, draw lines of support of P parallel to BC, CA, AB and not intersecting the triangle ABC They determine a convex hexagon Ua Va Ub Vb Uc Vc containing P, with Vb , Uc ∈ B1 C1 , Vc , Ua ∈ C1 A1 , Va , Ub ∈ A1 B1 Each of the line segments Ua Va , Ub Vb , Uc Vc contains points of P Choose such points A0 , B0 , C0 on Ua Va , Ub Vb , Uc Vc , respectively The convex hexagon AC0 BA0 CB0 is contained in P, because the latter is convex We prove that AC0 BA0 CB0 has area at least 3/4 the area of P Let x, y, z denote the areas of triangles Ua BC, Ub CA, and Uc AB Then S1 = SAC0 BA0 CB0 = S + x + y + z On the other hand, the triangle A1 Ua Va is similar to A1 BC with similitude τ = (S − x)/S, and hence its area is τ S = (S − x)2 /S Thus the area of quadrilateral Ua Va CB is S − (S − x)2 /S = 2z − z /S Analogous formulas hold for quadrilaterals Ub Vb AC and Uc Vc BA Therefore SP ≤ SUa Va Ub Vb Uc Vc = S + SUa Va CB + SUb Vb AC + SUc Vc BA x2 + y + z = S + 2(x + y + z) − S (x + y + z)2 ≤ S + 2(x + y + z) − 3S Now 4S1 −3SP ≥= S−2(x+y+z)+(x+y+z)2 /S = (S−x−y−z)2 /S ≥ 0; i.e., S1 ≥ 3SP /4, as claimed 22 The proof uses the following observation: 4.45 Shortlisted Problems 2004 725 Lemma In a triangle ABC, let K, L be the midpoints of the sides AC, AB, respectively, and let the incircle of the triangle touch BC, CA at D, E, respectively Then the lines KL and DE intersect on the bisector of the angle ABC Proof Let the bisector b of ∠ABC meet DE at T One can assume that AB = BC, or else T ≡ K ∈ KL Note that the incenter I of ABC is between B and T , and also T = E From the triangles BDT and DEC we obtain ∠IT D = α/2 = ∠IAE, which implies that A, I, T, E are concyclic Then ∠AT B = ∠AEI = 90◦ Thus L is the circumcenter of AT B from which ∠LT B = ∠LBT = ∠T BC ⇒ LT BC ⇒ T ∈ KL, which is what we were supposed to prove Let the incircles of ABX and ACX touch BX at D and F , respectively, and let them touch AX at E and G, respectively Clearly, DE and F G are parallel If the line P Q intersects BX and AX at M and N , respectively, then M D2 = M P · M Q = M F , i.e., M D = M F and analogously N E = N G It follows that P Q is parallel to DE and F G and equidistant from them The midpoints of AB, AC, and AX lie on the same line m, parallel to BC Applying the lemma to ABX, we conclude that DE passes through the common point U of m and the bisector of ∠ABX Analogously, F G passes through the common point V of m and the bisector of ∠ACX Therefore P Q passes through the midpoint W of the line segment U V Since U, V not depend on X, neither does W 23 To start with, note that point N is uniquely determined by the imposed properties Indeed, f (X) = AX/BX is a monotone function on both arcs AB of the circumcircle of ABM E Denote by P and Q respectively the second points of intersection of the line EF with the circumcircles C of ABE and ABF The probM lem is equivalent to showing that N ∈ P Q In fact, we shall prove D F that N coincides with the midpoint N of segment P Q B The cyclic quadrilaterals AP BE, AQBF , and ABCD yield ∠AP Q = A 180◦ − ∠AP E = 180◦ − ∠ABE = P ∠ADC and ∠AQP = ∠AQF = Ω ∠ABF = ∠ACD It follows that N AP Q ∼ ADC, and conseQ quently ANP ∼ AM D Analogously BN P ∼ BM C Therefore AN /AM = P Q/DC = BN /BM , i.e., AN /BN = AM/BM Moreover, ∠AN B = ∠AN P + ∠P NB = ∠AM D + ∠BM C = 180◦ − ∠AM B, which means that point N lies on 726 Solutions the circumcircle of AM B By the uniqueness of N , we conclude that N ≡ N , which completes the solution 24 Setting m = an we reduce the given equation to