IX OBM

Note first that draws are not possible so any position $(x,y)$ is either a win or a loss for the player receiving it. Let $\varphi$ be the positive root of $t^2-t-1=0$, so $\varphi =\frac{1+\sqrt{5}}{2}$. Let $m=\min (x,y)$, $M=\max (x,y)$. We show that a player receiving $(x,y)$ wins if and only if $m=M$ or $M>\varphi m$.

If $m\ne M$ and $M<\varphi m$, then the player must pass on $(m,M-m)\ne (m,0)$ and $m>\varphi (M-m)$. So it is sufficient to show that a player receiving a position with $M>\varphi m$ can either win or pass back a position with $\min m^{\prime }$ and $\max M^{\prime }$ such that $M^{\prime }<\varphi m^{\prime }$ and $M^{\prime }\ne m^{\prime }$.

If $M>\varphi m$, and $M$ is a multiple of $m$, then the player can pass on $(m,0)$ and win. So assume $M>\varphi m$ and $M=qm+r$ with $0<r<m$. If $q\ge 2$, then the player can choose whether to pass on $(m,r)$ or $(m,m+r)$. If $(m,r)$ is a losing position, then he wins by passing that. If it is a winning position, then

he passes $(m,m+r)$ to the other player. The other player is now forced to pass back $(m,r)$, which is a winning position. So we may assume $q=1$. Now the player passes $(m,r)$. We claim that $m<\varphi r$ and $m\ne r$. Certainly $m\le r$. But $m+r>\varphi m$, so $(\varphi -1)m<r$, so $m<\varphi r$.