3. In an equilateral triangular pyramid (regular tetrahedron) with volume $V$, a regular equilateral triangular prism is inscribed such that three of its vertices lie on the base, and the other three on the lateral edges of the pyramid. Calculate the volume of the prism!
Solution. Let $a, H, b$ be the lengths of: the edge of the tetrahedron, the height of the tetrahedron, and the edge of the prism, respectively. Let $DT$ be the height dropped from vertex $D$ to the base $ABC$ (see diagram). From the similarity of triangles $ATD$ and $NT'D$, it follows that

$$
\overline{AT} : \overline{NT'} = \overline{DT} : \overline{T'D}
$$

Since $AB \parallel KL$, triangles $ABT$ and $KTL$ are similar, so we have

$$
\overline{AT} : \overline{KT} = \overline{AB} : \overline{KL}
$$

From the above two equalities and $\overline{KT} = \overline{NT}$, we get that $a : b = H : (H - b)$, which together with

![](https://cdn.mathpix.com/cropped/2024_06_05_ddd0b825d3ca510ab0f2g-4.jpg?height=581&width=495&top_left_y=267&top_left_x=971)
$H = \frac{a \sqrt{6}}{3}$, implies that $b = \frac{aH}{a + H} = a(\sqrt{6} - 2)$.

Thus, the volume $V_1$ of the prism is

$$
V_1 = \frac{b^3 \sqrt{3}}{4} = \frac{a^3 \sqrt{3}}{4}(\sqrt{6} - 2)^3
$$

Since $V = \frac{a^3 \sqrt{2}}{12}$, we get that

$$
\frac{V_1}{V} = \frac{3 \sqrt{3}(\sqrt{6} - 2)^3}{\sqrt{2}} \text{, i.e., } V_1 = 6V(27 - 11 \sqrt{6})
$$