New PDF release: Mathematical Olympiads 2000-2001: Problems and Solutions

By Titu Andreescu

ISBN-10: 088385810X

ISBN-13: 9780883858103

This quantity is a continuation of Mathematical Olympiads 1999-2000: difficulties and recommendations From around the globe, republishing enormous quantities of arithmetic difficulties and recommendations from that booklet in addition to chosen difficulties (without strategies) from nationwide and nearby contests in 2001. the gathering offers perform fabric for severe high-school point mathematicians who desire to arrange for america Math Olympiad (USAMO) and group choice attempt (TST). The newly additional 2001 difficulties have been contributed via dozens of nations together with Korea, Belarus, China, Poland, and Romania

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Additional resources for Mathematical Olympiads 2000-2001: Problems and Solutions from around the World

Example text

Solution: Set AB = c, BC = a, CA = b, ∠A = 2x, ∠B = 2y, ∠C = 2z. Then 0 < x ≤ y ≤ z and x + y + z = π/2. Let s denote the given quantity BC + CA − 2R − 2r = a + b − 2R − 2r. Using the well known formulas a b c a b c 2R = = = = = = , sin ∠A sin ∠B sin ∠C sin 2x sin 2y sin 2z ∠A ∠B ∠C r = 4R sin sin sin = 4R sin x sin y sin z, 2 2 2 we find that s = 2R(sin 2x + sin 2y − 1 − 4 sin x sin y sin z). Note that in a right triangle ABC with ∠C = π/2, we have 2R = c and 2r = a + b − c, implying that s = 0.

At } such that Φ(S) = S2 . Suppose that there existed X, Y ⊆ S such that X = Y, |X| = |Y | = m, and σ(X) = σ(Y ). Then we would also have Φ(X), Φ(Y ) ⊆ Φ(S) = S2 , Φ(X) = Φ(Y ), |Φ(X)| = m = |Φ(Y )|, and σ(X) − ma1 σ(Y ) − ma1 σ(Φ(X)) = = = σ(Φ(Y )). α 2 2α However, this is impossible because S2 is weakly m-efficient. Therefore, S is strongly (n + 3)-efficient as well. This completes the inductive step and the proof. 8 India Problem 1 Let ABC be a nonequilateral triangle. Suppose there is an interior point P such that the three cevians through P all have the same length λ where λ < min{AB, BC, CA}.

N. For each ej , we can choose ei such that f (ei ) = ej . The j th terms of f (a), f (b), f (c) equal ai , bi , ci , respectively, implying that these three terms have even sum. Therefore, f (a) + f (b) + f (c) has j th term 0 for all j, and f (a) + f (b) + f (c) = 0. 3 Canada Canada Problem 1 Let a1 , a2 , . . , a2000 be a sequence of integers each lying 2000 in the interval [−1000, 1000]. Suppose that i=1 ai = 1. Show that the terms in some nonempty subsequence of a1 , a2 , . . , a2000 sum to zero.

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Mathematical Olympiads 2000-2001: Problems and Solutions from around the World by Titu Andreescu


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