Balkan Mathematical Olympiad

Place the given rectangle into the coordinate plane so that the center of the square at the intersection of $i$-th column and $j$-th row has the coordinates $(i,j)$. Suppose that a desired numeration of the red points exists; it corresponds to a path, i.e. a closed poligonal line consisting of 96 segments of length $\sqrt{13}$, passing through each red point exactly once. Note that points $(i,j)$ and $(k,l)$ are adjacent in the path if and only if $\{|i-k|,|j-l|\}=\{2,3\}$.

The center of symmetry must be at point $C\left(5\frac{1}{2},5\right)$. Consider the points $A(2,2)$, $B(11,8)$. These two points are symmetric with respect to $C$ and divide the path into two parts ${\gamma }_1$ and ${\gamma }_2$. Note that, if the rectangular board is colored alternately white and black (like a chessboard), $A$ and $B$ are of different colors, and each segment connects two squares of different colors. It follows that each of ${\gamma }_1,{\gamma }_2$ consists of an odd number of segments. Thus these two parts are of different lengths and cannot be symmetric to each other. Therefore each



of ${\gamma }_1,{\gamma }_2$ is centrally symmetric itself.

Being of an odd length, each of the parts ${\gamma }_1,{\gamma }_2$ must contain a segment which is centrally symmetric with respect to $C$. There are only two such segments one connecting $(5,4)$ and $(8,6)$, and one connecting $(5,6)$ and $(8,4)$, so these two segments must be parts of our path. Moreover, point $(2,2)$ is connected with only two points, namely $(4,5)$ and $(5,4)$, so these three points are directly connected. Analogous conclusions can be made about points $(2,8),(11,2)$ and $(11,8)$, so the closed path $(5,4)-(2,2)-(4,5)-(2,8)-(5,6)-(8,4)-(11,2)-(9,5)-(11,8)-(8,6)-(5,4)$ is entirely contained in our path, which is clearly a contradiction.