m/τ (m) = a Let us show that for a = pp−1 the above equation has no solutions in N if p > is a prime Assume to the contrary that m ∈ N is such that m = pp−1 τ (m) Then pp−1 | m, so we may set m = pα k, where α, k ∈ N, αr α ≥ p − 1, and p k Let k = pα · · · pr be the decomposition of k into primes Then τ (k) = (α1 + 1) · · · (αr + 1) and τ (m) = (α + 1)τ (k) Our equation becomes (1) pα−p+1 k = (α + 1)τ (k) We observe that α = p − 1: otherwise the RHS would be divisible by p and the LHS would not be so It follows that α ≥ p, which also easily implies p (α + 1) that pα−p+1 ≥ p+1 Furthermore, since α + cannot be divisible by pα−p+1 for any α ≥ p, it follows that p | τ (k) Thus if p | τ (k), then at least one αi +1 is divisible by p and consequently αi ≥ p−1 for some i Hence k ≥ But then we have pα−p+1 k ≥ α pi i αi +1 τ (k) ≥ 2p−1 p τ (k) 2p−1 p (α + 1) · τ (k) > (α + 1)τ (k), p+1 p contradicting (1) Therefore (1) has no solutions in N Remark There are many other values of a for which the considered equation has no solutions in N: for example, a = 6p for a prime p ≥ 25 Let n be a natural number For each k = 1, 2, , n, the number (k, n) is a divisor of n Consider any divisor d of n If (k, n) = n/d, then k = nl/d for some l ∈ N, and (k, n) = (l, d)n/d, which implies that l is coprime to d and l ≤ d It follows that (k, n) is equal to n/d for exactly ϕ(d) natural numbers k ≤ n Therefore n ψ(n) = (k, n) = k=1 ϕ(d) d|n n =n d d|n ϕ(d) d (1) (a) Let n, m be coprime Then each divisor f of mn can be uniquely expressed as f = de, where d | n and e | m We now have by (1) ψ(mn) = mn f |mn ϕ(f ) = mn f = mn d|n, e|m = ψ(m)ψ(n) d|n, e|m ϕ(de) de ⎛ ϕ(d) ϕ(e) ⎝ = n d e ⎞⎛ d|n ϕ(d) ⎠ ⎝ m d ⎞ e|m ϕ(e) ⎠ e 4.45 Shortlisted Problems 2004 727 (b) Let n = pk , where p is a prime and k a positive integer According to (1), ψ(n) = n k i=0 ϕ(pi ) k(p − 1) =1+ pi p Setting p = and k = 2(a − 1) we obtain ψ(n) = an for n = 22(a−1) (c) We note that ψ(pp ) = pp+1 if p is a prime Hence, if a has an odd prime factor p and a1 = a/p, then x = pp 22a1 −2 is a solution of ψ(x) = ax different from x = 22a−2 Now assume that a = 2k for some k ∈ N Suppose x = 2α y is a positive integer such that ψ(x) = 2k x Then 2α+k y = ψ(x) = ψ(2α )ψ(y) = (α+2)2α−1 ψ(y), i.e., 2k+1 y = (α+2)ψ(y) We notice that for each odd y, ψ(y) is (by definition) the sum of an odd number of odd summands and therefore odd It follows that ψ(y) | y On the other hand, ψ(y) > y for y > 1, so we must have y = Consequently α = 2k+1 −2 = 2a−2, giving us the unique solution x = 22a−2 Thus ψ(x) = ax has a unique solution if and only if a is a power of 26 For m = n = we obtain that f (1)2 + f (1) divides (12 + 1)2 = 4, from which we find that f (1) = Next, we show that f (p−1) = p−1 for each prime p By the hypothesis for m = and n = p − 1, f (p − 1) + divides p2 , so f (p − 1) equals either p − or p2 − If f (p − 1) = p2 − 1, then f (1) + f (p − 1)2 = p4 − 2p2 + divides (1+(p−1)2 )2 < p4 −2p2 +2, giving a contradiction Hence f (p−1) = p−1 Let us now consider an arbitrary n ∈ N By the hypothesis for m = p − 1, A = f (n) + (p − 1)2 divides (n + (p − 1)2 )2 ≡ (n − f (n))2 (mod A), and hence A divides (n−f (n))2 for any prime p Taking p large enough, we can obtain A to be greater than (n−f (n))2 , which implies that (n−f (n))2 = 0, i.e., f (n) = n for every n 27 Set a = and assume that b ∈ N is such that b2 ≡ b + (mod m) An easy induction gives us xn ≡ bn (mod m) for all n ∈ N0 Moreover, b is obviously coprime to m, and hence each xn is coprime