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51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad51st International Mathematical Olympiad Astana, Kazakhstan 2010 Shortlisted Problems with Solutions www facebook com7khmer fb Entertainment And Knowledge www facebook com7khmer fb Entertainment And.
51st International Mathematical Olympiad Astana, Kazakhstan 2010 Shortlisted Problems with Solutions www.facebook.com/7khmer fb : Entertainment And Knowledge www.facebook.com/7khmer fb : Entertainment And Knowledge Contents Note of Confidentiality Contributing Countries & Problem Selection Committee Algebra Problem Problem Problem Problem Problem Problem Problem Problem 7 10 12 13 15 17 19 Combinatorics Problem C1 Problem C2 Problem C3 Problem C4 Problem C41 Problem C5 Problem C6 Problem C7 23 23 26 28 30 30 32 35 38 Geometry Problem Problem Problem Problem Problem Problem Problem Problem A1 A2 A3 A4 A5 A6 A7 A8 G1 G2 G3 G4 G5 G6 G61 G7 44 44 46 50 52 54 56 56 60 Number Theory Problem N1 Problem N11 Problem N2 Problem N3 Problem N4 Problem N5 Problem N6 64 64 64 66 68 70 71 72 www.facebook.com/7khmer fb : Entertainment And Knowledge www.facebook.com/7khmer fb : Entertainment And Knowledge Note of Confidentiality The Shortlisted Problems should be kept strictly confidential until IMO 2011 Contributing Countries The Organizing Committee and the Problem Selection Committee of IMO 2010 thank the following 42 countries for contributing 158 problem proposals Armenia, Australia, Austria, Bulgaria, Canada, Columbia, Croatia, Cyprus, Estonia, Finland, France, Georgia, Germany, Greece, Hong Kong, Hungary, India, Indonesia, Iran, Ireland, Japan, Korea (North), Korea (South), Luxembourg, Mongolia, Netherlands, Pakistan, Panama, Poland, Romania, Russia, Saudi Arabia, Serbia, Slovakia, Slovenia, Switzerland, Thailand, Turkey, Ukraine, United Kingdom, United States of America, Uzbekistan Problem Selection Committee Yerzhan Baissalov Ilya Bogdanov G´eza K´os Nairi Sedrakyan Damir Yeliussizov Kuat Yessenov www.facebook.com/7khmer fb : Entertainment And Knowledge www.facebook.com/7khmer fb : Entertainment And Knowledge Algebra A1 Determine all functions f : R Ñ R such that the equality f prxsy q f pxqrf py qs holds for all x, y (1) P R Here, by rxs we denote the greatest integer not exceeding x Answer f pxq const C, where C (France) or ă C Solution First, setting x in (1) we get f p0q f p0qrf py qs (2) for all y P R Now, two cases are possible Case Assume that f p0q Then from (2) we conclude that rf py qs for all y P R Therefore, equation (1) becomes f prxsy q f pxq, and substituting y we have f pxq f p0q C Finally, from rf py qs rC s we obtain that ă C Case Now we have f p0q Here we consider two subcases Subcase 2a Suppose that there exists α such that f pαq Then setting x α in (1) we obtain f p0q f pαqrf py qs for all y P R Hence, rf py qs for all y P R Finally, substituting x in (1) provides f py q for all y P R, thus contradicting the condition f pαq Subcase 2b Conversely, we have f pq for all ă Consider any real z; there z exists an integer N such that α P r0, 1q (one may set N rzs if z © and N rzs N otherwise) Now, from (1) we get f pz q f prN sαq f pN qrf pαqs for all z P R Finally, a straightforward check shows that all the obtained functions satisfy (1) Solution Assume that rf py qs for some y; then the substitution x provides f py q f p1qrf py qs Hence, if rf py qs for all y, then f py q for all y This function obviously satisfies the problem conditions So we are left to consider the case when rf paqs for some a Then we have f prxsaq f pxqrf paqs, or f pxq f prxsaq rf paqs (3) This means that f px1 q f px2 q whenever rx1 s rx2 s, hence f pxq f prxsq, and we may assume that a is an integer Now we have f paq f 2a 12 f p2aq f 12 f p2aqrf p0qs; this implies rf p0qs 0, so we may even assume that a Therefore equation (3) provides f pxq f p0q rf p0qs C0 for each x Now, condition (1) becomes equivalent to the equation C exactly when rC s www.facebook.com/7khmer