Difference between revisions of "2002 AIME II Problems/Problem 14"
m (→Solution: minor) |
|||
| Line 13: | Line 13: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 19:37, 4 July 2013
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. 
