8. [7] Let $\triangle A B C$ be an equilateral triangle with height 13 , and let $O$ be its center. Point $X$ is chosen at random from all points inside $\triangle A B C$. Given that the circle of radius 1 centered at $X$ lies entirely inside $\triangle A B C$, what is the probability that this circle contains $O$ ?
Answer: $\frac{\sqrt{3} \pi}{100}$ The set of points $X$ such that the circle of radius 1 centered at $X$ lies entirely inside $\triangle A B C$ is itself a triangle, $A^{\prime} B^{\prime} C^{\prime}$, such that $A B$ is parallel to $A^{\prime} B^{\prime}, B C$ is parallel to $B^{\prime} C^{\prime}$, and $C A$ is parallel to $C^{\prime} A^{\prime}$, and furthermore $A B$ and $A^{\prime} B^{\prime}, B C$ and $B^{\prime} C^{\prime}$, and $C A$ and $C^{\prime} A^{\prime}$ are all 1 unit apart. We can use this to calculate that $A^{\prime} B^{\prime} C^{\prime}$ is an equilateral triangle with height 10 , and hence has area $\frac{100}{\sqrt{3}}$. On the other hand, the set of points $X$ such that the circle of radius 1 centered at $X$ contains $O$ is a circle of radius 1 , centered at $O$, and hence has area $\pi$. The probability that the circle centered at $X$ contains $O$ given that it also lies in $A B C$ is then the ratio of the two areas, that is, $\frac{\pi}{\frac{1 \pi 0}{\sqrt{3}}}=\frac{\sqrt{3} \pi}{100}$.