Difference between revisions of "2002 AIME II Problems/Problem 14"
(→Solution) |
(→Solution) |
||
| Line 8: | Line 8: | ||
<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>. | ||
| − | |||
| − | |||
== See also == | == See also == | ||
Revision as of 21:42, 1 January 2018
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 
and 
Substituting for 
, we see that 
, 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
