3.101. 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.
3.101. Given that $A B C A_{1} B_{1} C_{1}$ is a regular truncated triangular pyramid, $\angle B_{1} B C A=\alpha$, and $L$ is the center of the inscribed sphere (Fig. 3.101); we need to find the ratio $S_{\text {full trunc. pyr }}: S_{\text {sphere }}$. Let $A K$ and $A_{1} K_{1}$ be the heights of the bases; then $K_{1} K$ is the height of the lateral face, and $\angle A K K_{1}=\alpha$ as the linear angle of the dihedral angle $B_{1} B C A$. Further, let $O$ and $O_{1}$ be the centers of the bases of the truncated pyramid; then $O O_{1}$ is the diameter of the inscribed sphere. Let $L O=R$ and draw $K_{1} M \| O_{1} O$; we have

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-566.jpg?height=455&width=633&top_left_y=511&top_left_x=322)

Fig. 3.101

$K_{1} M=2 R, K_{1} K=\frac{2 R}{\sin \alpha}$ (from $\Delta K_{1} M K$). Assuming $A B=a, A_{1} B_{1}=b$, we find $O K=\frac{a \sqrt{3}}{6}, O_{1} K_{1}=\frac{b \sqrt{3}}{6}, M K=\frac{(a-b) \sqrt{3}}{6}=2 R \operatorname{ctg} \alpha$. According to the property of segments of tangents drawn from a single point to a circle, $K_{1} K=K_{1} O_{1}+K O=\frac{2 R}{\sin \alpha}$. Now, solving the system of equations

$$
\left\{\begin{array}{l}
\frac{(a-b) \sqrt{3}}{6}=2 R \operatorname{ctg} \alpha \\
\frac{(a+b) \sqrt{3}}{6}=\frac{2 R}{\sin \alpha}
\end{array}\right.
$$

we get $a=2 \sqrt{3} R \operatorname{ctg} \frac{\alpha}{2}, b=2 \sqrt{3} R \operatorname{tg} \frac{\alpha}{2}$. Therefore,

$$
\begin{gathered}
S_{\text {full trunc. pyr }}=\frac{3(a+b)}{2} \cdot \frac{2 R}{\sin \alpha}+\frac{a^{2} \sqrt{3}}{4}+\frac{b^{2} \sqrt{3}}{4}= \\
=\frac{3}{2} \cdot \frac{12 R}{\sqrt{3} \sin \alpha} \cdot \frac{2 R}{\sin \alpha}+3 R^{2} \sqrt{3}\left(\operatorname{ctg}^{2} \frac{\alpha}{2}+\operatorname{tg}^{2} \frac{\alpha}{2}\right)= \\
=\frac{12 R^{2} \sqrt{3}}{\sin ^{2} \alpha}+3 R^{2} \sqrt{3}\left(\frac{1+\cos \alpha}{1-\cos \alpha}+\frac{1-\cos \alpha}{1+\cos \alpha}\right)=\frac{3 R^{2} \sqrt{3}}{\sin ^{2} \alpha}\left(2+2 \cos ^{2} \alpha+4\right)= \\
=6 R^{2} \sqrt{3}\left(\frac{3}{\sin ^{2} \alpha}+\operatorname{ctg}^{2} \alpha\right)=6 R^{2} \sqrt{3}\left(3+4 \operatorname{ctg}^{2} \alpha\right) ; S_{\text {sphere }}=4 \pi R^{2} .
\end{gathered}
$$

Thus,

$$
\frac{S_{\text {full trunc. pyr }}}{S_{\text {sphere }}}=\frac{3 \sqrt{3}}{2 \pi}\left(3+4 \operatorname{ctg}^{2} \alpha\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-567.jpg?height=491&width=452&top_left_y=889&top_left_x=59)

Fig. 3.102