11.197 Find the volume of the common part of two cubes, if one of them is obtained by rotating the other cube by $90^{\circ}$ around an axis passing through the midline of one of its faces. The edge of the cube is $a$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-258.jpg?height=366&width=359&top_left_y=753&top_left_x=120)

Let $EF$ be the midline of the face $ABCD$ of the first cube, from which the second cube is obtained by a $90^{\circ}$ rotation around $FE$ clockwise (when viewed from $F$ to $E$). Then the face $ABCD$ will transition to the face $A_{1}B_{1}C_{1}D_{1}$. Clearly, the volume of the common part $A_{1}B_{1}EFNMCD$ of the two given cubes is equal to 1/4 of their volume, i.e., $\frac{a^{3}}{4} \quad (M$ and $N$ are the points of intersection of the edges of the cubes).

Answer: $\frac{a^{3}}{4}$.