What is the minimum number of three-cell corners that can be placed in an $8 \times 8$ square so that no more such corners can be placed in this square?
In each $2 \times 2$ square, at least two cells must be covered by corners (otherwise, another corner can fit into such a square).

A $8 \times 8$ square can be divided into 16 squares of size $2 \times 2$ each, meaning that at least 32 cells must be covered by corners, which requires no fewer than 11 corners.

![](https://cdn.mathpix.com/cropped/2024_05_06_e53884ac984930b8f3cbg-05.jpg?height=509&width=512&top_left_y=86&top_left_x=773)

## Answer

11.