2008 AIME II Problems/Problem 11

Problem

In triangle $ABC$ , $AB = AC = 100$ , and $BC = 56$ . Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$ . Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$ . No point of circle $Q$ lies outside of $\triangle ABC$ . The radius of circle $Q$ can be expressed in the form $m - n\sqrt {k}$ , where $m$ , $n$ , and $k$ are positive integers and $k$ is the product of distinct primes. Find $m + nk$ .

Solution

We inscribe the circles as in the problem statement. From there, we drop the perpendiculars from $P$ and $Q$ to $BC$ to the points $X$ and $Y$ respectively.

Let radius of $Q$ be $R$ . So we know that $PQ$ = $R + 16$ . Now from $P$ draw segment $\overline{PM} \parallel \overline{BC}$ such that $M$ is on $QY$ . Clearly, $PM = XY$ and $QM = R - 16$ . Also, we know $QPM$ is a right triangle.

So let's find $XY$ .

Consider the right triangle $PBX$ . Clearly $PB$ bisects angle $ABC$ . Let $\angle ABC = 2\theta$ . So $\angle PBX = \theta$ . We will apply tangent half-angle identites... dropping altitude from $A$ to $BC$ , we recognize the ever-popular $7 - 24 - 25$ right triangle (except it's scaled by 4).

So we get that $tan(2\theta) = 24/7$ . From half-angle identity, we can easily see that $tan(\theta) = \frac {3}{4}.$

So $PX = \frac {64}{3}$ . By similar reasoning in triangle $QCY$ , we see that $CY = \frac {4R}{3}$ (recall $ABC$ is isoceles!)

We conclude that $XY = 56 - \frac {4R + 64}{3} = \frac {104 - 4R}{3}$

So our right triangle has sides: $R + 16$ , $R - 16$ , $\frac {104 - 4R}{3}$

By pythagorean and lots of simplification and quadratic formula, we can get $R = 44 - 6\sqrt {35}$ , for a final answer of $\fbox{254}$

2008 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2008 AIME II Problems/Problem 11

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