4. Two identical chessboards ($8 \times 8$ squares) have a common center and one completely covers the other (placed one on top of the other). One of them is rotated around the center by $45^{\circ}$. Determine the area of the intersection of all the black squares of one board with all the black squares of the other board if the area of one square is 1.
Solution. Let us denote by $S_{CC}$ - the sum of all common areas of the black squares on the upper board and the black squares on the lower board (i.e., what is to be calculated in the problem), by $S_{cb}$ - the sum of all common areas of the black squares on the upper board and the white squares on the lower board, by $S_{bc}$ - the sum of all common areas of the white squares on the upper board and the black squares on the lower board, and by $S_{bb}$ - the sum of all common areas of the white squares on the upper board and the white squares on the lower board.

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 when the upper board is rotated by $45^{\circ}$ as shown in the diagram. Let this position of the boards be the initial one.

If the upper board is rotated by $90^{\circ}$ around its center, then the black squares on that board will move to the positions of its white squares, so $S_{cc}=S_{bc}$ and $S_{cb}=S_{bb}$.

If the lower board is rotated by $90^{\circ}$ around its center, then the black squares on that board will move to the positions of its white squares, so $S_{CC}=S_{cb}$ and $S_{bc}=S_{bb}$. Therefore, $S_{cc}=S_{cb}=S_{bc}=S_{bb} \Rightarrow S_{cc}=\frac{1}{4} S$

It is also true that the area of the common octagon is obtained when the area of the entire board is reduced by the area of the four right-angled triangles at the edges of the board, i.e., $S=64-4P$. It remains to calculate the area $P$. The triangle is an isosceles right triangle and if we denote the hypotenuse by $a$ and the height dropped to the hypotenuse by $h$, we have $a=2h$. If we denote the diagonal of the board by $d$, then we have $d=2h+8$, i.e., $h=\frac{d-8}{2}=\frac{8\sqrt{2}-8}{2}=4(\sqrt{2}-1)$. Then for the area of the triangle, we get: $P=\frac{ah}{2}=h^2=16(\sqrt{2}-1)^2=16(3-2\sqrt{2})$, i.e.

$$
S=64-4P=128(\sqrt{2}-10), \quad S_{CC}=32(\sqrt{2}-1)
$$

## 2nd Year