2017 AIME I Problem 15

Put the legs on the axes and let the side between the legs run from $(s\cos\theta, 0)$ to $(0, s\sin\theta)$

Requiring the third vertex to lie on the hypotenuse gives $s\,(7\sqrt{3}\cos\theta + 11\sin\theta) = 20\sqrt{3};$ maximize the left combination using $\sqrt{a^2 + b^2}$

Place the right angle at the origin with vertices $(0, 0),$ $(5, 0),$ and $(0, 2\sqrt{3}),$ so the hypotenuse lies on the line $2\sqrt{3}\,x + 5y = 10\sqrt{3}.$ Let the equilateral triangle’s side between the two legs have endpoints $(s\cos\theta, 0)$ and $(0, s\sin\theta),$ where $s$ is the side length. Its midpoint is $\frac{s}{2}(\cos\theta, \sin\theta),$ and moving a distance $\frac{\sqrt{3}}{2}s$ perpendicular to the side places the third vertex at $\frac{s}{2}$ $\cdot\small\left(\cos\theta + \sqrt{3}\sin\theta,\ \sin\theta + \sqrt{3}\cos\theta\right).$

Substituting this vertex into the hypotenuse equation and simplifying gives $$s = \frac{20\sqrt{3}}{7\sqrt{3}\cos\theta + 11\sin\theta}.$$ The denominator is at most $\sqrt{(7\sqrt{3})^2 + 11^2}$ $= \sqrt{268} = 2\sqrt{67},$ attained for an admissible $\theta,$ so the minimum side length satisfies $s^2 = \frac{(10\sqrt{3})^2}{67} = \frac{300}{67}.$

The minimum area is $\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},$ so $m + n + p = 75 + 67 + 3 = 145.$