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.