to m It remains to show the existence of b The congruence b2 ≡ b + (mod m) is equivalent to (2b − 1)2 ≡ (mod m) Taking 2b − ≡ 2k, i.e., b ≡ 2k + k − (mod m), does the job Remark A desired b exists whenever is a quadratic residue modulo m, in particular, when m is a prime of the form 10k ± 28 If n is divisible by 20, then every multiple of n has two last digits even and hence it is not alternate We shall show that any other n has an alternate multiple (i) Let n be coprime to 10 For each k there exists a number Ak (n) = mk −1 10 010 01 01 = 1010k −1 (m ∈ N) that is divisible by n (by Euler’s theorem, choose m = ϕ[n(10k − 1)]) In particular, A2 (n) is alternate 728 Solutions (ii) Let n = · 5r · n1 , where r ≥ and (n1 , 10) = We shall show by induction that, for each k, there exists an alternative k-digit odd number Mk that is divisible by 5k Choosing the number 10A2r (n1 )M2r will then solve this case, since it is clearly alternate and divisible by n We can trivially choose M1 = Let there be given an alternate r-digit multiple Mr of 5r , and let c ∈ {0, 1, 2, 3, 4} be such that Mr /5r ≡ −c · 2r (mod 5) Then the (r + 1)digit numbers Mr + c · 10r and Mr + (5 + c) · 10r are respectively equal to 5r (Mr /5r + 2r · c) and 5r (Mr /5r + 2r · c + · 2r ), and hence they are divisible by 5r+1 and exactly one of them is alternate: we set it to be Mr+1 (iii) Let n = 2r ·n1 , where r ≥ and (n1 , 10) = We show that there exists an alternate 2r-digit number Nr that is divisible by 22r+1 Choosing the number A2r (n1 )Nr will then solve this case We choose N1 = 16, and given Nr , we can prove that one of Nr + m · 102r , for m ∈ {10, 12, 14, 16}, is divisible by 22r+3 and therefore suitable for Nr+1 Indeed, for Nr = 22r+1 d we have Nr + m · 102r = 22r+1 (d + 5r m/2) and d + 5r m/2 ≡ (mod 4) has a solution m/2 ∈ {5, 6, 7, 8} for each d and r Remark The idea is essentially the same as in (SL94-24) 29 Let Sn = {x ∈ N | x ≤ n, n | x2 − 1} It is easy to check that Pn ≡ (mod n) for n = and Pn ≡ −1 (mod n) for n ∈ {3, 4}, so from now on we assume n > We note that if x ∈ Sn , then also n−x ∈ Sn and (x, n) = Thus Sn splits into pairs {x, n − x}, where x ∈ Sn and x ≤ n/2 In each of these pairs the product of elements gives remainder −1 upon division by n Therefore Pn ≡ (−1)m , where Sn has 2m elements It remains to find the parity of m Suppose first that n > is divisible by Whenever x ∈ Sn , the numbers |n/2−x|, n−x, n−|n/2−x| also belong to Sn (indeed, n | (n/2−x)2 −1 = n2 /4 − nx + x2 − because n | n2 /4, etc.) In this way the set Sn splits into four-element subsets {x, n/2 − x, n/2 + x, n − x}, where x ∈ Sn and x < n/4 (elements of these subsets are different for x = n/4, and n/4 doesn’t belong to Sn for n > 4) Therefore m = |Sn |/2 is even and Pn ≡ (mod m) Now let n be odd If n | x2 − = (x − 1)(x + 1), then there exist natural numbers a, b such that ab = n, a | x − 1, b | x + Obviously a and b are coprime Conversely, given any odd a, b ∈ N such that (a, b) = and ab = n, by the Chinese remainder theorem there exists x ∈ {1, 2, , n−1} such that a | x − and b | x + This gives a bijection between all ordered pairs (a, b) with ab = n and (a, b) = and the elements of Sn Now if αk n = pα · · · pk is the decomposition of n into primes, the number of pairs αi i (a, b) is equal to 2k (since for every i, either pα i | a or pi | b), and hence 4.45 Shortlisted Problems 2004 729 m = 2k−1 Thus Pn ≡ −1 (mod n) if n is a power of an odd prime, and Pn ≡ otherwise Finally, let n be even but not divisible