12.438 The side of the base of a regular triangular pyramid is equal to $a$, and the dihedral angle at the base is $\alpha$. A right triangular prism is inscribed in this pyramid; three of its vertices lie on the apothems of the pyramid, and the other three - in the plane of the pyramid's base. Find the volume of the prism, if the center of the sphere inscribed in the pyramid lies in the plane of the upper base of the prism.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-311.jpg?height=311&width=393&top_left_y=113&top_left_x=180)

$C$

Let $O$ be the center of the sphere inscribed in the given regular pyramid $FABC$, $FP$ be its height, $FE$ and $FD$ be its apothems. Let $M$ and $N$ be two vertices of the upper base of the prism $T$ inscribed in the pyramid. Since $BC \perp PD$, $BC \perp FD \Rightarrow BC \perp FDP$ (Theorem 11) $\Rightarrow FDP \perp FBC$ (Theorem 15) $\Rightarrow$ the radius $OL$ of the sphere $O$, drawn to the point of tangency $L$ with the face $FBC$, belongs to $\triangle FDP \Rightarrow$

$\Rightarrow OD$ is the bisector of angle $FDP$, equal to $\alpha$ (by condition) $\Rightarrow$

$$
\Rightarrow FP = PD \cdot \tan \alpha = \frac{a}{2 \sqrt{3}} \cdot \tan \alpha, OP = \frac{a}{2 \sqrt{3}} \cdot \tan \frac{\alpha}{2}
$$

Since by condition $O$ lies in the plane of the upper base of the prism $T$, we have from the similarity: $\frac{MN}{ED} = \frac{FO}{FP} = 1 - \frac{OP}{FP} \Rightarrow MN = \frac{a}{2} \left(1 - \frac{\tan \frac{\alpha}{2}}{\tan \alpha}\right)$

Since the other sides of the base of the prism $T$ are calculated by the same formula (the third vertex of the upper base also lies on the apothem of the face $FAB$), they are equal to $MN \Rightarrow$

Answer: $V_{T} = MN^{2} \cdot \frac{\sqrt{3}}{4} \cdot OP = \frac{a^{3}}{32} \cdot \tan \frac{\alpha}{2} \left(1 - \frac{\tan \frac{\alpha}{2}}{\tan \alpha}\right)^{2} = \frac{a^{3} \cdot \sin \frac{\alpha}{2}}{128 \cdot \cos^{5} \frac{\alpha}{2}}$.