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>. | + | 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 | + | 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> | ||
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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, | ||
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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 | |
| − | <cmath>2OP = PB+38 \text{ and } 2PB = OP+19</cmath> | + | <cmath>2OP = PB+38 \text{ and } 2PB = OP+19.</cmath> |
| − | + | 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>. | |
| − | < | ||
| − | |||
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== 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}} | ||
| + | |||
| + | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 21:39, 25 August 2011
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,
![\[\frac{PB+38}{OP}= 2 \text{ and } \frac{OP+19}{PB} = 2\]](http://latex.artofproblemsolving.com/b/a/4/ba48eb05b5e743167842f148a2c8b89ce53c6a76.png)
so
![\[2OP = PB+38 \text{ and } 2PB = OP+19.\]](http://latex.artofproblemsolving.com/5/c/c/5cc83867035528c0cc5b54905fe567d2ab982175.png)
Substituting for 
, 
, so 
and 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 | ||
