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
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, the inradius could be directly by using the formula , where and are the legs of the right triangle and is the hypotenuse. (This formula should be used only for right triangles.) 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 .
Solution 3
The radius of an incircle is . The area of the triangle is equal to and the semiperimeter is equal to . The radius, therefore, is equal to . Thus using similar triangles the dimensions of the triangle circumscribing the circle with center are equal to , , and . The radius of the circle inscribed in this triangle with dimensions is found using the formula mentioned at the very beginning. The radius of the incircle is equal to .
Defining as , is equal to or . Also using similar triangles, the dimensions of the triangle circumscribing the circle with center are equal to , , or . The radius of by using the formula mentioned at the beginning is . Using as , is equal to or . Using the distance formula, the distance between and : this equals or , thus is .
Note
The problem can be reduced to a triangle for the initial calculations, where is calculated to be (, and is calculated to be (. After we find the incenters the points can be scaled up by a factor of for the final distance calculation.
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 |
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