Let's replace the board with an equivalent $9 \times 9$ square board, where diagonals of the same direction are drawn in all cells (see Fig. 1).

Six chips are sufficient to cover all cells (see Fig. 1).

![](https://cdn.mathpix.com/cropped/2024_05_06_de7bb834a1e75b584fa6g-27.jpg?height=760&width=1655&top_left_y=432&top_left_x=133)

Suppose there are only five chips. Consider one of the columns where no chip is placed. In this column, there are at least four squares that are not covered horizontally. These correspond to eight cells (see Fig. 2), none of which can be covered by a single chip diagonally.

Thus, at least three cells will not be covered by any chip.

## Answer

6 chips.
10,11

The game board has the shape of a rhombus with an angle of $60^{\circ}$. Each side of the rhombus is divided into nine parts. Lines are drawn through the division points, parallel to the sides and the short diagonal of the rhombus, dividing the board into triangular cells. If a chip is placed on a certain cell, three lines are drawn through this cell, parallel to the sides and the short diagonal of the rhombus. The cells they intersect will be considered as being captured by the chip. What is the minimum number of chips needed to capture all the cells of the board?