2002 AIME II Problems/Problem 14
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 1
Let the circle intersect at . Then note and are similar. Also note that by power of a point. Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have Solving, . So the ratio of the side lengths of the triangles is 2. Therefore, so and Substituting for , we see that , so and the answer is .
Solution 2
Reflect triangle across line , creating an isoceles triangle. Let be the distance from the top of the circle to point , with as . Given the perimeter is 152, subtracting the altitude yields the semiperimeter of the isoceles triangle, as . The area of the isoceles triangle is:
Now use similarity, draw perpendicular from to , name the new point . Triangle is similar to triangle , by AA Similarity. Equating the legs, we get:
Solving for , it yields .
The cancels, yielding a quadratic. Solving yields . Add to find , yielding or .
Solution 3
Let the foot of the perpendicular from to be now Also let and This means that , since is on the angle bisector of
We have that so
However , so
We now use the fact that the perimeter of is : This quadratic factors as so , and
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 | ||
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