andreescu contests around the world 2000-2001
Mathematical Olympiads 2000–2001 Problems and Solutions From Around the World Copyright Information Mathematical Olympiads 2000–2001 Problems and Solutions From Around the World Edited by Titu Andreescu, Zuming Feng, and George Lee, Jr. Published and distributed by The Mathematical Association of America MAA PROBLEM BOOKS SERIES INFORMATION Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xi 1 2000 National Contests: Problems and Solutions 1 1.1 Belarus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Bulgaria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.3 Canada . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 1.4 China . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .27 1.5 Czech and Slovak Republics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 1.6 Estonia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 1.7 Hungary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 1.8 India . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 1.9 Iran . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 1.10 Israel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 1.11 Italy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 1.12 Japan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .69 1.13 Korea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .74 1.14 Mongolia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 1.15 Poland . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .85 1.16 Romania . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .91 1.17 Russia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 1.18 Taiwan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 1.19 Turkey . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 v vi 1.20 United Kingdom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 1.21 United State s of Am erica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 1.22 Vietnam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 2 2000 Regional Contests: Problems and Solutions 163 2.1 Asian Pacific Mathem atical Olympiad . . . . . . . . . . . . . . . . . 164 2.2 Austrian-Polish Mathematics Competition . . . . . . . . . . . 170 2.3 Balkan Mathematical Olympiad . . . . . . . . . . . . . . . . . . . . . . . 175 2.4 Mediterranean Mathematical Olympiad . . . . . . . . . . . . . . . . 179 2.5 St. Petersburg City Mathematical Olympiad (Russia) . . 182 3 2001 National Contests: Problems . . . . . . . . . . . . 207 3.1 Belarus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 3.2 Bulgaria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 3.3 Canada . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 3.4 China . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 3.5 Czech and Slovak Republics . . . . . . . . . . . . . . . . . . . . . . . . . . . .215 3.6 Hungary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 3.7 India . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 3.8 Iran . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 3.9 Japan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 3.10 Korea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 3.11 Poland . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 3.12 Romania . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 3.13 Russia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226 3.14 Taiwan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 3.15 United State s of Am erica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 3.16 Vietnam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 4 2001 Regional Contests: Problems . . . . . . . . . . . . 235 4.1 Asian Pacific Mathem atical Olympiad . . . . . . . . . . . . . . . . . 236 4.2 Austrian-Polish Mathematics Competition . . . . . . . . . . . 237 4.3 Balkan Mathematical Olympiad . . . . . . . . . . . . . . . . . . . . . . . 238 vii 4.4 Baltic Mathematics Competition . . . . . . . . . . . . . . . . . . . . . . .239 4.5 Czech-Slovak-Polish Match . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 Classification of Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .256 Preface This book is a continuation of Mathematical Olympiads 1999–2000: Problems and Solutions From Around the World, published by the Mathematical Association of America. It contains solutions to the problems from 27 national and regional contests featured in the earlier book, together with selected problems (without solutions) from national and regional contests given during 2001. In many cases multiple solutions are provided in order to encourage students to compare different problem-solving strategies. This collection is intended as practice for the serious student who wishes to improve his or her performance on the USA Math Olympiad (USAMO) and Team Selection Test (TST). Some of the problems are comparable to the USAMO in that they came from national contests. Others are harder, as some countries first have a national Olympiad, and later one or more exams to select a team for the IMO. And some problems come from regional international contests (“m ini-IMOs”). Different nations have different mathematical cultures, so you will find some of these problems extremely hard and some rather eas y. We have tried to present a wide variety of problems, especially from those countries that have often done well at the IMO. Each contest has its own time limit. We have not furnished this information, because we have not always included complete exams. As a rule of thumb, most contests allow time ranging between one-half to one full hour per problem. The problems themselves should provide much enjoyment for all those fascinated by solving challenging mathematics questions. ix Acknowledgments Thanks to the following participants of the Mathematical Olympiad Summer Program who helped in preparing and pro ofreading so- lutions: Reid Barton, Steve Byrnes, Gabriel Carroll, Kamaldeep Gandhi, Stephen Guo, Luke Gustafson, Michael Hamburg, Daniel Jerison, Daniel Kane, Kiran Kedlaya, Ian Le, Tiankai Liu, Po-Ru Loh, Sean Markan, Alison Miller, Christopher Moore, Gregory Price, Michael Rothenberg, Inna Zakharevich, Tony Zhang, and Yan Zhang. Without the ir efforts this work would not have been possible. Titu Andreescu Zuming Feng George Lee, Jr. x [...]