# Difference between revisions of "2002 AIME II Problems/Problem 14"

## Problem

The perimeter of triangle $APM$ is $152$, and the angle $PAM$ is a right angle. A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$. Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$.

## Solution

Let the circle intersect $\overline{PM}$ at $B$. Then note $\triangle OPB$ and $\triangle MPA$ are similar. Also note that $AM = BM$ by power of a point. So we have:

$$\frac{19}{AM} = \frac{152-2AM}{152}$$

Solving, $AM = 38$. So the ratio of the side lengths of the triangles is 2. Therefore,

$\frac{PB+38}{OP}= 2$ and $\frac{OP+19}{PB} = 2$

$2OP = PB+38$ and $2PB = OP+19$

$4OP-76 = OP+19$

Finally, $OP = \frac{95}3$, so the answer is $098$.