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, $\frac{PB+38}{OP}= 2 \text{ and } \frac{OP+19}{PB} = 2$ so $2OP = PB+38$ and $2PB = OP+19.$ Substituting for $PB$ , we see that $4OP-76 = OP+19$ , so $OP = \frac{95}3$ and the answer is $\boxed{098}$ .

2002 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2002 AIME II Problems/Problem 14

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