A Similar Median Right Triangle

Let $\angle B$ be the right angle and $b$ be the length of the side $AC$ . Then the medians are of lengths $\dfrac{b}{2}$ , $\dfrac{b}{2}\sqrt {1+3cos^2 A}$ , $\dfrac{b}{2}\sqrt {1+3sin^2 A}$ . Since the two triangles are similar and $\angle A$ is the smallest angle, therefore $\dfrac{b^2}{4}(1+3cos^2 A)=\dfrac{b^2}{4}+\dfrac{b^2}{4}(1+3sin^2 A)$ or $cos^2 A=\dfrac{2}{3}$ . So $p=2,q=3$ , and $p+q=\boxed 5$

To simplify things (while still preserving $\cos^2 A$ ), let $a = 1$ . By Pythagorean's Theorem, $1 + b^2 = c^2$ .

According to Appolonius' Theorem for medians, $m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}$ , $m_b^2 = \frac{2a^2 + 2c^2 - b^2}{4}$ , and $m_c^2 = \frac{2a^2 + 2b^2 - c^2}{4}$ .

For the medians to make a similar triangle to its original, $\frac{2b^2 + 2c^2 - a^2}{4c^2} = \frac{2a^2 + 2c^2 - b^2}{4b^2} = \frac{2a^2 + 2b^2 - c^2}{4a^2}$ . Solving this with $a = 1$ and $1 + b^2 = c^2$ gives $b = \sqrt{2}$ and $c = \sqrt{3}$ .

Therefore, $\cos^2 A = \frac{b^2}{c^2} = \frac{2}{3}$ , so $p = 2$ , $q = 3$ , and $p + q = \boxed{5}$ .