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.
Solution 4 (easy but hard to see)
We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the hexagon is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with as its hypotenuse. The right angle will be at point . We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of is , as seen by the inradius of and less than the square's side length. is , which is less than the square plus the inradius of . Our final answer is .
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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