2023 WSMO Tiebreaker Round Problems/Problem 3

Problem

Let be the set of all values of such that the area of a triangle with side lengths is a positive integer. Find

Solution

Letting be the base of the triangle, we have a height of for . Note for each height, we have two posible values of , if the triangle if acute or obtuse. We have \begin{align*} a_1^2 &= \left(5-\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\right)^2+\left(\frac{2h}{5}\right)^2\\ &= 5^2+\left(7^2-\left(\frac{2h}{5}\right)^2\right)-10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}+\left(\frac{2h}{5}\right)^2\\ &= 74-10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\\ \end{align*} for when the triangle is acute and \begin{align*} a_2^2 &= \left(5+\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\right)^2+\left(\frac{2h}{5}\right)^2\\ &= 5^2+\left(7^2-\left(\frac{2h}{5}\right)^2\right)+10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}+\left(\frac{2h}{5}\right)^2\\ &= 74+10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\\ \end{align*} for when the triangle is obtuse. We have So,

~pinkpig