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

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== Problem ==
 
== Problem ==
The perimeter of triangle <math>APM</math> is <math>152</math>, and the angle <math>PAM</math> is a right angle. A circle of radius <math>19</math> with center <math>O</math> on <math>\overline{AP}</math> is drawn so that it is tangent to <math>\overline{AM}</math> and <math>\overline{PM}</math>. Given that <math>OP=m/n</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m+n</math>.
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The [[perimeter]] of triangle <math>APM</math> is <math>152</math>, and the angle <math>PAM</math> is a [[right angle]]. A [[circle]] of [[radius]] <math>19</math> with center <math>O</math> on <math>\overline{AP}</math> is drawn so that it is [[Tangent (geometry)|tangent]] to <math>\overline{AM}</math> and <math>\overline{PM}</math>. Given that <math>OP=m/n</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers, find <math>m+n</math>.
  
 
== Solution ==
 
== Solution ==
Let the circle intersect <math>\overline{PM}</math> at <math>B</math>. Then note <math>\triangle OPB</math> and <math>\triangle MPA</math> are similar. Also note that <math>AM = BM</math> by power of a point. So we have:
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Let the circle intersect <math>\overline{PM}</math> at <math>B</math>. Then note <math>\triangle OPB</math> and <math>\triangle MPA</math> are similar. Also note that <math>AM = BM</math> by power of a point. So we have
 
 
 
<cmath>\frac{19}{AM} = \frac{152-2AM}{152}</cmath>
 
<cmath>\frac{19}{AM} = \frac{152-2AM}{152}</cmath>
 
 
Solving, <math>AM = 38</math>. So the ratio of the side lengths of the triangles is 2. Therefore,
 
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>
 
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so
<cmath>2OP = PB+38 \text{ and } 2PB = OP+19</cmath>
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<cmath>2OP = PB+38 \text{ and } 2PB = OP+19.</cmath>
 
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Substituting for <math>PB</math>, <math>4OP-76 = OP+19</math>, so <math>OP = \frac{95}3</math> and the answer is <math>\boxed{098}</math>.
<cmath>4OP-76 = OP+19</cmath>
 
 
 
Finally, <math>OP = \frac{95}3</math>, so the answer is <math>098</math>.
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=II|num-b=13|num-a=15}}
 
{{AIME box|year=2002|n=II|num-b=13|num-a=15}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 20:39, 25 August 2011

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

See also

2002 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions