Difference between revisions of "2002 AIME II Problems/Problem 14"
Brut3Forc3 (talk | contribs) m (→Solution) |
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<math>4OP-76 = OP+19</math> | <math>4OP-76 = OP+19</math> | ||
− | Finally, <math>OP = \frac{95}3</math>, so the answer is <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 22:51, 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:
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 |