11.219 The diagonals of two identical cubes with an edge equal to $a$ lie on the same straight line. A vertex of the second cube coincides with the center of the first, and the second cube is rotated around the diagonal by $60^{\circ}$ relative to the first. Find the volume of the common part of these cubes.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-271.jpg?height=385&width=435&top_left_y=561&top_left_x=161)

Let the first cube be denoted as $T_{1}$, the second cube as $T_{2}=A B C D K L M N$, and after rotation, the second cube as $T_{2}^{n}$. The common part of the cubes $T_{1}$ and $T_{2}$ is denoted as $T_{3}$, and the common part of $T_{1}$ and $T_{2}^{n}$ as $T_{3}^{n}$.

Let points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}$ be the midpoints of the edges $M N, M C, B C, B A, K A, K N$, respectively. Then $P_{1} L = P_{1} D = \frac{a \sqrt{5}}{2}$.

If $R$ is the center of the cube $T_{2}$, then $P_{1} R \perp L D$, since $\triangle L P_{1} D$ is isosceles.

From this, $R_{1} R = \sqrt{R D^{2} - R D^{2}} = \frac{a}{\sqrt{2}}$. Similarly, $P_{i} R \perp L D, P_{i} R = \frac{a}{\sqrt{2}}$, where $i=2,3,4,5,6$. Moreover, $P_{1} P_{2} = \sqrt{2 \cdot M P_{1}^{2}} = \frac{a}{\sqrt{2}} \Rightarrow \Delta R R_{1} P_{2}$ is equilateral $\Rightarrow$ $\angle R R P_{2} = 60^{\circ}$. Similarly, $\angle P_{i} R P_{i+1} = 60^{\circ}, i=2,3,4,5$ and $\angle P_{6} R P_{1} = 60^{\circ}$.

From this, we conclude: when $T_{2}$ is rotated by $60^{\circ}$ around $L D$ clockwise (if viewed from $L$ to $D$), point $P_{i}$ will move to $P_{i+1}, i=1,2,3,4,5$, and $P_{6}$ will move to $P_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-271.jpg?height=387&width=435&top_left_y=1295&top_left_x=491)

Since a plane is uniquely determined by its three non-collinear points, considering the immobility of points $L$ and $D$, we conclude: the face $K L M N$ will pass through points $L, P_{1}, P_{2}$ (since $L, P_{6}, P_{1}$ will move to $L, P_{1}, P_{2}$), the face $D C M N$ will pass through points $D, P_{2}, P_{3}$, the face $L M C B$ will pass through $L, P_{3}, P_{4}$, the face $D A B C$ will pass through $D, P_{4}, P_{5}$, the face $L K A B$ will pass through $L, P_{5}, P_{6}$, and the face $D A K N$ will pass through $D, P_{6}, P_{1}$.

Obviously, three faces of the cube $T_{2}^{n}$ with a common vertex $D$ do not intersect the cube $T_{1}$. Therefore, the three faces of the cube $T_{2}^{n}$ with a common vertex $L$, which we denote as $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$, are of interest.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-272.jpg?height=550&width=542&top_left_y=601&top_left_x=381)

Now consider the cube $T_{1}$ and the cube $T_{3}=A_{1} B_{1} C_{1} D_{1} K_{1} L_{1} M_{1} N_{1}$. From the similarity, it follows that the faces $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$ of the cube $T_{2}^{n}$ intersect the midpoints of the corresponding edges of the cube $T_{3}$ at points $P_{i}^{\prime}, i=1,2,3,4,5,6$. Extend the segments $P_{1}^{\prime} P_{2}^{\prime}, P_{5}^{\prime} P_{6}^{\prime}, P_{3}^{\prime} P_{4}^{\prime}$ until they intersect the edges of the cube $T_{1}$ at points $E, F, I$. Then the body $L_{1} D_{1} E I F$ is the common part $T_{3}^{n}$. Since the trihedral angles $L_{1} E I F$ and $D_{1} E I F$ are formed by the faces of the cubes $T_{2}^{n}$ and $T_{1}$, respectively, their planar angles are right angles. Further, $D_{1} E = D_{1} I = D_{1} F = \frac{3}{4} a$ (since $\left.A_{1} F = C_{1} E = N_{1} I = B_{1} P_{3}^{\prime} = \frac{a}{4}\right) \Rightarrow V_{D_{1} E I F} = \frac{1}{6} D_{1} E \quad D_{1} F \cdot D_{1} I = \frac{9 a^{3}}{128}$.

$\Delta D_{1} F I = \Delta L_{1} F I$ (isosceles right triangles with equal hypotenuses) $\Rightarrow L_{1} F = D_{1} F = \frac{3 a}{4} \Rightarrow \Delta L_{1} F E = \Delta D_{1} F E \Rightarrow$

$$
\Rightarrow L_{1} E = D_{1} E = \frac{3 a}{4} \Rightarrow V_{L_{1} E I F} = V_{D_{1} E I F} = \frac{9 a^{3}}{128} \Rightarrow V_{T_{3}^{n}} = \frac{9 a^{3}}{64}
$$

Answer: $\frac{9 a^{3}}{64}$.