... 2000 terms b1 , b2 , , b2000 n By construction, each of the 2000 partial sums σn = i=1 bi (for 1 ≤ n ≤ 2000) equals one of the 2000 integers in [−999, 1000] Therefore, either σi = σj for some i < j or else σi = 0 for some i In the first case, the terms in the subsequence bi+1 , bi+2 , , bj sum to zero; in the second, the terms in the subsequence b1 , b2 , , bi sum to zero It follows that the. .. values of the balls in the box This product is initially i2000 = 1 If three balls were left in the box, none of them green, then the product of their values would be ±i, a contradiction Hence, if three balls remain, at least one is green, proving the claim in part (a) Furthermore, because no ball has value 1, the box must contain at least two balls at any time Therefore, the answer to the question in part... make the side opposite that corner horizontal In each of the n − 1 horizontal rows of two or more coins, choose two adjacent pennies and flip all the coins in that row; all the coins will then show tails Therefore, the desired initial arrangements are those in which the coin showing tails is in the corner 2000 National Contests: Problems 3 Problem 3 We are given triangle ABC with ∠C = π/2 Let M be the. .. two of its sides are on y = 0 and y = 1 Then the region R1 ∪ R2 − R1 ∩ R2 is the union of eight congruent triangular regions Let T and U be the left and right vertices of R2 on y = 1, and let V be the vertex of R1 above the line y = 1 Finally, let K and L be the uppermost points on the vertical 19 2000 National Contests: Problems sides of R2 that also belong to the boundary of R1 We have KT R ∼ = S... operations there are three balls left in the box Prove that at least one of them is a green ball (b) Is it possible after finitely many operations to have only one ball left in the box? 2000 National Contests: Problems 13 Solution: Assign the value i to each white ball, −i to each red ball, and −1 to each green ball A quick check verifies that the given operations preserve the product of the values of the balls... parity of the number of heads in the three corners is preserved If the coin showing tails is not in a corner, all three coins in the corners initially show heads, so there will always be an odd number of heads in the corners Hence, the three corners will never simultaneously show tails Conversely, if the coin showing tails is in a corner, we prove that we can make all the coins show tails Orient the triangle... bk+1,k+1 equal these numbers Each of these k + 1 numbers equals 1 or is the product of some terms of the sequence a1 , , ap−2 , and the induction is complete Consider the resulting list bp−1,1 , , bp−1,p−1 Exactly one of these numbers is congruent to 2 modulo p; because this number is not equal to 1, it is congruent to the product of some of the ak , as desired Problem 11 Let D be the midpoint... addition to these 83 unit squares, consider the proper unit squares whose left sides lie on lines of the form x = xi or x = xi − 1 Order all these unit squares S1 , , Sk from left to right, where the left side of Si lies on the line x = zi For i = 1, 2, , k − 1, at most one of the given points lies in the region determined by zi ≤ x < zi+1 , and at most one of the given points lies in the region... property: If we project any two of the points onto the corresponding side, the midpoint of the projected segment is the third point (b) Prove that triangle A1 B1 C1 is similar to the triangle formed by the medians of triangle ABC Solution: (a) We work backward by first assuming such a triangle exists Let T be the midpoint of A1 B1 By definition, C1 T ⊥ AB Let P be the centroid of triangle A1 B1 C1 ... (x + y + z)1/3 ≥ a + b + c 2000 National Contests: Problems 11 Cubing both sides and then dividing both sides by 3(x + y + z) gives the desired result Problem 11 Let P be the intersection point of the diagonals AC and BD of the convex quadrilateral ABCD in which AB = AC = BD Let O and I be the circumcenter and incenter of triangle ABP, respectively Prove that if O = I, then lines OI and CD are perpendicular . Mathematical Olympiads 2000 2001 Problems and Solutions From Around the World Copyright Information Mathematical Olympiads 2000 2001 Problems and Solutions From Around the World Edited. continuation of Mathematical Olympiads 1999 2000: Problems and Solutions From Around the World, published by the Mathematical Association of America. It contains solutions to the problems from. is the altitude to the base of isosceles triangle ACD, H is the midpoint of CD and hence lies in ω. By the Power of a Point Theorem, P H · HQ = CH · HD = CH 2 . Because CH is the altitude to the