3.94. A cube is inscribed in a regular quadrilateral pyramid; the vertices of its upper base lie on the lateral edges, and the vertices of its lower base are in the plane of the pyramid's base. Find the ratio of the volume of the cube to the volume of the pyramid, if the lateral edge of the pyramid makes an angle $\alpha$ with the base plane.
3.94. Given that $P A B C D$ is a regular quadrilateral pyramid, $A_{1} B_{1} C_{1} D_{1} A_{2} B_{2} C_{2} D_{2}$ is a cube inscribed in it, points $A_{1}, B_{1}, C_{1}, D_{1}$ lie on the lateral edges, and $A_{2}, B_{2}, C_{2}, D_{2}$ lie on the base of the pyramid, $P O \perp(A B C), \angle P A O=\alpha$ (Fig. 3.94); we need to find $V_{\text {cube }}: V_{\text {pyr }}$. Let $D_{1} D_{2}=x$; then $V_{\text {cube }}=x^{3}, O D_{2}=\frac{x \sqrt{2}}{2}$. From $\Delta D_{1} D_{2} D$ we find $D D_{2}=x \operatorname{ctg} \alpha ;$ further, we have $O D=\frac{x \sqrt{2}}{2}+x \operatorname{ctg} \alpha, P O=O D \operatorname{tg} \alpha=$ $=\frac{x(\sqrt{2}+2 \operatorname{ctg} \alpha)}{2} \cdot \operatorname{tg} \alpha$. Therefore,

$$
\begin{aligned}
V_{\text {pyr }}=\frac{1}{3} \cdot \frac{1}{2} D B^{2} \cdot P O & =\frac{2}{3} \cdot \frac{x^{2}(\sqrt{2}+2 \operatorname{ctg} \alpha)^{2} \cdot x(\sqrt{2}+2 \operatorname{ctg} \alpha) \operatorname{tg} \alpha}{4 \cdot 2}= \\
& =\frac{\sqrt{2} x^{3}(1+\sqrt{2} \operatorname{ctg} \alpha)^{3}}{6 \operatorname{ctg} \alpha}
\end{aligned}
$$

Thus,

$$
\frac{V_{\text {cube }}}{V_{\text {pyr }}}=\frac{x^{3} \cdot 6 \operatorname{ctg} \alpha}{\sqrt{2} x^{3}(1+\sqrt{2} \operatorname{ctg} \alpha)^{3}}=\frac{3 \sqrt{2} \operatorname{ctg} \alpha}{(1+\sqrt{2} \operatorname{ctg} \alpha)^{3}}
$$