12.459 Find the radius of the sphere inscribed in a regular triangular pyramid, the height of which is $H$, and the angle between the lateral edge and the base plane is $\alpha$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-325.jpg?height=355&width=394&top_left_y=273&top_left_x=192)

Let $O$ be the center of the sphere of radius $x$ inscribed in the given regular pyramid $F A B C$, $F K$ the height of the pyramid, and $D$ the midpoint of $B C$. Then $B C \perp F D, B C \perp K D \Rightarrow B C \perp F D K \quad$ (Theorem 11) $\Rightarrow F B C \perp F D K$ (Theorem 15) $\Rightarrow$ the point of tangency $M$ of the sphere $O$ with the face $F B C$ lies on the apothem $F D$.

Thus,

$$
\begin{aligned}
& K D=\frac{1}{2} B K=\frac{1}{2} H \cdot \operatorname{ctg} \alpha \Rightarrow \\
& \Rightarrow \dot{F D}=\sqrt{K D^{2}+F K^{2}}=\frac{H}{2} \sqrt{4+\operatorname{ctg}^{2} \alpha}
\end{aligned}
$$

$\Delta F M O \sim \triangle F D K \Rightarrow \frac{x}{K D}=\frac{H-x}{F D} \Rightarrow x=\frac{H}{1+\sqrt{4 \operatorname{tg}^{2} \alpha+1}}=\frac{H}{4 \operatorname{tg}^{2} \alpha}\left(\sqrt{4 \operatorname{tg}^{2} \alpha+1}-1\right)$. Answer: $\frac{H}{4 \operatorname{tg}^{2} \alpha}\left(\sqrt{4 \operatorname{tg}^{2} \alpha+1}-1\right)$.