59. The edge of a cube and the edge of a regular tetrahedron lie on the same line, and the midpoints of the opposite edges of the cube and the tetrahedron coincide. Find the volume of the common part of the cube and the tetrahedron, if the edge of the cube is equal to \(a\).
59. Let $K$ be the midpoint of edge $AB$ of the cube $ABCD A_{1} B_{1} C_{1} D_{1}$, and $M$ be the midpoint of edge $D_{1} C_{1}$. $K$ and $M$ are simultaneously the midpoints of edges $PQ$ and $RS$ of the regular tetrahedron $PQRS$. $D_{1} C_{1}$ lies on $RS$. If the edge of the tetrahedron is $b$, then $|MK| = b \sqrt{2} / 2 = a \sqrt{2}$. Therefore, $b = 2a$.

Project the tetrahedron onto the plane $ABCD$ (Fig. 3), $P_{1}, Q_{1}, R_{1}, S_{1}$ are the projections of $P, Q, R, S$. Since $PQ$ forms a $45^{\circ}$ angle with this plane, the length of $P_{1} Q_{1}$ is $a \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-054.jpg?height=315&width=389&top_left_y=141&top_left_x=701)

Fig. 3. 

Let $L$ be the intersection point of line $AB$ and line $P_{1} R_{1}$. From the similarity of triangles $P_{1} L K$ and $P_{1} R_{1} M_{1}$, we find

$$
|LK| = \frac{|R_{1} M_{1}| \cdot |P_{1} K|}{|P_{1} M_{1}|} = \frac{a}{1 + \sqrt{2}} < \frac{a}{2}
$$

Thus, the edge of the tetrahedron $PR$ (and, consequently, the other edges: $PS, QR$, and $QS$) intersects the cube.

To compute the volume of the resulting body, it is convenient to consider this body as a tetrahedron with truncated corners.

Answer: $\frac{a^{3} \sqrt{2}}{12}(16 \sqrt{2} - 17)$.