# Difference between revisions of "2004 AIME I Problems/Problem 10"

## Problem

A circle of radius 1 is randomly placed in a 15-by-36 rectangle $ABCD$ so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal $AC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

## Solution

### Solution 2

The location of the center of the circle must be in the $34 \times 13$ rectangle that is one unit away from the sides of rectangle $ABCD$. We want to find the area of the right triangle with hypotenuse one unit away from $\overline{AC}$.

Let $A$ be at the origin, $B$ at $(36,0)$, $C$ at $(36,15)$, $D$ at $(0,15)$. The slope of diagonal $\overline{AC}$ is $\frac{15}{36} = \frac{5}{12}$. Since the hypotenuse is parallel to the diagonal, it has the same slope, and its equation is $y = \frac{5}{12}x + c$. Manipulating, $5x - 12y + 12c = 0$. We need to find the value of $c$, which can be determined using the fact that the hypotenuse is one unit away from the diagonal. Since the diagonal contains the origin, we can use the distance from a point to the line formula at the origin:

$$\left|\frac{Ax + By + C}{\sqrt{A^2+B^2}}\right| = 1$$ $$\left|\frac{(5)(0) + (-12)(0) + 12c}{\sqrt{5^2 + (-12)^2}}\right| = 1$$ $$c = \pm \frac{13}{12}$$

It makes sense that we have two values of $c$, one for the triangle on top of the diagonal, and one for the bottom. We will just consider the bottom triangle, so $c = -\frac{13}{12}$. Then the equation of the line is $y = \frac{5}{12}x - \frac{13}{12}$. Solving for its intersections with the lines $y = 1, x = 35$ we find the coordinates of the triangles are at $(5,1)(35,1)(35,\frac{27}{2})$. The area is $\frac{1}{2}bh = \frac{1}{2}(35-5)\left(\frac{27}{2} - 1\right) = \frac{375}{2}$.

Finally, the probability is $\frac{2\cdot \mathrm{area\ of\ triangle}}{34 \times 13} = \frac{375}{442}$, and $m + n = 817$.