2001 AIME II Problems/Problem 7
Problem
Let be a right triangle with , , and . Let be the inscribed circle. Construct with on and on , such that is perpendicular to and tangent to . Construct with on and on such that is perpendicular to and tangent to . Let be the inscribed circle of and the inscribed circle of . The distance between the centers of and can be written as . What is ?
Contents
[hide]Solution
Solution 1 (analytic)
Let be at the origin. Using the formula on , where is the inradius (similarly define to be the radii of ), is the semiperimeter, and is the area, we find . Thus lie respectively on the lines , and so .
Note that . Since the ratio of corresponding lengths of similar figures are the same, we have
Let the centers of be , respectively; then by the distance formula we have . Therefore, the answer is .
Solution 2 (synthetic)
We compute as above. Let respectively the points of tangency of with .
By the Two Tangent Theorem, we find that , . Using the similar triangles, , , so . Thus .
See also
2001 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |