11.143. A regular triangular prism is placed inside a regular tetrahedron such that the vertices of one of its bases lie on the lateral edges of the tetrahedron, and the other base lies in the plane of the tetrahedron's base. The edge of the tetrahedron is equal to \(a\). Determine the volume of the prism, given that all its edges are equal.
## Solution.

Let $SO$ be the height of the tetrahedron $SABC$, and $A_1B_1C_1A_2B_2C_2$ be the regular prism mentioned in the problem (Fig. 11.40). From $\triangle SOC (\angle SOC = 90^\circ): SO = \sqrt{SC^2 - OC^2} = \sqrt{a^2 - \frac{a^2}{3}} = \frac{a \sqrt{2}}{\sqrt{3}}$. Let the edge of the prism be $x$. Then $C_1O = \frac{x \sqrt{3}}{3}, CC_1 = CO - C_1O = \frac{a \sqrt{3}}{3} - \frac{x \sqrt{3}}{3} = \frac{a - x}{\sqrt{3}}, \Delta C_2C_1C \sim \Delta SOC$.

From the similarity, it follows: $\frac{C_1C_2}{C_1C} = \frac{SO}{OC}; C_1C_2 \cdot OC = C_1C \cdot SO; x \cdot \frac{a \sqrt{3}}{3} = \frac{a - x}{\sqrt{3}} \cdot \frac{a \sqrt{2}}{\sqrt{3}}; x \sqrt{3} = a \sqrt{2} - x \sqrt{2}; x = \frac{a \sqrt{2}}{\sqrt{3} + \sqrt{2}} = a \sqrt{2}(\sqrt{3} - \sqrt{2})$.

The volume of the prism

$$
\begin{gathered}
V = S_{\Delta A_1B_1C_1} \cdot C_1C_2 = \frac{x^2 \sqrt{3}}{4} \cdot x = \frac{x^3 \sqrt{3}}{4} = \frac{a^3 2 \sqrt{2}(\sqrt{3} - \sqrt{2})^3 \sqrt{3}}{4} = \\
= \frac{a^3 \sqrt{6}(3 \sqrt{3} - 9 \sqrt{2} + 6 \sqrt{3} - 2 \sqrt{2})}{2} = \frac{a^3 \sqrt{6}(9 \sqrt{3} - 11 \sqrt{2})}{2} = \frac{a^3 (27 \sqrt{2} - 22 \sqrt{3})}{2}.
\end{gathered}
$$

Answer: $\frac{a^3 (27 \sqrt{2} - 22 \sqrt{3})}{2}$.