Evdokimov M.A.

In each cell of a $5 \times 5$ board, there is either a cross or a zero, and no three crosses stand in a row either horizontally, vertically, or diagonally. What is the maximum number of crosses that can be on the board?
Example of placing 16 crosses according to the condition, see the figure on the left.
![](https://cdn.mathpix.com/cropped/2024_05_06_0e910282974b56b80e3ag-10.jpg?height=222&width=646&top_left_y=1909&top_left_x=707)

Evaluation. Divide the board into the central cell and 8 rectangles of size $3 \times 1$ (see the figure on the right). In each rectangle, there must be at least one zero. Therefore, there are at least eight zeros on the board. If there is also a zero in the central cell, then there are no more than 16 crosses, and the problem is solved. It remains to consider the case where there is a cross in the central cell.

First method. Consider all the cells of the main diagonals (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_0e910282974b56b80e3ag-10.jpg?height=225&width=211&top_left_y=2380&top_left_x=932)

Each pair of adjacent shaded cells, together with the central cell, forms a triplet, in which there must be at least one zero. This means that in the eight shaded cells, there are at least four zeros. Therefore, in each of the $3 \times 1$ rectangles located at the edge of the board, there must be at least one zero. In total, there are already at least eight zeros. But if the remaining cells contain crosses, then three crosses in a row will form crosses adjacent to the central one horizontally and vertically, which contradicts the condition. Therefore, there must be at least nine zeros.

Second method. Consider 8 pairs of cells highlighted in the figure on the left. In one of the cells of each pair, there must be at least one zero, so there are at least eight zeros.
![](https://cdn.mathpix.com/cropped/2024_05_06_0e910282974b56b80e3ag-11.jpg?height=248&width=1130&top_left_y=92&top_left_x=472)

Assume that in each highlighted pair of cells, there is exactly one zero and there are no zeros outside these pairs. Then the remaining cells contain crosses (see the figure in the center). Consider the four triplets of cells highlighted on this figure. Each of them already contains two crosses, so there must be zeros between them. Again, consider four of the eight previously highlighted pairs of cells (see the figure on the right). We have already determined that in each pair, there is exactly one cell with a zero, so the other cell contains a cross. If we place these crosses, then in the center of the board, there will be triplets of crosses in a row. Therefore, this arrangement also does not satisfy the condition.

## Answer

16 crosses.