Let $A_{1}, \ldots, A_{2022}$ be the vertices of a regular 2022-gon in the plane. Alice and Bob play a game. Alice secretly chooses a line and colors all points in the plane on one side of the line blue, and all points on the other side of the line red. Points on the line are colored blue, so every point in the plane is either red or blue. (Bob cannot see the colors of the points.)

In each round, Bob chooses a point in the plane (not necessarily among $A_{1}, \ldots$, $A_{2022}$ ) and Alice responds truthfully with the color of that point. What is the smallest number $Q$ for which Bob has a strategy to always determine the colors of points $A_{1}, \ldots, A_{2022}$ in $Q$ rounds?
The answer is 22 . To prove the lower bound, note that there are $2022 \cdot 2021+2>2^{21}$ possible colorings. If Bob makes less than 22 queries, then he can only output $2^{21}$ possible colorings, which means he is wrong on some coloring.

Now we show Bob can always win in 22 queries. A key observation is that the set of red points is convex, as is the set of blue points, so if a set of points is all the same color, then their convex hull is all the same color.
Lemma

Let $B_{0}, \ldots, B_{k+1}$ be equally spaced points on a circular arc such that colors of $B_{0}$ and $B_{k+1}$ differ and are known. Then it is possible to determine the colors of $B_{1}$, $\ldots, B_{k}$ in $\left\lceil\log _{2} k\right\rceil$ queries.

Proof. There exists some $0 \leq i \leq k$ such that $B_{0}, \ldots, B_{i}$ are the same color and $B_{i+1}$, $\ldots, B_{k+1}$ are the same color. (If $i\lceil k / 2\rceil$, then the segment $P B_{i}$ is blue and intersect the segment $B_{0} B_{[k / 2\rceil}$, which is red, contradiction.

Now the strategy is: Bob picks $A_{1}$. WLOG it is red. Now suppose Bob does not know the colors of $\leq 2^{k}-1$ points $A_{i}, \ldots, A_{j}$ with $j-i+1 \leq 2^{k}-1$ and knows the rest are red. I claim Bob can win in $2 k-1$ queries. First, if $k=1$, there is one point and he wins by querying the point, so the base case holds, so assume $k>1$. Bob queries $A_{i+\lceil(j-i+1) / 2\rceil}$. If it is blue, he finishes in $2 \log _{2}\lceil(j-i+1) / 2\rceil \leq 2(k-1)$ queries by the first lemma, for a total of $2 k-1$ queries. If it is red, he can query one more point and learn some half of $A_{i}, \ldots, A_{j}$ that are red by the second lemma, and then he has reduced it to the case with $\leq 2^{k-1}-1$ points in two queries, at which point we induct.