Writeup: Putnam 2012 A1

18 Sep, 2025

Problem Statement:

Let $d_{1}, d_{2}, \dots d_{12}$ be real numbers in the open interval (1, 12). Show that there exist distinct indices $i, j, k$ such that $d_{i}, d_{j}, d_{k}$ form the vertices of an acute triangle.

Solution:

WLOG, we may assume that $d_{1} \leq d_{2} \leq \dots \leq d_{12}$ . For the sake of contradiction, assume that there do not exist indices $i, j, k$ such that $d_{i}, d_{j}, d_{k}$ form the vertices of an acute triangle. Then note that $\forall i \geq 3$ , $d_{i}^{2} \geq d_{i - 1}^{2} + d_{i - 2}^{2}$ .

Let the sequence $d_{i}^{2}$ be $s_{i}$ . Then note that $s_{i}$ , is also ordered, i.e. $s_{1} \leq s_{2} \leq \dots$ . Note that $s_{1}, s_{2} > 1$ , and $s_{i} \geq s_{i - 1} + s_{i - 2}$ . Therefore, we must have $s_{i} > F_{i}$ , where $F_{i}$ is the $n$ th Fibonacci number and $F_{1} = F_{2} = 1$ .

Computing, we know that $F_{12} = 144$ , so we know that $s_{12} > 144$ and therefore $d_{12} > 12$ . But since we assumed all $d_{i} \in (1, 12)$ , this is a contradiction! Therefore we must have some acute triangle whose side lengths lie in $d_{i}$ .