49th Mathematical Olympiad in Ukraine

The second player can follow the following rules for his steps.

1) If he can, he paints the side which is the fourth yellow side in some square. He does it and he wins in this game.

2) If the first player paints some side, the second player paints the side which is symmetric to the one just painted about the centre of the field $2009\times 2009$. On the Fig. 4 we can see the corresponding steps.



Fig. 4

Thus we have that 1) in this game certainly the first or the second player wins; 2) the second player can always make the next step which satisfies his rules. Suppose the first player can win. Then his last step is painting the side which is the fourth yellow side in some square. Let it be the segment $AB$ (Fig. 5). Before that, the second player had to paint one of the segments $BC$, $CD$ or $DA$. Otherwise, as soon as these three segments were painted by the first player, the second player would have painted the segment $AB$ following the first rule and the game would be finished. So he had to paint one of the segments $BC$, $CD$ or $DA$. It means that the square $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ which is symmetric to the square $ABCD$ about the centre of the board had painted sides $B^{\prime }{C}^{\prime }$, $C^{\prime }{D}^{\prime }$ and $D^{\prime }{A}^{\prime }$ before a move of the second player. And so the second player would have painted the segment $A^{\prime }{B}^{\prime }$ to win this game. Obviously, this argument also works for the central square.



Fig. 5