12.375. The base of the pyramid is a square with side $a$; two lateral faces of the pyramid are perpendicular to the base, and the larger lateral edge is inclined to the base plane at an angle $\beta$. A rectangular parallelepiped is inscribed in the pyramid; one of its bases lies in the plane of the pyramid's base, and the vertices of the other base lie on the lateral edges of the pyramid. Find the volume of the parallelepiped if its diagonal makes an angle $\alpha$ with the base plane.
Solution.

Square $ABCD$ is the base of the pyramid $SABCD$ (Fig. 12.375), the lateral faces $SBA$ and $SBC$ are perpendicular to the base plane, so their common edge $SB$ is the height of the pyramid.

$BA, BC, BD$ are the projections of the edges $SA, SC, SD$ on the base plane, $BA=BC<BD$. Thus, $SD$ is the larger lateral edge and $\angle SDB=\beta$.

$E_1, B_1, F_1, K_1$ are the vertices of the rectangular parallelepiped from the problem statement, lying on the edges $SA, SB, SC, SD$ of the pyramid. The lateral edge of the parallelepiped passing through vertex $B_1$ (segment $B_1B$) is part of the edge $SB$ of the pyramid.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1137.jpg?height=751&width=735&top_left_y=40&top_left_x=337)

Fig. 12.235

In the plane $SBD$, draw the perpendicular $K_1K$ to $BD$. Since the plane $SBD$ is perpendicular to the base plane of the pyramid, $K_1K$ is also perpendicular to this plane, so $K_1K$ is a lateral edge of the parallelepiped. The lateral edges $E_1E$ and $F_1F$ of the parallelepiped lie in the faces $SBA$ and $SBC$ of the pyramid.

$BK$ is the diagonal of the base of the parallelepiped, and since point $K$ lies on segment $BD$, $\angle ABK=45^\circ$ and $EBFK$ is a square.

$\angle B_1KB=\alpha$.

If $BB_1=KK_1=b$, then from $\Delta B_1BK\left(\angle B_1BK=90^\circ\right): BK=b \operatorname{ctg} \alpha$.

\[
\begin{aligned}
& BD=a \sqrt{2} ; K D=BD-BK=a \sqrt{2}-b \operatorname{ctg} \alpha . \\
& \Delta SBD \sim \Delta K_1KD \Rightarrow \frac{KK_1}{KD}=\frac{SB}{BD}=\operatorname{tg} \beta \Leftrightarrow \frac{b}{a \sqrt{2}-b \operatorname{ctg} \alpha}=\operatorname{tg} \beta \Leftrightarrow \\
& \Leftrightarrow b \operatorname{ctg} \beta=a \sqrt{2}-b \operatorname{ctg} \alpha \Leftrightarrow b=\frac{a \sqrt{2}}{\operatorname{ctg} \alpha+\operatorname{ctg} \beta}=\frac{a \sqrt{2} \sin \alpha \sin \beta}{\sin (\alpha+\beta)} \\
& EK=\frac{BK}{\sqrt{2}}=\frac{b \operatorname{ctg} \alpha}{\sqrt{2}}=\frac{a \sin \alpha \sin \beta}{\sin (\alpha+\beta)} \operatorname{ctg} \alpha=\frac{a \cos \alpha \sin \beta}{\sin (\alpha+\beta)}
\end{aligned}
\]

Volume of the parallelepiped

\[
V=EK^2 \cdot KK_1=\frac{a^2 \cos^2 \alpha \sin^2 \beta}{\sin^2(\alpha+\beta)} \cdot \frac{a \sqrt{2} \sin \alpha \sin \beta}{\sin (\alpha+\beta)}=
\]

\[
\begin{aligned}
& =\frac{a^3 \sqrt{2} \sin \alpha \cos^2 \alpha \sin^3 \beta}{\sin^3(\alpha+\beta)} \\
& \text{Answer: } \frac{a^3 \sqrt{2} \sin \alpha \cos^2 \alpha \sin^3 \beta}{\sin^3(\alpha+\beta)}
\end{aligned}
\]