HMMT-Feb 2011 gen, Problem 13

Sarah and Hagar play a game of darts. Let $O_{0}$ be a circle of radius 1 . On the $n$th turn, the player whose turn it is throws a dart and hits a point $p_{n}$ randomly selected from the points of $O_{n-1}$. The player then draws the largest circle that is centered at $p_{n}$ and contained in $O_{n-1}$, and calls this circle $O_{n}$. The player then colors every point that is inside $O_{n-1}$ but not inside $O_{n}$ her color. Sarah goes first, and the two players alternate turns. Play continues indefinitely. If Sarah's color is red, and Hagar's color is blue, what is the expected value of the area of the set of points colored red?

Solution: Let $f(r)$ be the average area colored red on a dartboard of radius $r$ if Sarah plays first. Then $f(r)$ is proportional to $r^{2}$. Let $f(r)=(\pi x) r^{2}$ for some constant $x$. We want to find $f(1)=\pi x$. In the first throw, if Sarah's dart hits a point with distance $r$ from the center of $O_{0}$, the radius of $O_{1}$ will be $1-r$. The expected value of the area colored red will be $\left(\pi-\pi(1-r)^{2}\right)+\left(\pi(1-r)^{2}-f(1-r)\right)=$ $\pi-f(1-r)$. The value of $f(1)$ is the average value of $\pi-f(1-r)$ over all points in $O_{0}$. Using polar coordinates, we get \[ \begin{aligned} & f(1)=\frac{\int_{0}^{2 \pi} \int_{0}^{1}(\pi-f(1-r)) r d r d \theta}{\int_{0}^{2 \pi} \int_{0}^{1} r d r d \theta} \\ & \pi x=\frac{\int_{0}^{1}\left(\pi-\pi x(1-r)^{2}\right) r d r}{\int_{0}^{1} r d r} \\ & \frac{\pi x}{2}=\int_{0}^{1} \pi r-\pi x r(1-r)^{2} d r \\ & \frac{\pi x}{2} \frac{\pi}{2}-\pi x\left(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right) \\ & \frac{\pi x}{2}=\frac{\pi}{2}-\frac{\pi x}{12} \end{aligned} \] Algebra \& Calculus Individual Test \[ \pi x=\frac{6 \pi}{7} \] $\fbox{\frac{6 \pi}{7}}$.

Answer: \frac{6 \pi}{7}