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 $m+n$ .

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:

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

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