Subfield: Combinatorics


Question: For an integer $n \geq 5$, two players play the following game on a regular $n$-gon. Initially, three consecutive vertices are chosen, and one counter is placed on each. A move consists of one player sliding one counter along any number of edges to another vertex of the $n$-gon without jumping over another counter. A move is legal if the area of the triangle formed by the counters is strictly greater after the move than before. The players take turns to make legal moves, and if a player cannot make a legal move, that player loses. For which values of $n$ does the player making the first move have a winning strategy?

Solution:
We shall prove that the first player wins if and only the exponent of 2 in the prime decomposition of $n-3$ is odd.



Since the game is identical for both players, has finitely many possible states and always terminates, we can label the possible states Wins od Losses according as whether a player faced with that position has a winning strategy or not. A state is a Win if and only if there is some legal move taking the state to a Loss, and a state is a Loss if and only if all moves take that state to a Win (including the case where there are no legal moves).



Lemma. Any configuration in which the triangle formed by the three counters is not isosceles is necessarily a Win.



Proof. Label the positions of the counters $X, Y, Z$ so that the $\operatorname{arc} Y Z$ of the circumcircle is shortest and the $\operatorname{arc} Z X$ is longest. Begin by moving the counter at $Z$ around the polygon on the arc $Y Z X$ until it forms an isosceles triangle $X Y Z^{\prime}$ with apex at $Y$ (note that the $\operatorname{arc} X Y$ is less than half the circle, so that $Z$ does not jump over the counter at $X$ ). If this configuration is a Loss, we are done.



If instead this configuration is a Win, then the counters can be moved legally from triangle $X Y Z^{\prime}$ to reach a losing state. This cannot involve the counter at $Y$, so by symmetry a Loss state can be reached by moving the counter at $Z^{\prime}$ to a new location $Z^{\prime \prime}$. But then the counter at $Z$ could have been moved to $Z^{\prime \prime}$ in the first place, so the original configuration was a Win as well.



For every nonzero integer $x$, denote by $v_{2}(x)$ the exponent of 2 in the prime decomposition of $x$. Now, given a configuration in which the triangle formed by the three counters is isosceles, the arcs between the vertices having lengths $a, a, b$ respectively (in appropriate units so that $2 a+b=n$ ), we show that the configuration is a Win if and only if $a \neq b$ and $v_{2}(a-b)$ is odd.



Write $b=a \pm|a-b|$ and notice that the only other isosceles triangle that can be reached from the original configuration is one with arc lengths $a, a \pm|a-b| / 2, a \pm|a-b| / 2$. If $|a-b|$ is odd, this is of course impossible, so the configuration is a Loss, since all non-isosceles configurations are Wins, by the lemma.



If instead $|a-b|$ is even, then all states that can be reached from the original configuration are Wins, except possibly the state with arc lengths $a, a \pm|a-b| / 2, a \pm|a-b| / 2$. Consequently, $(a, a, b)$ is a Win if and only if $(a, a \pm|a-b| / 2, a \pm|a-b| / 2)$ is a Loss. Since the side lengths of this new triangle differ by $|a-b| / 2$, the conclusion follows inductively once the exceptional and trivial case $a=b$ is dealt with.



As an immediate corollary, the configuration with arc lengths 1, 1, $n-2$ (the starting configuration of the question) is a Win if and only if $v_{2}(n-3)$ is odd.

Final answer: the first player wins if and only the exponent of 2 in the prime decomposition of $n-3$ is odd