Difference between revisions of "2002 AIME II Problems/Problem 14"
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== 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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| + | <cmath>\frac{19}{AM} = \frac{152-2AM}{152}</cmath> | ||
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| + | Solving, <math>AM = 38</math>. So the ratio of the side lengths of the triangles is 2. Therefore, | ||
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| + | <math>\frac{PB+38}{OP}= 2</math> and <math>\frac{OP+19}{PB} = 2</math> | ||
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| + | <math>2OP = PB+38</math> and <math>2PB = OP+19</math> | ||
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| + | <math>4OP-76 = OP+19</math> | ||
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| + | Finally, <math>OP = \frac{95}3</math>, so the answer is <math>98</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 22:49, 15 February 2009
Problem
The perimeter of triangle 
is 
, and the angle 
is a right angle. A circle of radius 
with center 
on 
is drawn so that it is tangent to 
and 
. Given that 
where 
and 
are relatively prime positive integers, find 
.
Solution
Let the circle intersect at 
. Then note 
and 
are similar. Also note that 
by power of a point. So we have:
![\[\frac{19}{AM} = \frac{152-2AM}{152}\]](http://latex.artofproblemsolving.com/a/6/0/a60315140ef1b249f249f56e6ac9ca7af9cc2562.png)
Solving, 
. So the ratio of the side lengths of the triangles is 2. Therefore,
and 
and 
Finally, 
, so the answer is 
.
See also
| 2002 AIME II (Problems • Answer Key • Resources) | ||
| 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 | ||
