4. Two equal squares are superimposed on each other such that a vertex of the upper square coincides with the point $O$ where the diagonals of the lower square intersect. In this configuration, the upper square can freely rotate around point $O$. How should the upper square be positioned to cover the largest possible area together with the lower square? If there is more than one such position, describe all of them.
# Solution

Let's illustrate an arbitrary position of the two squares.

Drop perpendiculars from the center of the stationary square to two sides (in the figure - the right and bottom sides). Then, the two shaded right triangles are equal by a leg and an acute angle.

Therefore, regardless of the angle of rotation of the moving square, the area of the common part will always be the same. Thus, the two squares together always cover the same area.

Answer: any position.

![](https://cdn.mathpix.com/cropped/2024_05_06_3704f63868e768c015c9g-2.jpg?height=605&width=458&top_left_y=1080&top_left_x=1386)