2004 AIME I Problem 10

Attempt Problem 10 of the 2004 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AIME I solutions, or check the answer key.

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10.

A circle of radius 11 is randomly placed in a 1515-by-3636 rectangle ABCDABCD so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal AC\overline{AC} is m/n,m/n, where mm and nn are relatively prime positive integers, find m+n.m + n.

Answer: 817
Concepts:geometric probabilitycoordinate geometrydistance formula
Difficulty rating: 2790
Solution:

Place A=(0,0),A = (0, 0), B=(36,0),B = (36, 0), C=(36,15).C = (36, 15). For the circle to lie in the rectangle, its center must lie in the rectangle [1,35]×[1,14],[1, 35] \times [1, 14], of area 3413=442,34 \cdot 13 = 442, and the center is uniformly distributed there. The diagonal AC\overline{AC} lies on the line 5x12y=0,5x - 12y = 0, and the circle misses it exactly when the center's distance 5x12y13\frac{|5x - 12y|}{13} exceeds 1,1, that is, 5x12y>13.|5x - 12y| \gt 13.

The line 5x12y=135x - 12y = 13 meets y=1y = 1 at x=5x = 5 and x=35x = 35 at y=272,y = \frac{27}{2}, so below the diagonal the favorable region is the right triangle with vertices (5,1),(5, 1), (35,1),(35, 1), (35,272),(35, \tfrac{27}{2}), with legs 3030 and 252\frac{25}{2} and area 1230252=3752.\frac{1}{2} \cdot 30 \cdot \frac{25}{2} = \frac{375}{2}. Rotating 180180^\circ about the rectangle's center (18,152),(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 375442,\frac{375}{442}, and since 442=21317442 = 2 \cdot 13 \cdot 17 shares no factor with 375=353,375 = 3 \cdot 5^3, we get m+n=375+442=817.m + n = 375 + 442 = 817.

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