# 2012 USAMO Problems/Problem 1

## Problem

Find all integers $n \ge 3$ such that among any $n$ positive real numbers $a_1$, $a_2$, $\dots$, $a_n$ with $$\max(a_1, a_2, \dots, a_n) \le n \cdot \min(a_1, a_2, \dots, a_n),$$ there exist three that are the side lengths of an acute triangle.

## Solution

Without loss of generality, assume that the set $\{a\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \le n \cdot a_1.$ Now set $b_i \equiv \frac{a_i}{a_1},$ and since a triangle with sidelengths from $\{a\}$ will be similar to the corresponding triangle from $\{b\},$ we simply have to show the existence of acute triangles in $\{b\}.$ Note that $b_1 = 1$ and for all $i$, $b_i \le n.$

Now three arbitrary sidelengths $x$, $y$, and $z$, with $x \le y \le z,$ will form a valid triangle if and only if $x+y>z.$ Furthermore, this triangle will be acute if and only if $x^2 + y^2 > z^2.$ However, the first inequality can actually be inferred from the second, since $x+y>z \longrightarrow x^2 + y^2 +2xy > z^2$ and $2xy$ is trivially greater than $0.$ So we just need to find all $n$ such that there is necessarily a triplet of $b$'s for which $b_i^2 + b_j^2 > b_k^2$ (where $b_i < b_j < b_k$).

We now make another substitution: $c_i \equiv b_i ^2.$ So $c_1 = 1$ and for all $i$, $c_i \le n^2.$ Now we examine the smallest possible sets $\{c\}$ for small $n$ 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 $n=3$, then the smallest possible set, call it $\{s_3\},$ is trivially $\{1,1,2\}$, since $c_1$ and $c_2$ are obviously minimized and $c_3$ follows as minimal. Using this as the base case, we see inductively that in general $\{s_n\}$ is the set of the first $n$ Fibonacci numbers. To show this note that if $\{s_n\} = \{F_0, F_1, ... F_n\}$, then $\{s_{n+1}\} = \{F_0, F_1, ... F_n, c_{n+1}\}.$ The smallest possible value for $c_{n+1}$ is the sum of the two greatest values of $\{s_n\}$ which are $F_{n-1}$ and $F_n$. But these sum to $F_{n+1}$ so $\{s_{n+1}\} = \{F_0, F_1, ... F_{n+1}\}$ 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 $\{c\}$ whose greatest term is less than $F_{n-1}$ must satisfy the conditions. And since $\{c\}$ is bounded between $1$ and $n^2$, then the conditions of the problem are met if and only if $F_{n-1} > n^2$. The first $n$ for which this restriction is satisfied is $n=13$ and the exponential behavior of the Fibonacci numbers ensure that every $n$ greater than $13$ will also satisfy this restriction. So the final solution set is $\boxed{\{n \ge 13\}}$.

## Solution 2

Outline:

1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \ge 3$.

2. If the chosen $n$ is such that $F_n \le n^2$, then choose the sequence $a_n$ such that $a_k = \sqrt{F_k}$ for $1 \le k \le n$. It is easy to verify that such a sequence satisfies the condition that the largest term is less than or equal to $n$ times the smallest term. Also, because for any three terms $x = \sqrt{F_a}, y = \sqrt{F_b}, z = \sqrt{F_c}$ with $a, $x^2 + y^2 = F_a + F_b \le F_{b-1} + F_b = F_{b+1} \le F_c = z^2$, x, y, z do not form an acute triangle. Thus, all $n$ such that $F_n \le n^2$ do not work.

3. It is easy to observe via a contradiction argument that all $n$ such that $F_n > n^2$ produce an acute triangle. (If, without loss of generality, $a_n$ is an increasing sequence, such that no three (in particular, consecutive) terms form an acute triangle, then $a_2^2 \ge F_1a_1^2, a_3^2 \ge a_2^2 + a_1^2 \ge F_2a^2$, and by induction $a_n^2 > F_na_1^2$, a contradiction to the condition's inequality.)

4. Note that $F_{12} = 144 = 12^2$ and $F_{13} = 233 > 169 = 13^2$. It is easily to verify through strong induction that all $n$ greater than 12 make $F_n > n^2$. Thus, $\boxed{n \ge 13}$ is the desired solution set.