by Then x ∈ Sn if and only if x or n − x belongs to Sn/2 and x is odd Since n/2 is odd, for each x ∈ Sn/2 either x or x + n/2 belongs to Sn , and by the case of n odd we have Sn ≡ ±1 (mod n/2), depending on whether or not n/2 is a power of a prime Since Sn is odd, it follows that Pn ≡ −1 (mod n) if n/2 is a power of a prime, and Pn ≡ otherwise Second solution Obviously Sn is closed under multiplication modulo n This implies that Sn with multiplication modulo n is a subgroup of Zn , and therefore there exist elements a1 = −1, a2 , , ak ∈ Sn that generate Sn In other words, since the are of order two, Sn consists of products k i∈A , where A runs over all subsets of {1, 2, , k} Thus Sn has 2k−1 elements, and the product of these elements equals Pn ≡ (a1 a2 · · · ak ) (mod n) Since a2i ≡ (mod n), it follows that Pn ≡ if k ≥ 2, i.e., if |Sn | > Otherwise Pn ≡ −1 (mod n) We note that |Sn | > is equivalent to the existence of a ∈ Sn with < a < n − It is easy to find that such an a exists if and only if neither of n, n/2 is a power of an odd prime 30 We shall denote by k the given circle with diameter pn Let A, B be lattice points (i.e., points with integer coordinates) We shall denote by µ(AB) the exponent of the highest power of p that divides the integer AB We observe that if S is the area of a triangle ABC where A, B, C are lattice points, then 2S is an integer According to Heron’s formula and the formula for the circumradius, a triangle ABC whose circumcenter has diameter pn satisfies 2AB BC + 2BC CA2 + 2CA2 AB − AB − BC − CA4 = 16S (1) and AB · BC · CA2 = (2S)2 p2n (2) Lemma Let A, B, and C be lattice points on k If none of AB , BC , CA2 is divisible by pn+1 , then µ(AB), µ(BC), µ(CA) are 0, n, n in some order Proof Let k = min{µ(AB), µ(BC), µ(CA)} By (1), (2S)2 is divisible by p2k Together with (2), this gives us µ(AB) + µ(BC) + µ(CA) = 2k + 2n On the other hand, if none of AB , BC , CA2 is divisible by pn+1 , then µ(AB) + µ(BC) + µ(CA) ≤ k + 2n Therefore k = and the remaining two of µ(AB), µ(BC), µ(CA) are equal to n Lemma Among every four lattice points on k, there exist two, say M, N , such that µ(M N ) ≥ n + Proof Assume that this doesn’t hold for some points A, B, C, D on k By Lemma 1, µ for some of the segments AB, AC, , CD is 0, say µ(AC) = It easily follows by Lemma that then µ(BD) = and µ(AB) = µ(BC) = µ(CD) = µ(DA) = n Let a, b, c, d, e, f ∈ N be 730 Solutions such that AB = pn a, BC = pn b, CD2 = pn c, DA2 = pn d, AC = e, √ √ √ BD2 = f By Ptolemy’s theorem we have ef = pn ac + bd √ √ √ ac + bd = ac + bd + abcd Taking squares, we get that pef 2n = is rational and hence an integer It follows that ef is divisible by p2n , a contradiction Now we consider eight lattice points A1 , A2 , , A8 on k We color each segment Ai Aj red if µ(Ai Aj ) > n and black otherwise, and thus obtain a graph G The degree of a point X will be the number of red segments with an endpoint in X We distinguish three cases: (i) There is a point, say A8 , whose degree is at most We may suppose w.l.o.g that A8 A7 is red and A8 A1 , , A8 A6 black By a well-known fact, the segments joining vertices A1 , A2 , , A6 determine either a red triangle, in which case there is nothing to prove, or a black triangle, say A1 A2 A3 But in the latter case the four points A1 , A2 , A3 , A8 not determine any red segment, a contradiction to Lemma (ii) All points have degree Then the set of red segments partitions into cycles If one of these cycles has length 3, then the proof is complete If all the