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 $m + nk$ .

Let $X$ and $Y$ be the feet of the perpendiculars from $P$ and $Q$ to $BC$ , respectively. Let the radius of $\odot Q$ be $r$ . We know that $PQ = r + 16$ . From $Q$ draw segment $\overline{QM} \parallel \overline{BC}$ such that $M$ is on $PX$ . Clearly, $QM = XY$ and $PM = r - 16$ . Also, we know $QPM$ is a right triangle.

To find $XY$ , consider the right triangle $PCX$ . Since $\odot P$ is tangent to $\overline{AC},\overline{BC}$ , then $PC$ bisects $\angle ACB$ . Let $\angle ACB = 2\theta$ ; then $\angle PCX = \angle QBX = \theta$ . Dropping the altitude from $A$ to $BC$ , we recognize the $7 - 24 - 25$ right triangle, except scaled by $4$ .

So our right triangle $QPM$ has sides $r + 16$ , $r - 16$ , and $\frac {104 - 4r}{3}$ .

By the Pythagorean Theorem, simplification, and the quadratic formula, we can get $r = 44 - 6\sqrt {35}$ , for a final answer of $\fbox{254}$ .