8. [6] Let $A B C$ be an equilateral triangle with side length 2 , and let $\Gamma$ be a circle with radius $\frac{1}{2}$ centered at the center of the equilateral triangle. Determine the length of the shortest path that starts somewhere on $\Gamma$, visits all three sides of $A B C$, and ends somewhere on $\Gamma$ (not necessarily at the starting point). Express your answer in the form of $\sqrt{p}-q$, where $p$ and $q$ are rational numbers written as reduced fractions.
Answer: $\sqrt{\frac{28}{3}}-1$ Suppose that the path visits sides $A B, B C, C A$ in this order. Construct points $A^{\prime}, B^{\prime}, C^{\prime}$ so that $C^{\prime}$ is the reflection of $C$ across $A B, A^{\prime}$ is the reflection of $A$ across $B C^{\prime}$, and $B^{\prime}$ is the reflection of $B$ across $A^{\prime} C^{\prime}$. Finally, let $\Gamma^{\prime}$ be the circle with radius $\frac{1}{2}$ centered at the center of $A^{\prime} B^{\prime} C^{\prime}$. Note that $\Gamma^{\prime}$ is the image of $\Gamma$ after the three reflections: $A B, B C^{\prime}, C^{\prime} A^{\prime}$.
When the path hits $A B$, let us reflect the rest of the path across $A B$ and follow this reflected path. When we hit $B C^{\prime}$, let us reflect the rest of the path across $B C^{\prime}$, and follow the new path. And when we hit $A^{\prime} C^{\prime}$, reflect the rest of the path across $A^{\prime} C^{\prime}$ and follow the new path. We must eventually end up at $\Gamma^{\prime}$.
It is easy to see that the shortest path connecting some point on $\Gamma$ to some point on $\Gamma^{\prime}$ lies on the line connecting the centers of the two circles. We can easily find the distance between the two centers to be $\sqrt{3^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}}=\sqrt{\frac{28}{3}}$. Therefore, the length of the shortest path connecting $\Gamma$ to $\Gamma^{\prime}$ has length $\sqrt{\frac{28}{3}}-1$. By reflecting this path three times back into $A B C$, we get a path that satisfies our conditions.