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 . Or, you could find the inradius directly by using the formula , where and are the legs of the right triangle and is the hypotenuse. (Can you see why? Plus, use this formula only for right triangles because these are actually the only cases.) 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 | |
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