12.334. A sphere is inscribed in a regular quadrilateral pyramid. The distance from the center of the sphere to the vertex of the pyramid is $a$, and the angle between the lateral face and the base plane is $\alpha$. Find the total surface area of the pyramid.
Solution.

Let $L O$ be the height of the regular pyramid $L A B C D$, and $E$ be the midpoint of $B C$ (Fig. 12.195). Then $\angle M E O=\alpha$, the center $F$ of the inscribed sphere lies on the segment $L O$, $\angle O E F=\angle L E F=\frac{\alpha}{2} \cdot \angle L F E$ is the external angle of triangle $F O E$, $\angle L F E=\angle F O E+\angle O E F=90^{\circ}+\frac{\alpha}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1093.jpg?height=802&width=781&top_left_y=51&top_left_x=284)

Fig. 12.196

In $\triangle L F E$:

$\frac{L F}{\sin \angle L E F}=\frac{L E}{\sin \angle L F E} \Rightarrow L E=\frac{L F \sin \angle L F E}{\sin \angle L E F}=\frac{a \sin \left(90^{\circ}+\frac{\alpha}{2}\right)}{\sin \frac{\alpha}{2}}=a \operatorname{ctg} \frac{\alpha}{2}$.

In $\triangle L O E\left(\angle L O E=90^{\circ}\right): O E=L E \cos \angle L E O=a \operatorname{ctg} \frac{\alpha}{2} \cos \alpha$.

The side of the base of the pyramid $A B=2 \cdot O E=2 a \operatorname{ctg} \frac{\alpha}{2} \cos \alpha$, the area of its base $S_{\text {base}}=A B^{2}=4 a^{2} \operatorname{ctg}^{2} \frac{\alpha}{2} \cos ^{2} \alpha$.

The total surface area of the pyramid:

$$
\begin{aligned}
& S=S_{\text {base}}+S_{\sigma}=S_{\text {base}}+\frac{S_{\text {base}}}{\cos \alpha}=S_{\text {base}} \cdot \frac{\cos \alpha+1}{\cos \alpha}=4 a^{2} \operatorname{ctg}^{2} \frac{\alpha}{2} \cos ^{2} \alpha \cdot \frac{2 \cos ^{2} \frac{\alpha}{2}}{\cos \alpha}= \\
& =8 a^{2} \cos \alpha \cos ^{2} \frac{\alpha}{2} \operatorname{ctg}^{2} \frac{\alpha}{2}
\end{aligned}
$$

Answer: $8 a^{2} \cos \alpha \cos ^{2} \frac{\alpha}{2} \operatorname{ctg}^{2} \frac{\alpha}{2}$.