HMMT-Nov 2009 guts, Problem 32

A circle $\omega_{1}$ of radius 15 intersects a circle $\omega_{2}$ of radius 13 at points $P$ and $Q$. Point $A$ is on line $P Q$ such that $P$ is between $A$ and $Q . R$ and $S$ are the points of tangency from $A$ to $\omega_{1}$ and $\omega_{2}$, respectively, such that the line $A S$ does not intersect $\omega_{1}$ and the line $A R$ does not intersect $\omega_{2}$. If $P Q=24$ and $\angle R A S$ has a measure of $90^{\circ}$, compute the length of $A R$.

Solution: Let $O_{1}$ be the center of $\omega_{1}$ and $O_{2}$ be the center of $\omega_{2}$. Then $O_{1} O_{2}$ and $P Q$ are perpendicular. Let their point of intersection be $X$. Using the Pythagorean theorem, the fact that $P Q=24$, and our knowledge of the radii of the circles, we can compute that $O_{1} X=9$ and $O_{2} X=5$, so $O_{1} O_{2}=14$. Let $S O_{1}$ and $R O_{2}$ meet at $Y$. Then $S A R Y$ is a square, say of side length $s$. Then $O_{1} Y=s-15$ and $O_{2} Y=s-13$. So, $O_{1} O_{2} Y$ is a right triangle with sides $14, s-15$, and $s-13$. By the Pythagorean theorem, $(s-13)^{2}+(s-15)^{2}=14^{2}$. We can write this as $2 s^{2}-4 \cdot 14 s+198=0$, or $s^{2}-28 s+99=0$. The quadratic formula then gives $s=\frac{28 \pm \sqrt{388}}{2}=14 \pm \sqrt{97}$. Since $14-\sqrt{97}<15$ and $Y O_{1}>15$, we can discard the root of $14-\sqrt{97}$, and the answer is therefore $14+\sqrt{97}$. $\fbox{\sqrt{14+\sqrt{97}}}$.

Answer: \sqrt{14+\sqrt{97}}