2012 USAMO Problems/Problem 1
Contents
Problem
Find all integers such that among any
positive real numbers
,
,
,
with
there exist three that are the side lengths of an acute triangle.
Solution
Without loss of generality, assume that the set is ordered from least to greatest so that the bounding condition becomes
Now set
and since a triangle with sidelengths from
will be similar to the corresponding triangle from
we simply have to show the existence of acute triangles in
Note that
and for all
,
Now three arbitrary sidelengths ,
, and
, with
will form a valid triangle if and only if
Furthermore, this triangle will be acute if and only if
However, the first inequality can actually be inferred from the second, since
and
is trivially greater than
So we just need to find all
such that there is necessarily a triplet of
's for which
(where
).
We now make another substitution: So
and for all
,
Now we examine the smallest possible sets
for small
for which the conditions of the problem are not met. Note that by smallest, we mean the set whose greatest element is as small as possible. If
, then the smallest possible set, call it
is trivially
, since
and
are obviously minimized and
follows as minimal. Using this as the base case, we see inductively that in general
is the set of the first
Fibonacci numbers. To show this note that if
, then
The smallest possible value for
is the sum of the two greatest values of
which are
and
. But these sum to
so
and our induction is complete.
Now since we know that the Fibonacci set is the smallest possible set which does not satisfy the conditions of the problem, then any set whose greatest term is less than
must satisfy the conditions. And since
is bounded between
and
, then the conditions of the problem are met if and only if
. The first
for which this restriction is satisfied is
and the exponential behavior of the Fibonacci numbers ensure that every
greater than
will also satisfy this restriction. So the final solution set is
.
Solution 2
Outline:
1. Define the Fibonacci numbers to be and
for
.
2. If the chosen is such that
, then choose the sequence
such that
for
. It is easy to verify that such a sequence satisfies the condition that the largest term is less than or equal to
times the smallest term. Also, because for any three terms
,
, x, y, z do not form an acute triangle. Thus, all
such that
do not work.
3. It is easy to observe via a contradiction argument that all such that
produce an acute triangle.
4. Note that and
. It is easily to verify through strong induction that all
greater than 12 make
. Thus,
is the desired solution set.
See also
2012 USAMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
2012 USAJMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
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