2008 AIME II Problems/Problem 11
In triangle , , and . Circle has radius and is tangent to and . Circle is externally tangent to and is tangent to and . No point of circle lies outside of . The radius of circle can be expressed in the form , where , , and are positive integers and is the product of distinct primes. Find .
Let and be the feet of the perpendiculars from and to , respectively. Let the radius of be . We know that . From draw segment such that is on . Clearly, and . Also, we know is a right triangle.
So we get that . From the half-angle identity, we find that . Therefore, . By similar reasoning in triangle , we see that .
We conclude that .
So our right triangle has sides , , and .
Solution 2(pure synthetic)
Refer to the above diagram. Let the larger circle have center , the smaller have center , and the incenter be . We can easily calculate that the area of , and and , where is the inradius.
Now, Line is the perpendicular bisector of , as is isosceles. Letting the point of intersection be , we get that and , and are collinear as is equidistant from and . By Pythagoras, , and we notice that is a 3-4-5 right triangle.
Letting be the desired radius and letting be the projection of onto , we find that . Similarly, we find that the distance between the projection from onto , , and , is . From there, we let the projection of onto be , and we have , , and . We finish with Pythagoras on , whence we get the desired answer of . - Spacesam
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