cycles have length at least 4, then we have two possibilities: two 4-cycles, say A1 A2 A3 A4 and A5 A6 A7 A8 , or one 8-cycle, A1 A2 A8 In both cases, the four points A1 , A3 , A5 , A7 not determine any red segment, a contradiction (iii) There is a point of degree at least 3, say A1 Suppose that A1 A2 , A1 A3 , and A1 A4 are red We claim that A2 , A3 , A4 determine at least one red segment, which will complete the solution If not, by Lemma 1, µ(A2 A3 ), µ(A3 A4 ), µ(A4 A2 ) are n, n, in some order Assuming w.l.o.g that µ(A2 A3 ) = 0, denote by S the area of triangle A1 A2 A3 Now by formula (1), 2S is not divisible by p On the other hand, since µ(A1 A2 ) ≥ n + and µ(A1 A3 ) ≥ n + 1, it follows from (2) that 2S is divisible by p, a contradiction A Notation and Abbreviations A.1 Notation We assume familiarity with standard elementary notation of set theory, algebra, logic, geometry (including vectors), analysis, number theory (including divisibility and congruences), and combinatorics We use this notation liberally We assume familiarity with the basic elements of the game of chess (the movement of pieces and the coloring of the board) The following is notation that deserves additional clarification ◦ B(A, B, C), A − B − C: indicates the relation of betweenness, i.e., that B is between A and C (this automatically means that A, B, C are different collinear points) ◦ A = l1 ∩ l2 : indicates that A is the intersection point of the lines l1 and l2 ◦ AB: line through A and B, segment AB, length of segment AB (depending on context) ◦ [AB: ray starting in A and containing B ◦ (AB: ray starting in A and containing B, but without the point A ◦ (AB): open interval AB, set of points between A and B ◦ [AB]: closed interval AB, segment AB, (AB) ∪ {A, B} ◦ (AB]: semiopen interval AB, closed at B and open at A, (AB) ∪ {B} The same bracket notation is applied to real numbers, e.g., [a, b) = {x | a ≤ x < b} ◦ ABC: plane determined by points A, B, C, triangle ABC ( ABC) (depending on context) ◦ [AB, C: half-plane consisting of line AB and all points in the plane on the same side of AB as C 732 A Notation and Abbreviations ◦ (AB, C: [AB, C without the line AB → → − − → − → → → ◦ − a, b , − a · b : scalar product of − a and b ◦ a, b, c, α, β, γ: the respective sides and angles of triangle ABC (unless otherwise indicated) ◦ k(O, r): circle k with center O and radius r ◦ d(A, p): distance from point A to line p ◦ SA1 A2 An : area of n-gon A1 A2 An (special case for n = 3, SABC : area of ABC) ◦ N, Z, Q, R, C: the sets of natural, integer, rational, real, complex numbers (respectively) ◦ Zn : the ring of residues modulo n, n ∈ N ◦ Zp : the field of residues modulo p, p being prime ◦ Z[x], R[x]: the rings of polynomials in x with integer and real coefficients respectively ◦ R∗ : the set of nonzero elements of a ring R ◦ R[α], R(α), where α is a root of a quadratic polynomial in R[x]: {a + bα | a, b ∈ R} / X ◦ X0 : X ∪ {0} for X such that ∈ ◦ X + , X − , aX + b, aX + bY : {x | x ∈ X, x > 0}, {x | x ∈ X, x < 0}, {ax + b | x ∈ X}, {ax + by | x ∈ X, y ∈ Y } (respectively) for X, Y ⊆ R, a, b ∈ R ◦ [x], x : the greatest integer smaller than or equal to x ◦ x : the smallest integer greater than or equal to x The following is notation simultaneously used in different concepts (depending on context) ◦ |AB|, |x|, |S|: the distance between two points AB, the absolute value of the number x, the number of elements of the set S (respectively) ◦ (x, y), (m, n), (a, b): (ordered) pair x and y, the