2008 AIME II Problems/Problem 11
Problem
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 .
Solution
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.
To find , consider the right triangle . Since is tangent to , then bisects . Let ; then . Dropping the altitude from to , we recognize the right triangle, except scaled by .
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 .
By the Pythagorean Theorem, simplification, and the quadratic formula, we can get , for a final answer of .
Solution
first let A = <PCB ; now connect the points as shown in the first solution's diagram ; realise that tanA = r/x = 16/y = r + 16/(x+y) where x = BY and y = CX ( the 2 tangents) ; then we have that QM = 64r = 56 - x - y => (x+y) = 56 - 64r hence r/x = 16+r/(56-64r) ; now drop altitude AY to solve for tan2A ; now since we know tan2A we know tan A = r/x in terms of r hence solve the resulting equation in r
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
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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