# 2003 Indonesia MO Problems/Problem 8

## Problem

Given a triangle $ABC$ with $C$ as the right angle, and the sidelengths of the triangle are integers. Determine the sidelengths of the triangle if the product of the legs of the right triangle equals to three times the perimeter of the triangle.

## Solution

Let $a$ and $b$ be the legs of the triangle. We can express the comparison as an equation. $$ab = 3(a+b+\sqrt{a^2 + b^2})$$ We can manipulate the equation to find a relationship between $a$ and $b.$ \begin{align*} ab-3(a+b) &= 3\sqrt{a^2 + b^2} \\ a^2b^2 - 6a^2b - 6ab^2 + 9a^2 + 18ab + 9b^2 &= 9a^2 + 9b^2 \\ a^2b^2 - 6a^2b - 6ab^2 + 18ab &= 0 \end{align*} Since $ab \ne 0,$ we can divide both sides by $ab$ without affecting the solution. $$ab-6a-6b+18 = 0$$ We can use Simon's Favorite Factoring Trick to find integral solutions. $$(a-6)(b-6) =18$$ From the equation, we find that the three possible triangles can have the following side lengths: $\boxed{(7,24,25), (8,15,17), (9,12,15)}.$