12.420 Find the radius of the sphere that touches the base and the lateral edges of a regular triangular pyramid, where the side of the base is equal to $a$, and the dihedral angle at the base is equal to $\alpha$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-299.jpg?height=342&width=381&top_left_y=109&top_left_x=133)

Let $O$ be the center of the sphere that touches the lateral edges $F A, F B, F C$ and the base of the given regular pyramid $F A B C$ at points $M, N, K$ and $P$ respectively. $\triangle O A M = \triangle O B N = \triangle O C K \Rightarrow O A = O B = O C \Rightarrow$ $\Rightarrow P$ is the center of the base $A B C$. Let $L$ be the midpoint of $B C$, then $B C \perp A L, B C \perp F L \Rightarrow \angle A L F = \alpha$ (by the problem statement).

Let $\angle F A L = 2 \beta$. Then $F P = A P \cdot \operatorname{tg} 2 \beta = P L \cdot \operatorname{tg} \alpha \Rightarrow$

$\Rightarrow 2 \operatorname{tg} 2 \beta = \operatorname{tg} \alpha (A P = 2 \cdot P L) \Rightarrow \frac{4 \operatorname{tg} \beta}{1 - \operatorname{tg}^{2} \beta} = \frac{1}{\operatorname{ctg} \alpha} \Rightarrow \operatorname{tg}^{2} \beta + 4 \operatorname{ctg} \alpha \cdot \operatorname{tg} \beta - 1 = 0 \Rightarrow$

$\Rightarrow \operatorname{tg} \beta = -2 \operatorname{ctg} \alpha + \sqrt{4 \operatorname{ctg}^{2} \alpha + 1} \Rightarrow O P = A P \cdot \operatorname{tg} \beta = \frac{a \sqrt{3}}{3} \left( -2 \operatorname{ctg} \alpha + \sqrt{4 \operatorname{ctg}^{2} \alpha + 1} \right)$.

Answer: $\frac{a \sqrt{3}}{3} \left( -2 \operatorname{ctg} \alpha + \sqrt{4 \operatorname{ctg}^{2} \alpha + 1} \right)$.