Balti Tee 2020 · Ülesanne 10

Let $S_0$ be the set of all points in the square, and for each $1 \le k \le N$, let $S_k$ be the set of possible points $B$ consistent with everything Bob has said. For each $k$, we then have that $S_k$ is the disjoint union of the two possible values $S_{k+1}$ can take for each of Bob's possible answers. Hence one of these must have area $\le \frac{|S_{k+1}|}{2}$, and the other must have area $\ge \frac{|S_{k+1}|}{2}$. Suppose now that Alice always receives the answer resulting in the greater half. After receiving $N$ answers, then, $|S_N| \ge \frac{1}{2^N}$. If Alice has a winning strategy, there must be a point $A$ in $S_N$ so that the circle of radius $\frac{1}{2020}$ centered at $A$ contains $S_N$. Hence $\frac{\pi}{2020^2} \ge \frac{1}{2^N}$. It therefore suffices to show that this inequality does not hold for $N=18$. This follows from the estimates $\pi \le 2^2$ and $2020 > 1024 = 2^{10}$, meaning that $\frac{\pi}{2020^2} > \frac{2^2}{2^{20}} = \frac{1}{2^{18}}$.