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

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(Solution)
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<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>
 
so <math>2OP = PB+38</math> and <math>2PB = OP+19.</math> Substituting for <math>PB</math>, we see that <math>4OP-76 = OP+19</math>, so <math>OP = \frac{95}3</math> and the answer is <math>\boxed{098}</math>.
 
so <math>2OP = PB+38</math> and <math>2PB = OP+19.</math> Substituting for <math>PB</math>, we see that <math>4OP-76 = OP+19</math>, so <math>OP = \frac{95}3</math> and the answer is <math>\boxed{098}</math>.
 
lolz shrayus pro
 
  
 
== See also ==
 
== See also ==

Revision as of 22:42, 1 January 2018

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 $\boxed{098}$.

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

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