Task B-3.5. (20 points) Two equal chessboards ($8 \times 8$ squares) have a common center and one completely covers the other. One of them is rotated around the center by $45^{\circ}$. Determine the area of the intersection of all black squares of one board with all black squares of the other board if the area of one square is 1.
Solution. Let's denote:

- $S_{cc}$ - the sum of all common areas of black squares of the upper and lower board,
- $S_{cb}$ - the sum of all common areas of black squares of the upper and white squares of the lower board,
- $S_{bc}$ - the sum of all common areas of white squares of the upper and black squares of the lower board,
- $S_{bb}$ - the sum of all common areas of white squares of the upper and lower board.

![](https://cdn.mathpix.com/cropped/2024_05_30_978576e245d98ac92b08g-17.jpg?height=607&width=607&top_left_y=478&top_left_x=609)

Then, $S=S_{cc}+S_{cb}+S_{bc}+S_{bb}$, where $S$ is the area of the regular octagon which is the intersection of the two boards.

Let's call the position of the boards in the picture the initial position.

If we rotate the upper board by $90^{\circ}$ around its center, the black squares of this board will move to the positions of its white squares from the initial position, so $S_{cc}=S_{bc}$ and $S_{cb}=S_{bb}$.

If we rotate the lower board by $90^{\circ}$ around its center, its black squares will move to the positions of its white squares from the initial position, so $S_{cc}=S_{cb}, S_{bc}=S_{bb} \mathrm{tj}$.

$$
S_{cc}=S_{cb}=S_{bc}=S_{bb} \quad \text { or } \quad S_{cc}=\frac{1}{4} S
$$

Furthermore, the area of the octagon is equal to the area of the square (chessboard) minus four times the area of the cut-off triangle, i.e., $S=64-4 P$.

The cut-off triangle is a right-angled isosceles triangle, so its base is equal to twice the height, i.e., $a=2 v$.

If we denote the length of the diagonal of the chessboard by $d$, the height of the triangle is $v=\frac{d-a}{2}=4(\sqrt{2}-1)$.

The area of the triangle is $P=\frac{a v}{2}=v^{2}=16(3-2 \sqrt{2})$.

Thus, $S=64-4 P=128(\sqrt{2}-1)$.

Finally, $S_{cc}=32(\sqrt{2}-1)$.

## NATIONAL MATHEMATICS COMPETITION

## 4th grade - high school - B category

Pula, March 30, 2009.