fb : Entertainment And Knowledge C rC s which holds A2 Let the real numbers a, b, c, d satisfy the relations a b c d and a2 b2 c2 d2 12 Prove that 36 ă 4pa3 b3 d3 q pa4 c3 b4 d4 q ă 48 c4 (Ukraine) Solution Observe that 4pa3 b3 c3 d3 q pa4 c4 d4 q pa 1q4 pb 1q4 pc 1q4 6pa2 b2 c2 d2 q 4pa b c dq b4 pd 1q4 pa 1q4 pb 1q4 pc 1q4 pd 1q4 52 Now, introducing x a 1, y b 1, z c 1, t d 1, we need to prove the inequalities 16 © x4 y z t4 © 4, under the constraint x2 y2 z2 t2 pa2 b2 d q pa c2 b dq c 44 (1) (we will not use the value of x y z t though it can be found) Now the rightmost inequality in (1) follows from the power mean inequality: x y z 4 t p x2 © y2 t2 q2 z2 For the other one, expanding the brackets we note that px2 y2 z2 t2 q2 px4 y4 t4 q z4 q, where q is a nonnegative number, so x4 y4 z4 t4 ă px2 y2 t2 q2 z2 16, and we are done Comment The estimates are sharp; the lower and upper bounds are attained at p3, 1, 1, 1q and p0, 2, 2, 2q, respectively Comment After the change of variables, one can finish the solution in several different ways The latter estimate, for instance, it can be performed by moving the variables – since we need only the second of the two shifted conditions Solution First, we claim that ă a, b, c, d ă Actually, we have a c d, b a2 b2 c2 12 d2, hence the power mean inequality a2 rewrites as 12 d2 b2 © p6 3 dq c2 www.facebook.com/7khmer fb : Entertainment And Knowledge © pa ðđ b cq2 2dpd 3q ă 0, which implies the desired inequalities for d; since the conditions are symmetric, we also have the same estimate for the other variables Now, to prove the rightmost inequality, we use the obvious inequality x2 px 2q2 © for each real x; this inequality rewrites as 4x3 x4 ă 4x2 It follows that p4a3 a4 q p4b3 b4 q p4c3 c4q p4d3 d4q ă 4pa2 b2 d2 q 48, c2 as desired Now we prove the leftmost inequality in an analogous way For each x P r0, 3s, we have px 1qpx 1q2px 3q ă which is equivalent to 4x3 x4 © 2x2 4x This implies that p4a3 a4 q p4b3 b4 q p4c3 c4 q p4d3 d4q © 2pa2 b2 c2 d2 q pa b c dq 12 36, as desired Comment It is easy to guess the extremal points p0, 2, 2, 2q and p3, 1, 1, 1q for this inequality This provides a method of finding the polynomials used in Solution Namely, these polynomials should have the form x4 4x3 ax2 bx c; moreover, the former polynomial should have roots at (with an even multiplicity) and 0, while the latter should have roots at (with an even multiplicity) and These conditions determine the polynomials uniquely Solution First, expanding 48 4pa2 we have a4 b4 c4 d4 48 pa4 b2 c2 d2 q and applying the AM–GM inequality, pb4 4b2 q pc4 4c2q pd4 4d2 q ? ? ? ? © a4 4a2 b4 4b2 c4 4c2 d4 4d2 4p|a3| |b3| |c3| |d3|q © 4pa3 b3 c3 d3q, 4a2 q which establishes the rightmost inequality To prove the leftmost inequality, we first show that a, b, c, d P r0, 3s as in the previous solution Moreover, we can assume that ă a ă b ă c ă d Then we have a b ă b c ă pb c dq ă 23 Next, we show that 4b b2 ¨ 4c c2 Actually, this inequality rewrites as pc bqpb c 4q ă 0, which follows from the previous estimate The inequality 4a a2 ă 4b b2 can be proved analogously Further, the inequalities a ă b ă c together with 4a a2 ¨ 4b b2 ¨ 4c c2 allow us to apply the Chebyshev inequality obtaining a2 p4a a2 q b2 p4b b2 q c2 p4c c2 q © pa2 b2 c2 q 4pa b cq pa2 b2 c2 q 2 p12 d qp4p6 3dq p12 d qq This implies that 2 p4a3 a4 q p4b3 b4 q p4c3 c4q p4d3 d4q © p12 d qpd3 4d 144 48d 16d 4d 36 p3 dqpd 1qpd2 3q 12q 4d3 d4 (2) 3 Finally, we have d2 © 14 pa2 b2 c2 d2 q (which implies d ¡ 1); so, the expression p3 dqpd 1qpd2 3q in the right-hand part of (2) is nonnegative, and the desired inequality is proved Comment The rightmost inequality is easier than the leftmost one In particular, Solutions and show that only the condition a2 b2 c2 d2 12 is needed for the former one www.facebook.com/7khmer fb : Entertainment And Knowledge .. .51st International Mathematical Olympiad Astana, Kazakhstan 2010 Shortlisted Problems with Solutions www.facebook.com/7khmer