1995 USAMO Problems
Problem 1
Let be an odd prime. The sequence is defined as follows: and, for all is the least positive integer that does not form an arithmetic sequence of length with any of the preceding terms. Prove that, for all is the number obtained by writing in base and reading the result in base .
Problem 2
A calculator is broken so that the only keys that still work are the and buttons. The display initially shows 0. Given any positive rational number show that pressing some finite sequence of buttons will yield . Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians.
Problem 3
Given a nonisosceles, nonright triangle let denote the center of its circumscribed circle, and let and be the midpoints of sides and respectively. Point is located on the ray so that is similar to . Points and on rays and respectively, are defined similarly. Prove that lines and are concurrent, i.e. these three lines intersect at a point.
Problem 4
Suppose is an infinite sequence of integers satisfying the following two conditions:
(i) divides for
(ii) there is a polynomial such that for all .
Prove that there is a polynomial such that for all .
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 persons and 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 or fewer amicable pairs.
See Also
1995 USAMO (Problems • Resources) | ||
Preceded by 1994 USAMO |
Followed by 1996 USAMO | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.