1997 AIME Problem 15

The largest triangle is tilted, with one vertex at a corner of the rectangle and the other two on the far sides

With side $s$ tilted $\theta$ above the length-$11$ side, $s\cos\theta = 11$ and $s\sin(\theta + 60^\circ) = 10;$ solve for $\tan\theta$

By reflecting or rotating the configuration, take one side of the equilateral triangle to make an angle $0\le\theta\le30^\circ$ with the length-$11$ side of the rectangle. A triangle of side $s$ in this orientation has horizontal and vertical spans $s\cos\theta$ and $s\sin(\theta+60^\circ),$ respectively. Hence $$s\le\min\left(\frac{11}{\cos\theta}, \frac{10}{\sin(\theta+60^\circ)}\right).$$ The first bound increases with $\theta$ and the second decreases, so their minimum is largest when they are equal. This bound is attainable by putting one vertex at a corner and the other two on the far sides, giving $$\begin{aligned} s\cos\theta &= 11, \\ s\sin(\theta + 60^\circ) &= 10. \end{aligned}$$

Dividing, $11\sin(\theta + 60^\circ) = 10\cos\theta,$ and expanding the left side gives $\frac{11}{2}\sin\theta + \frac{11\sqrt{3}}{2}\cos\theta = 10\cos\theta,$ so $\tan\theta = \frac{20 - 11\sqrt{3}}{11}$ (about $4.9^\circ,$ a legal tilt). Then $$\begin{aligned} s^2 &= \frac{121}{\cos^2\theta} \\ &= 121\left(1 + \tan^2\theta\right) \\ &= 121 + \left(20 - 11\sqrt{3}\right)^2 \\ &= 884 - 440\sqrt{3}. \end{aligned}$$

The area is $\frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}\left(884 - 440\sqrt{3}\right)$ $= 221\sqrt{3} - 330 \approx 52.8,$ which indeed beats the untilted triangle of side $10.$ Thus $p + q + r$ $= 221 + 3 + 330 = 554.$