12.233. 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.
Solution.

Let $SO$ be the height of the regular pyramid $SABCD$ (Fig. 12.100), $\angle SAO=\alpha, MNKLM_1N_1K_1L_1$ be the cube from the problem's condition, $MNKL$ be its lower, and $M_1N_1K_1L_1$ be its upper base, $O_1$ be the intersection point of $SO$ and $M_1K_1$.

Let the edge of the cube be $a$.

$M_1K_1 \| AC, \angle SM_1O_1 = \angle SAO = \alpha$.

Then the volume of the cube $V_1 = a^3, M_1O_1 = \frac{a \sqrt{2}}{2}, SO_1 = \frac{a \sqrt{2}}{2} \tan \alpha$.

$SO = SO_1 + O_1O = a + \frac{a \sqrt{2}}{2} \tan \alpha = \frac{a \sqrt{2}}{2} \tan \alpha (\sqrt{2} \cot \alpha + 1)$

In $\triangle SOA (\angle SOA = 90^\circ)$:

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0982.jpg?height=751&width=828&top_left_y=75&top_left_x=256)

Fig. 12.100

\[
\begin{aligned}
& AO = SO \cot \angle SAO = \frac{a \sqrt{2}}{2} \tan \alpha (\sqrt{2} \cot \alpha + 1) \cot \alpha = \frac{a}{\sqrt{2}} (\sqrt{2} \cot \alpha + 1) \Rightarrow \\
& \Rightarrow S_{ABCD} = \frac{1}{2} AC^2 = 2 AO^2 = a^2 (\sqrt{2} \cot \alpha + 1)^2
\end{aligned}
\]

The volume of the pyramid

\[
\begin{aligned}
& V_2 = \frac{1}{3} S_{ABCD} \cdot SO = \frac{1}{3} a^2 (\sqrt{2} \cot \alpha + 1)^2 \cdot \frac{a (\sqrt{2} \cot \alpha + 1)}{\sqrt{2} \cot \alpha} = \\
& = \frac{a^3 (\sqrt{2} \cot \alpha + 1)^3}{3 \sqrt{2} \cot \alpha}
\end{aligned}
\]

Then \(\frac{V_1}{V_2} = \frac{3 \sqrt{2} \cot \alpha}{(\sqrt{2} \cot \alpha + 1)^3}\).

Answer: \(\frac{3 \sqrt{2} \cot \alpha}{(\sqrt{2} \cot \alpha + 1)^3}\).