AoPSWiki
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

1995 USAMO Problems

From AoPSWiki

Problems of the 1995 USAMO.

Contents

Problem 1

Let \, p \, be an odd prime. The sequence (a_n)_{n \geq 0} is defined as follows: \, a_0 = 0, a_1 = 1, \, \ldots, \, a_{p-2} = p-2 \, and, for all \, n \geq p-1, \, \, a_n \, is the least positive integer that does not form an arithmetic sequence of length \, p \, with any of the preceding terms. Prove that, for all \, n, \, \, a_n \, is the number obtained by writing \, n \, in base \, p-1 \, and reading the result in base \, p.

Solution

Problem 2

A calculator is broken so that the only keys that still work are the \, \sin, \; \cos, \tan, \; \sin^{-1}, \; \cos^{-1}, \, and \, \tan^{-1} \, buttons. The display initially shows 0. Given any positive rational number \, q, \, show that pressing some finite sequence of buttons will yield \, q. Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians.

Solution

Problem 3

Given a nonisosceles, nonright triangle \, ABC, \, let \, O \, denote the center of its circumscribed circle, and let \, A_1, \, B_1, \, and \, C_1 \, be the midpoints of sides \, BC, \, CA, \, and \, AB, \, respectively. Point \, A_2 \, is located on the ray \, OA_1 \, so that \, \triangle OAA_1 \, is similar to \, \triangle OA_2A. Points \, B_2 \, and \, C_2 \, on rays \, OB_1 \, and \, OC_1, \, respectively, are defined similarly. Prove that lines \, AA_2, \, BB_2, \, and \, CC_2 \, are concurrent, i.e. these three lines intersect at a point.

Solution

Problem 4

Suppose \, q_0, \, q_1, \,  q_2, \ldots \; \, is an infinite sequence of integers satisfying the following two conditions:
(i) \, m-n \, divides \, q_m - q_n \, for \, m > n \geq 0,
(ii) there is a polynomial \, P \, such that \, |q_n| < P(n) \, for all \, n.
Prove that there is a polynomial \, Q \, such that \, q_n = Q(n) \, for all \, n.

Solution

Problem 5

Suppose that in a certain society, each pair of persons can be classified as either amicable or hostile. We shall say that each member of an amicable pair is a friend of the other, and each member of a hostile pair is a foe of the other. Suppose that the society has \, n \, persons and \, q \, amicable pairs, and that for every set of three persons, at least one pair is hostile. Prove that there is at least one member of the society whose foes include \, q(1 - 4q/n^2) \, or fewer amicable pairs.

Solution

Resources

1995 USAMO (Problems)
Preceded by
1994 USAMO
1 2 3 4 5 Followed by
1996 USAMO
All USAMO Problems and Solutions
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us