OIM 2007 Problem 4

In a $19\times 19$ board, a piece called dragon moves as follows: It travels by four squares (either horizontally or vertically) and then it moves one square more in a direction perpendicular to its previous direction. It is known that, moving so, a dragon can reach every square of the board.

The draconian distance between two squares is defined as the least number of moves a dragon needs to move from one square to the other.

Let $C$ be a corner square, and $V$ the square neighbor of $C$ that has only a point in common with $C$ . Show that there exists a square $X$ of the board, such that the draconian distance between $C$ and $X$ is greater than the draconian distance between $C$ and $V$ .