12.465 The ratio of the volume of a regular truncated triangular pyramid to the volume of a sphere inscribed in it is $k$. Find the angle between the lateral face of the pyramid and the plane of the base and the permissible values of $k$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-328.jpg?height=326&width=342&top_left_y=1299&top_left_x=82)

Let $P$ and $P_1$ be the centers of the bases of the given regular truncated pyramid, $M M_1$ be the apothem of one of its faces, $O$ be the center of the sphere inscribed in the pyramid with radius $R$. Let $a$ and $b$ be the sides of the bases of the pyramid, $x = \angle M_1 M P$, $P_1 M_1 = \frac{a}{2 \sqrt{3}}$, $P M = \frac{b}{2 \sqrt{3}}$,

$P_1 P = 2R$. Denote also $V_{\text{c}}$ as the volume of the sphere $O$, and then $\frac{V_{\text{c}}}{V_{\text{w}}} = k \Rightarrow \frac{\sqrt{3}(a^2 + ab + b^2)}{8 \pi R^2} = k \Rightarrow \frac{\sqrt{3}}{8 \pi R^2}((b-a)^2 + 3ab) = k$.

Draw the radius $OA$ of the sphere $O$ to the apothem $M_1 M$. Then $\angle M_1 O M = 90^\circ$ (Theorem 9) $\Rightarrow$
$\Rightarrow R^2 = OA^2 = M_1 A \cdot M A \quad$ (Theorem 5$) = P_1 M_1 \cdot P M \quad$ (Theorem 3$) = \frac{ab}{12}$.

Let $M_1 K$ be the height of the trapezoid $P_1 M_1 M P$. Then $\operatorname{ctg} x = \frac{MK}{M_1 K} = \frac{b-a}{4 \sqrt{3} R} \quad \left(MK = \frac{b-a}{2 \sqrt{3}}\right)$.

In the end, we have: $k = \frac{\sqrt{3}}{8 \pi} \left( \left( \frac{b-a}{R} \right)^2 + \frac{3ab}{R^2} \right) = \frac{36 \operatorname{ctg}^2 x + 27}{2 \sqrt{3} \pi} \Rightarrow \operatorname{ctg} x = \frac{\sqrt{2 \sqrt{3} \pi k - 27}}{6}$, where $2 \sqrt{3} \pi k - 27 > 0$.

Answer: $\operatorname{arcctg} \frac{\sqrt{2 \sqrt{3} \pi k - 27}}{6}$, where $k > \frac{9 \sqrt{3}}{2 \pi}$.

## APPLICATION OF EQUATIONS TO SOLVING PROBLEMS