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

m (Solution)
Line 15: Line 15:
 
<math>4OP-76 = OP+19</math>
 
<math>4OP-76 = OP+19</math>
  
Finally, <math>OP = \frac{95}3</math>, so the answer is <math>98</math>.
+
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}}

Revision as of 23:51, 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,

$\frac{PB+38}{OP}= 2$ and $\frac{OP+19}{PB} = 2$

$2OP = PB+38$ 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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
Invalid username
Login to AoPS