139. The edge of a cube is $a$. The cube is rotated about a diagonal by an angle $\alpha$. Find the volume of the common part of the original cube and the rotated one.
139. Let the cube $A B C D A_{1} B_{1} C_{1} D_{\text {i }}$ rotate by an angle $\alpha$ around the diagonal $A C_{j}$ (Fig. 24). Take points $K$ and $L$ on the edges $A_{1} B_{1}$ and $A_{1} D_{1}$ such that $\left|A_{1} K\right| = \left|A_{1} L\right| = x$, and drop perpendiculars from $K$ and $L$ to the diagonal $A C_{1}$. Due to the symmetry of the cube with respect to the plane $A C C_{1} A_{1}$, these perpendiculars will pass through the same point $M$ on the diagonal $A C_{1}$. Suppose the value of $x$ is chosen such that $\widehat{K M L} = \alpha$. Then, after rotating around the diagonal $A C_{i}$ counterclockwise (if viewed from the direction from $A$ to $C_{1}$) by an angle $\alpha$, the point $K$ will move to $L$. We also take points $P$ and $Q$ on the edges $B_{1} A_{1}$ and $B_{1} B$ at the same distance $x$ from the vertex $B_{i}$. After the same rotation, the point $Q$ will move to $P$. Therefore, the face $A B B_{1} A_{i}$, after the rotation, will pass through the points $A, L$, and $P$ and cut off a pyramid $A A_{1} P L$ from our cube, the volume of which is $\frac{1}{6} a x(a-x)$. The same reasoning applies to all faces. Thus, the volume of the common part is $a^{3} - a x(a-x)$. We need to find the value of $x$ from the condition $\widehat{K M L} = \alpha$. For this, connect $M$ with the midpoint of the segment $L K$ - point $R$. We have $|M R| = x \frac{\sqrt{2}}{2} \operatorname{ctg} \frac{\alpha}{2}, \left|C_{1} R\right| = a \sqrt{2} - x \frac{\sqrt{2}}{2}$, and from the similarity of triangles $C_{1} R M$ and $C_{1} A_{1} A$, we find $x = \frac{2 a}{1 + \sqrt{3} \operatorname{ctg} \frac{\alpha}{2}}$.

Thus, the volume of the common part is $\frac{3 a^{9}\left(1 + \operatorname{ctg}^{2} \frac{\alpha}{2}\right)}{\left(1 + \sqrt{3} \operatorname{ctg} \frac{\alpha}{2}\right)^{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-082.jpg?height=364&width=389&top_left_y=671&top_left_x=716)

Fig. 24.