22nd Korean Mathematical Olympiad Final Round

Let $\{(x,y):x=0,1,\dots ,m,y=1,2,\dots ,m\}$ be the set of all points on the grid. Let $(a,0)$ be the initial position of the pebble. Let $$P=(a,0),Q=(0,a),R=(m+1-a,m+1)\text{ and }S=(m+1,m+1-a).$$

Let $X$ be the set of all points of the grid on the lines $PQ$, $QR$, $RS$, $SP$ or the interior of the quadrangle formed by those four lines. The first player initially moves the pebble to $(a,1)$ and removes the straightline segment joining $(a,0)$ and $(a,1)$. We color the point $(x,y)$ red if $x+y-a$ is even, and blue if $x+y-a$ is odd. Then the first player will be given a choice only if the pebble is on the red point, and the second player will be given a choice only if the pebble is on the blue point. Moreover, all blue points in $X$ are not adjacent to any point out of $X$. Furthermore, every red point in $X$ has even number of edges to a blue point. Therefore, whenever the first player is given a choice of moving the pebble, there are odd number of unused edges inside $X$ available for him for moving the pebble to a point in $X$. Since the game must end and the first player can not lose, the first player can always win. $\square$