Source: IMO-2018-P4
Category: Operations and Strategies

Problem:
A \emph{site} is any point $(x,y)$ in the plane for which $x,y \in \{1, \dots, 20\}$. Initially, each of the $400$ sites is unoccupied. Amy and Ben take turns placing stones on unoccupied sites, with Amy going first; Amy has the additional restriction that no two of her stones may be at a distance equal to $\sqrt5$. They stop once either player cannot move. Find the greatest $K$ such that Amy can ensure that she places at least $K$ stones.

Solution:
The answer is $K = 100$.

First, we show Amy can always place at least $100$ stones. Indeed, treat the problem as a grid with a checkerboard coloring. Since a distance of $\sqrt{5}$ corresponds to a knight jump, it always connects squares of different colors. Thus, Amy can choose to always play on one of the $200$ black squares without violating the distance condition. In this way, she can guarantee half the black squares, i.e., she can get $\frac{1}{2}\cdot 200 = 100$ stones.

Second, we show Ben can prevent Amy from placing more than $100$ stones. Divide the grid into several $4\times4$ squares and then further partition each $4\times4$ square according to the following labeling:
\[
\begin{array}{cccc}
1 & 2 & 3 & 4 \\
3 & 4 & 1 & 2 \\
2 & 1 & 4 & 3 \\
4 & 3 & 2 & 1
\end{array}
\]
The squares with each label form $4$-cycles by knight jumps. For each such cycle, whenever Amy plays in the cycle, Ben plays in the opposite point of the cycle, preventing Amy from playing any more stones in that original cycle. Hence Amy can play on at most $\frac{1}{4}$ of the sites, meaning she can place at most $100$ stones, as desired.

Answer: $K = 100$

Grading guidelines:
(Partial) 1. Observed that the forbidden distance √5 is exactly a knight move on the grid, and in particular connects opposite colors in a checkerboard coloring. 2. Gave a valid strategy for Amy guaranteeing at least 100 moves, e.g. “always play on black squares”, with a correct argument that this never violates the √5 condition and yields at least 100 stones. 3. Set up a correct partition/mirroring idea for Ben to limit Amy (e.g. decomposition into 4-cycles under knight moves, or into 4×4 blocks with a repeated 1–4 pattern), even if the cycle structure or the final counting is not fully finished. (Almost) 1. Proved both bounds K≥100 and K≤100 using the intended ideas, but left a minor gap in the game-theoretic justification of Ben’s reply strategy (e.g. why the “opposite point in the 4-cycle” is always unoccupied and always legal when needed). 2. Completed Ben’s blocking strategy up to showing Amy gets at most one stone per 4-cycle, but made a localized counting/tiling slip (e.g. not fully justifying that the 20×20 grid is partitioned into exactly 100 disjoint 4-cycles). 3. Solution is almost complete, but made minor mistakes which are not negligible.