11.207 Two cubes with an edge equal to $a$ have a common segment connecting the centers of two opposite faces, but one cube is rotated by $45^{\circ}$ relative to the other. Find the volume of the common part of these cubes.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-264.jpg?height=340&width=350&top_left_y=306&top_left_x=100)

Let $E$ and $F$ be the centers of the faces $ABCD$ and $KLMN$ of the given cube $T_1 = ABCD KLMN$, and $T_2$ be the cube obtained from $T_1$ by rotating it $45^\circ$ around $EF$ (clockwise when viewed from $E$ to $F$).

Let points $P_1, P_2, P_3$ on the edges of cube $T_1$ be such that

$BP_1 = CP_2 = CP_3 = \frac{a(\sqrt{2}-1)}{2}$.

Then $P_2 P_3 = \sqrt{2 CP_2^2} = a(\sqrt{2}-1) = R_1 P_2 = a - 2 CP_2$. Since $EP_1 = EP_2 = EP_3$ (by construction), $\Delta EP_1 P_2 = \Delta EP_2 P_3 \Rightarrow \angle EP_1 P_2 + \angle EP_2 P_1 = \angle EP_2 P_3 + \angle EP_2 P_1 = 135^\circ \Rightarrow \angle P_1 EP_2 = \angle P_2 EP_3 = 45^\circ$. Therefore, when rotating around $EF$ by $45^\circ$, $P_1$ will move to $P_2$, and $P_2$ will move to $P_3$. The same is true for points $Q_1, Q_2, Q_3$, where $LQ_1 = MQ_2 = MQ_3 = BP_1$. Consequently, the face $BLMC$ of cube $T_1$ after the rotation will pass through points $P_2, Q_2, Q_3, P_3$, i.e., it will cut off from $T_1$ a rectangular prism

$$
\Pi = P_2 C P_3 Q_2 M Q_3 \text{ with volume } \frac{1}{2} MC \cdot CP_2^2 = \frac{a^3}{4}(3-2\sqrt{2}).
$$

The volume of the common part of cubes $T_1$ and $T_2$ will be equal to the volume of $T_1$ minus the volumes of four prisms equal to prism $\Pi$, i.e., equal to $a^3 - 4 V_\Pi = 2a^3(\sqrt{2}-1)$.

Answer: $2a^3(\sqrt{2}-1)$.