2002 AIME II Problems/Problem 14: Difference between revisions

Revision as of 22:57, 15 February 2009

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 $098$ .

@@ Line 9: / Line 9: @@
 Solving, <math>AM = 38</math>. So the ratio of the side lengths of the triangles is 2. Therefore,
-<cmath>\frac{PB+38}{OP}= 2 \text{and} \frac{OP+19}{PB} = 2</cmath>
+<cmath>\frac{PB+38}{OP}= 2 \text{ and } \frac{OP+19}{PB} = 2</cmath>
-<cmath>2OP = PB+38 \text{and} 2PB = OP+19</cmath>
+<cmath>2OP = PB+38 \text{ and } 2PB = OP+19</cmath>
 <cmath>4OP-76 = OP+19</cmath>

2002 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2002 AIME II Problems/Problem 14: Difference between revisions

Revision as of 22:57, 15 February 2009

Problem

Solution

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