Iranian Mathematical Olympiad

Let $A$ be the first player and $B$ the second one. Since $A$ plays first, they can reach point $C$ sooner, making $n$ moves downwards and then $n$ moves to the left. Then they can get back to their start point, making $n$ moves upwards and $n$ moves to the right. In this case, the first player wins the upper right square with an area of $n^2$.

Now we prove that the second player can play in a way that the first player won't be able to win an area of more than $n^2$.

Without loss of generality, assume that $A$ moves down in their first move. Their distance from $X$ will be $2n-1$, while $B$ is $2n$ moves away from $X$. Clearly $B$ can make their $2n$ first moves upwards and $A$ won't be able to stop them. After $B$ has reached $X$, he/she shall move right. If they can make $2n$ moves in that direction, $A$ will not be able to go back to their start point. So let's assume $B$

$$\begin{array}{cc}n& A\\ C& \end{array}$$

has come to a halt after moving $x$ times on the upper side of the square. Thus the area that $A$ has won, will be a subset of the hatching area and therefore, less than it. The circumference of the hatching area equals $4n$, since $A$ has reached point $K$ after $2n+x+1$ moves and $|AK|=2n-x+1$.

Thus it suffices to prove that the area of a shape, which its circumference equals $4n$, would be $n^2$ at most. Assume one has started moving from $S$, making $x$ moves upwards, $x$ moves downwards, $y$ moves to the right and $y$ moves to the left (in some order) where $2x+2y=4n$, and then they has reached $S$ after all. So the final shape fits in a rectangle with $x,y$ as its length and height; which means the area of the final shape is less than $xy$

$$\begin{array}{cc}A& \\ & \\ B& \end{array}$$

and we have

$$\begin{array}{rl}2x+2y& =4n⟹x+y=2n⟹2n=x+y\ge 2\sqrt{xy}\\ ⟹xy& \le n^2\end{array}$$

The proof is completed.