2002 AIME II Problems/Problem 14

Revision as of 23:57, 15 February 2009 by Duelist (talk | contribs) (Solution)

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,

\[\frac{PB+38}{OP}= 2 \text{ and } \frac{OP+19}{PB} = 2\]

\[2OP = PB+38 \text{ and } 2PB = OP+19\]

\[4OP-76 = OP+19\]

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

See also

2002 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AIME Problems and Solutions
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