12.200. The side of the base of a regular quadrilateral pyramid is equal to $a$, the lateral edge forms an angle $\alpha$ with the base plane. A cube is inscribed in this pyramid such that four of its vertices lie on the apothems of the pyramid, and four lie on the base of the pyramid. Find the edge of the cube.
## Solution.

Let in the regular pyramid $S A B C D$ (Fig. 12.67) $A B=a, S O-$ be the height, $\angle S C O=\alpha, S E$ be the apothem, and $F$ and $F_{1}$ be the vertices of the cube mentioned in the problem. Then the edge of the cube $F F_{1}$ is perpendicular to $O E$. $C O=\frac{a \sqrt{2}}{2}, O E=\frac{a}{2} \cdot \mathrm{B} \triangle \operatorname{SOC}(\angle S O C=\pi 2): S O=C O \operatorname{tg} \angle S O C=\frac{a \sqrt{2}}{2} \operatorname{tg} \alpha$. Let the edge of the cube be $x$. Then $O F=\frac{x \sqrt{2}}{2}, E F=O E-O F=\frac{a-x \sqrt{2}}{2}$.

The right triangles $S O E$ and $F_{\mathrm{I}} F E$ are similar, therefore

$$
\begin{aligned}
& \frac{F F_{1}}{S O}=\frac{F E}{O E} \Rightarrow \frac{x}{\frac{a \sqrt{2}}{2} \operatorname{tg} \alpha}=\frac{\frac{a-x \sqrt{2}}{2}}{\frac{a}{2}} \Leftrightarrow x=\frac{\sqrt{2} \operatorname{tg} \alpha}{2}(a-x \sqrt{2}) \Leftrightarrow \\
& \Leftrightarrow x=\frac{a \sqrt{2} \operatorname{tg} \alpha}{2(1+\operatorname{tg} \alpha)}
\end{aligned}
$$

$$
\frac{a \sqrt{2} \operatorname{tg} \alpha}{2(1+\operatorname{tg} \alpha)}=\frac{a \sqrt{2} \operatorname{tg} \alpha}{2\left(\operatorname{tg} \frac{\pi}{4}+\operatorname{tg} \alpha\right)}=\frac{a \sqrt{2} \operatorname{tg} \alpha \cos \frac{\pi}{4} \cos \alpha}{2 \sin \left(\frac{\pi}{4}+\alpha\right)}=\frac{a \sin \alpha}{2 \sin \left(\frac{\pi}{4}+\alpha\right)}
$$

$$
\text { Answer: } \frac{a \sin \alpha}{2 \sin \left(\frac{\pi}{4}+\alpha\right)}
$$