11.205 Two regular tetrahedra are joined by two faces to form a double pyramid. The centers of the six lateral faces of this double pyramid are taken as the vertices of a right triangular prism. Calculate the volume of this prism if the edge of the tetrahedron is $a$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-263.jpg?height=420&width=391&top_left_y=129&top_left_x=147)

Let $M, K, N$ be the centers of three faces of the given double pyramid $F F^{\prime} A B C$, and $O$ be the center of $\triangle A B C$. Let $P$ and $D$ be the midpoints of $A C$ and $B C$. Then $M K$ is a side of the base of the desired prism, and $M N$ is its lateral edge. $M N \| F F^{\prime} \Rightarrow M N \perp A B C \Rightarrow$

$\Rightarrow$ the desired prism is a right prism (its bases are parallel to $A B C$). From the similarity of triangles, we have:

$\frac{M N}{F F^{\prime}}=\frac{P M}{P F}=\frac{1}{3}, \quad \frac{M K}{P D}=\frac{M F}{P F}=\frac{2}{3} . \quad$ But $P D=\frac{a}{2}$

(median of $\triangle A B C$),

$F F^{\prime}=2 F O=2 \sqrt{P F^{2}-P O^{2}}=\frac{2 a \sqrt{2}}{\sqrt{3}}$.

$$
\text { Therefore, the volume of the desired prism is } \frac{M N \cdot M K^{2} \sqrt{3}}{4}=\frac{\sqrt{2} a^{3}}{54} \text {. }
$$

Answer: $\frac{\sqrt{2} a^{3}}{54}$.