2004 AIME I Problem 10

The circle’s center is uniform over a $34 \times 13$ rectangle, and the circle misses the diagonal exactly when the center is more than $1$ from that line

The favorable region is two right triangles cut off by the lines parallel to the diagonal at distance $1;$ each has legs $30$ and $\frac{25}{2}$

Place $A = (0, 0),$ $B = (36, 0),$ $C = (36, 15).$ For the circle to lie in the rectangle, its center must lie in the rectangle $[1, 35] \times [1, 14],$ of area $34 \cdot 13 = 442,$ and the center is uniformly distributed there. The diagonal $\overline{AC}$ lies on the line $5x - 12y = 0,$ and the circle misses it exactly when the center’s distance $\frac{|5x - 12y|}{13}$ exceeds $1,$ that is, $|5x - 12y| \gt 13.$

The line $5x - 12y = 13$ meets $y = 1$ at $x = 5$ and $x = 35$ at $y = \frac{27}{2},$ so below the diagonal the favorable region is the right triangle with vertices $(5, 1),$ $(35, 1),$ $(35, \tfrac{27}{2}),$ with legs $30$ and $\frac{25}{2}$ and area $\frac{1}{2} \cdot 30 \cdot \frac{25}{2} = \frac{375}{2}.$ Rotating $180^\circ$ about the rectangle’s center $(18, \tfrac{15}{2}),$ which lies on the diagonal, maps the inner rectangle and the diagonal to themselves, so the region above the diagonal has the same area.

The probability is $\frac{375}{442},$ and since $442 = 2 \cdot 13 \cdot 17$ shares no factor with $375 = 3 \cdot 5^3,$ we get $m + n = 375 + 442 = 817.$