12.430 The surface of the sphere inscribed in a regular truncated triangular pyramid relates to the total surface area of the pyramid as $\pi: 6 \sqrt{3}$. Find the angle between the lateral face and the base plane of the pyramid.
Solution. Let $P_{1}$ and $P_{2}$ be the centers of the bases of the given regular truncated pyramid $A_{1} B_{1} C_{1} A_{2} B_{2} C_{2}, \quad O$ be the center of the sphere with radius $R$ inscribed in the pyramid, and $D_{1}$ and $D_{2}$ be the midpoints of the edges $B_{1} C_{1}$ and $B_{2} C_{2}$. Denote the sides of the bases by $a$ and $b$.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-306.jpg?height=291&width=346&top_left_y=116&top_left_x=131)

\[
\begin{aligned}
& A_{2} D_{2} \perp B_{2} C_{2} \Rightarrow B_{2} C_{2} \perp D_{1} D_{2} \text { (Theorem 12) } \Rightarrow \\
& \Rightarrow B_{2} C_{2} \perp A_{1} D_{1} D_{2} A_{2} \text { (Theorem 11) } \Rightarrow \\
& \left.\Rightarrow A_{1} D_{1} D_{2} A_{2} \perp C_{1} B_{1} B_{2} C_{2} \text { (Theorem 15) } \Rightarrow \text { the radius }\right.
\end{aligned}
\]

of the sphere $O K$, drawn to the point of tangency of the sphere with the face $C_{1} B_{1} B_{2} C_{2}$, lies in the plane $A_{1} D_{1} D_{2} A_{2}$.

\[
\text { Thus, } \frac{a}{2 \sqrt{3}}=R_{1} D_{1}=D_{1} K, \frac{b}{2 \sqrt{3}}=P_{2} D_{2}=D_{2} K \quad \text { (Theorem 3) } \Rightarrow D_{1} D_{2}=\frac{a+b}{2 \sqrt{3}}
\]

\[
S_{C_{1} B_{1} B_{2} C_{2}}=\frac{(a+b)^{2}}{4 \sqrt{3}} \text {. Since } \angle D_{1} O D_{2}=90^{\circ} \text { (Theorem 9), then } R^{2}=O K^{2}=\frac{a b}{12} \text { (Theorem 5). }
\]

From this, the area $S_{n}$ of the surface of the pyramid is

\[
\frac{a^{2} \sqrt{3}}{4}+\frac{b^{2} \sqrt{3}}{4}+3 \cdot \frac{(a+b)^{2}}{4 \sqrt{3}}=\frac{\sqrt{3}}{2}\left[(b-a)^{2}+36 R^{2}\right]
\]

Considering the equality $4 \pi R^{2}: S_{n}=\pi: 6 \sqrt{3}$, we conclude:

\[
\frac{b-a}{2 R}=\sqrt{3} \Rightarrow \operatorname{ctg} \angle D_{1} D_{2} P_{2}=\frac{P_{2} D_{2}-P_{1} D_{1}}{P_{1} P_{2}}=\frac{b-a}{4 \sqrt{3} R}=\frac{1}{2} \Rightarrow \angle D_{1} D_{2} P_{2}=\operatorname{arctg} 2
\].

Answer: $\operatorname{arctg} 2$.