12.243. The lateral face of a regular truncated triangular pyramid forms an angle $\alpha$ with the plane of the base. Find the ratio of the total surface area of the pyramid to the surface area of a sphere inscribed in it.
Solution.

Let $M$ be the center of the lower, $M_{1}$ the center of the upper bases of the regular truncated pyramid $A B C A_{1} B_{1} C_{1}$ (Fig. 12.110), $O$ the center of the inscribed sphere, $O$ lies on $M M_{1}$, $K$ is the midpoint of $B C$, $K_{1}$ is the midpoint of $B_{1} C_{1}$. $\angle K_{1} K M=\alpha$, the point $F$ of tangency of the sphere with the face $B C C_{1} B_{1}$ lies on $K_{1} K$.

Consider the section of this set of bodies by the plane $A A_{1} K_{1} K$ (Fig. $12.110, \mathrm{a})$.

$\angle M K O=\frac{1}{2} \angle K_{1} K M=\frac{\alpha}{2} ; \angle M_{1} K_{1} O=\frac{1}{2} \angle M_{1} K_{1} F=\frac{1}{2}\left(180^{\circ}-\alpha\right)=90^{\circ}-\frac{\alpha}{2}$.

Let the radius of the sphere $O M=O M_{1}=r, B C=\alpha, B_{1} C_{1}=\mathrm{b}$.

$$
B \triangle O M K\left(\angle O M K=90^{\circ}\right): M K=r \operatorname{ctg} \frac{\alpha}{2} \Rightarrow a=2 \sqrt{3} \cdot M K=2 \sqrt{3} r \operatorname{ctg} \frac{\alpha}{2}
$$

$$
\begin{aligned}
& B \triangle O M_{1} K_{1}\left(\angle O M_{1} K_{1}=90^{\circ}\right): M_{1} K_{1}=r \operatorname{ctg}\left(90^{\circ}-\frac{\alpha}{2}\right)=r \operatorname{tg} \frac{\alpha}{2} \Rightarrow b= \\
& =2 \sqrt{3} \cdot M_{1} K_{1}=2 \sqrt{3} r \operatorname{tg} \frac{\alpha}{2} . \\
& S_{\triangle A B C}=\frac{a^{2} \sqrt{3}}{4}=3 \sqrt{3} r^{2} \operatorname{ctg}^{2} \frac{\alpha}{2} \leftrightarrow S_{\triangle A_{1} B_{1} C_{1}}=\frac{b^{2} \sqrt{3}}{4}=3 \sqrt{3} r^{2} \operatorname{tg}^{2} \frac{\alpha}{2} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0992.jpg?height=474&width=900&top_left_y=74&top_left_x=204)

Fig. 12.110,a

$$
\begin{aligned}
& K_{1} K=K_{1} F+F K=K_{1} M_{1}+K M=r \operatorname{tg} \frac{\alpha}{2}+r \operatorname{ctg} \frac{\alpha}{2}=r\left(\operatorname{tg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\alpha}{2}\right)= \\
& =\frac{r}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}=\frac{2 r}{\sin \alpha}
\end{aligned}
$$

The lateral surface area

$$
\begin{aligned}
& S_{\text {lateral }}=\frac{3}{2}(a+b) \cdot K_{1} K=\frac{3}{2}\left(2 \sqrt{3} r \operatorname{ctg} \frac{\alpha}{2}+2 \sqrt{3} r \operatorname{tg} \frac{\alpha}{2}\right) \cdot \frac{2 r}{\sin \alpha}= \\
& =\frac{6 \sqrt{3} r^{2}\left(\operatorname{ctg} \frac{\alpha}{2}+\operatorname{tg} \frac{\alpha}{2}\right)}{\sin \alpha}=\frac{12 \sqrt{3} r^{2}}{\sin ^{2} \alpha}
\end{aligned}
$$

The total surface area of the truncated pyramid

$$
\begin{aligned}
& S=S_{\text {lateral }}+S_{\triangle A B C}+S_{\Delta A_{1} B_{1} C_{1}}=\frac{12 \sqrt{3} r^{2}}{\sin ^{2} \alpha}+3 \sqrt{3} r^{2}\left(\operatorname{ctg}^{2} \frac{\alpha}{2}+\operatorname{tg}^{2} \frac{\alpha}{2}\right)= \\
& =\frac{12 \sqrt{3} r^{2}}{\sin ^{2} \alpha}+3 \sqrt{3} r^{2} \cdot \frac{\cos ^{4} \frac{\alpha}{2}+\sin ^{4} \frac{\alpha}{2}}{\cos ^{2} \frac{\alpha}{2} \sin ^{2} \frac{\alpha}{2}}= \\
& =\frac{12 \sqrt{3} r^{2}}{\sin ^{2} \alpha}+12 \sqrt{3} r^{2} \cdot \frac{\left(\cos ^{2} \frac{\alpha}{2}+\sin ^{2} \frac{\alpha}{2}\right)^{2}-2 \cos ^{2} \frac{\alpha}{2} \sin ^{2} \frac{\alpha}{2}}{\sin ^{2} \alpha}= \\
& =\frac{12 \sqrt{3} r^{2}\left(2-\frac{1}{2} \sin ^{2} \alpha\right)}{\sin ^{2} \alpha}=\frac{6 \sqrt{3} r^{2}\left(4-\sin ^{2} \alpha\right)}{\sin ^{2} \alpha}
\end{aligned}
$$

The surface area of the sphere $S_{u}=4 \pi r^{2}$.

Then

$$
\begin{aligned}
& \frac{S}{S_{u l}}=\frac{3 \sqrt{3}}{2 \pi} \cdot \frac{4-\sin ^{2} \alpha}{\sin ^{2} \alpha}=\frac{3 \sqrt{3}}{2 \pi}\left(\frac{4}{\sin ^{2} \alpha}-1\right)=\frac{3 \sqrt{3}}{2 \pi}\left(4\left(1+\operatorname{ctg}^{2} \alpha\right)-1\right)= \\
& =\frac{3 \sqrt{3}}{2 \pi}\left(4 \operatorname{ctg}^{2} \alpha+3\right)
\end{aligned}
$$

Answer: $\frac{3 \sqrt{3}}{2 \pi}\left(4 \operatorname{ctg}^{2} \alpha+3\right)$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0993.jpg?height=717&width=567&top_left_y=558&top_left_x=96)

Fig. 12.111