greatest common divisor of integers m and n, the open interval between real numbers a and b (respectively) A.2 Abbreviations We tried to avoid using nonstandard notations and abbreviations as much as possible However, one nonstandard abbreviation stood out as particularly convenient: A.2 Abbreviations 733 ◦ w.l.o.g.: without loss of generality Other abbreviations include: ◦ RHS: right-hand side (of a given equation) ◦ LHS: left-hand side (of a given equation) ◦ QM, AM, GM, HM: the quadratic mean, the arithmetic mean, the geometric mean, the harmonic mean (respectively) ◦ gcd, lcm: greatest common divisor, least common multiple (respectively) ◦ i.e.: in other words ◦ e.g.: for example B Codes of the Countries of Origin ARG ARM AUS AUT BEL BLR BRA BUL CAN CHN COL CUB CYP CZE CZS EST FIN FRA FRG GBR GDR GEO GER Argentina Armenia Australia Austria Belgium Belarus Brazil Bulgaria Canada China Colombia Cuba Cyprus Czech Republic Czechoslovakia Estonia Finland France Germany, FR United Kingdom Germany, DR Georgia Germany GRE HKG HUN ICE INA IND IRE IRN ISR ITA JAP KAZ KOR KUW LAT LIT LUX MCD MEX MON MOR NET NOR NZL Greece Hong Kong Hungary Iceland Indonesia India Ireland Iran Israel Italy Japan Kazakhstan Korea, South Kuwait Latvia Lithuania Luxembourg Macedonia Mexico Mongolia Morocco Netherlands Norway New Zealand PHI POL POR PRK PUR ROM RUS SAF SIN SLO SMN SPA SWE THA TUN TUR TWN UKR USA USS UZB VIE YUG Philippines Poland Portugal Korea, North Puerto Rico Romania Russia South Africa Singapore Slovenia Serbia and Montenegro Spain Sweden Thailand Tunisia Turkey Taiwan Ukraine United States Soviet Union Uzbekistan Vietnam Yugoslavia References M Aassila, 300 D´efis Math´ematiques, Ellipses, 2001 M Aigner, G.M Ziegler, Proofs from THE BOOK, Springer; 3rd edition, 2003 G.L Alexanderson, L.F Klosinski, L.C Larson, The William Lowell Putnam Mathematical Competition, Problems and Solutions: 1965-1984, The Mathematical Association of America, 1985 T Andreescu, D Andrica, 360 Problems for Mathematical Contests, GIL Publishing House, Zal˘ au, 2003 T Andreescu, D Andrica, An Introduction to the Diophantine Equations, GIL Publishing House, 2002 T Andreescu, D Andrica, Complex Numbers from A to Z, Birkh¨ auser, Boston, 2005 T Andreescu, B Enescu, Mathematical Treasures, Birkh¨ auser, Boston, 2003 T Andreescu, Z Feng, 102 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edition, Cambridge University Press, 2001 74 I.M Vinogradov, Elements of Number Theory, Dover Publications, 2003 75 H.S Wilf, Generatingfunctionology, Academic Press, Inc.; 2nd edition, 1994 76 A.M Yaglom, I.M Yaglom, Challenging Mathematical Problems with Elementary Solutions, Dover Publications, 1987 77 I.M Yaglom, Geometric Transformations, Vols I, II, III, The Mathematical Association of America (MAA), 1962, 1968, 1973 78 P Zeitz, The Art and Craft of Problem Solving, Wiley; International Student edition, 1999 ... proposed for the IMO It consists of all problems selected for the IMO competitions, shortlisted problems from the 10th IMO 1.2 The IMO Compendium and from the 12th through 44th IMOs, and longlisted... presented to the IMO jury, consisting of all the team leaders From the short-listed problems the jury chooses six problems for the IMO Apart from its mathematical and competitive side, the IMO is also... longlisted problems from nineteen IMOs We not have shortlisted problems from the 9th and the 11th IMOs, and we could not discover whether competition problems at those two IMOs were selected from the