1991 AIME Problem 12

The center of the rhombus is also the center of the rectangle, and its half-diagonals have lengths $15$ and $20$

Use coordinates for $P$ and $Q$; their vectors from the common center are perpendicular

Let the rectangle have width $w$ and height $h,$ with $A=(0,0)$ and $B=(w,0).$ Then $P=(w-15,0)$ and $Q=(w,20).$ The diagonals of the rhombus bisect each other at the rectangle’s center $O=(\frac{w}{2},\frac{h}{2}).$ Put $p=\overrightarrow{OP}=(\frac{w}{2}-15,-\frac{h}{2}).$ Since $OP=15,$ $OQ=20,$ and the rhombus diagonals are perpendicular, while $\overrightarrow{PQ}=(15,20),$ we have $$\begin{aligned}|p|&=15,\\p\mathbin{\cdot}(15,20)&=-225.\end{aligned}$$ Solving these two equations, with the second coordinate negative because $h>0,$ gives $p=(\frac{21}{5},-\frac{72}{5}).$ Hence $$\begin{aligned}w&=2\left(15+\frac{21}{5}\right)=\frac{192}{5},\\h&=\frac{144}{5}.\end{aligned}$$ The perimeter is $2(w+h)=\frac{672}{5},$ so $m+n=672+5=677.$