12.258. The side of the square lying at the base of a regular quadrilateral pyramid is equal to $a$. A regular quadrilateral prism is inscribed in the pyramid; the vertices of the upper base lie on the lateral edges, and the vertices of the lower base lie in the plane of the pyramid's base. The diagonal of the prism forms an angle $\varphi$ with the base plane. Find the volume of the prism if the lateral edge of the pyramid forms an angle $\alpha$ with the base plane.
Solution.

Let $EO$ be the height of the regular pyramid $EABCD$, $AD=\alpha$, $\angle ECA=\alpha$, $MNKLM_1N_1K_1L_1$ be a regular prism, $\angle M_1KM=\varphi$ (Fig. 12.124). The vertices of the lower base of the prism lie on the diagonals of the square $ABCD$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1011.jpg?height=726&width=929&top_left_y=75&top_left_x=347)

Fig. 12.124

If $MN=b$, then $MK=b\sqrt{2}$.

$\Delta AMM_1 = \Delta CKK_1$ - by the leg and acute angle ($MM_1 = KK_1$, $\angle M_1AM = \angle K_1CK = \alpha$).

Hence, $CK = \frac{AC - MK}{2} = \frac{(a - b)\sqrt{2}}{2}$.

In $\Delta M_1MK (\angle M_1MK = 90^\circ): MM_1 = MK \tan \angle M_1KM = b\sqrt{2} \tan \varphi$.

In $\Delta K_1KC (\angle K_1KC = 90^\circ): CK = K_1K \cot \angle K_1CK = b\sqrt{2} \tan \varphi \cot \alpha \Rightarrow$

$\Rightarrow \frac{a - b}{2} = b \tan \varphi \cot \alpha; b = \frac{a}{1 + 2 \tan \varphi \cot \alpha} = \frac{a \cot \varphi}{\cot \varphi + 2 \cot \alpha}$.

Volume of the prism

\[
\begin{aligned}
& V = S_{MNKL} \cdot MM_1 = b^3 \sqrt{2} \tan \varphi = \frac{a^3 \cot^3 \varphi}{(\cot \varphi + 2 \cot \alpha)^3} \cdot \sqrt{2} \tan \varphi = \\
& = \frac{a^3 \sqrt{2} \cot^2 \varphi}{(\cot \varphi + 2 \cot \alpha)^3}. \\
& \text{Answer: } \frac{a^3 \sqrt{2} \cot^2 \varphi}{(\cot \varphi + 2 \cot \alpha)^3}.
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1012.jpg?height=682&width=874&top_left_y=71&top_left_x=255)

Fig. 12.125