2008 AMC 12A Problems/Problem 20
Contents
Problem
Triangle has , , and . Point is on , and bisects the right angle. The inscribed circles of and have radii and , respectively. What is ?
Solution 1
By the Angle Bisector Theorem, By Law of Sines on , Since the area of a triangle satisfies , where the inradius and the semiperimeter, we have Since and share the altitude (to ), their areas are the ratio of their bases, or The semiperimeters are and . Thus,
Solution 2
We start by finding the length of and as in solution 1. Using the angle bisector theorem, we see that and . Using Stewart's Theorem gives us the equation , where is the length of . Solving gives us , so .
Call the incenters of triangles and and respectively. Since is an incenter, it lies on the angle bisector of . Similarly, lies on the angle bisector of . Call the point on tangent to , and the point tangent to . Since and are right, and , . Then, .
We now use common tangents to find the length of and . Let , and the length of the other tangents be and . Since common tangents are equal, we can write that , and . Solving gives us that . Similarly, .
We see now that
(Thanks to above solution writer for the framework of my diagram)
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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