HMMT-Feb 2013 guts, Problem 16

The walls of a room are in the shape of a triangle $A B C$ with $\angle A B C=90^{\circ}, \angle B A C=60^{\circ}$, and $A B=6$. Chong stands at the midpoint of $B C$ and rolls a ball toward $A B$. Suppose that the ball bounces off $A B$, then $A C$, then returns exactly to Chong. Find the length of the path of the ball.

Solution: Let $C^{\prime}$ be the reflection of $C$ across $A B$ and $B^{\prime}$ be the reflection of $B$ across $A C^{\prime}$; note that $B^{\prime}, A, C$ are collinear by angle chasing. The image of the path under these reflections is just the line segment $M M^{\prime}$, where $M$ is the midpoint of $B C$ and $M^{\prime}$ is the midpoint of $B^{\prime} C^{\prime}$, so our answer is just the length of $M M^{\prime}$. Applying the Law of Cosines to triangle $M^{\prime} C^{\prime} M$, we have $M M^{\prime 2}=27+243-2 \cdot 3 \sqrt{3} \cdot 9 \sqrt{3} \cdot \frac{1}{2}=189$, so $M M^{\prime}=3 \sqrt{21}$. $\fbox{\quad 3 \sqrt{21}}$.

Answer: \quad 3 \sqrt{21}