12.349. The radius of the sphere inscribed in a regular triangular pyramid is four times smaller than the side of the pyramid's base. Find the cosine of the dihedral angle at the pyramid's vertex.
## Solution.

Let $E O_{2}$ be the height of the regular pyramid $E A B C$ (Fig. 12.210), and $E D$ be the apothem of its lateral face $B E C$. If $\angle B E C=\alpha$, then $\angle D E C=\frac{\alpha}{2}$.

Since the pyramid is regular, the center $O_{1}$ of the inscribed sphere lies on its height $E O_{2}, O_{1} O_{2}$ is the radius of the sphere. If $B C=a, \angle O_{2} D O_{1}=\beta$, then $O_{2} D=\frac{a \sqrt{3}}{6} ; O_{1} O_{2}=\frac{a}{4}$, from $\triangle O_{1} O_{2} D\left(\angle O_{1} O_{2} D=90^{\circ}\right)$ :

$\operatorname{tg} \beta=\frac{O_{1} O_{2}}{O_{2} D}=\frac{\sqrt{3}}{2}$.

$\angle E D O_{2}=2 \angle O_{1} D O_{2}=2 \beta ; \cos 2 \beta=\frac{1-\operatorname{tg}^{2} \beta}{1+\operatorname{tg}^{2} \beta}=\frac{1}{7}$.

In $\triangle E O_{2} D\left(\angle E O_{2} D=90^{\circ}\right): E D=\frac{O_{2} D}{\cos \angle E D O_{2}}=\frac{a \sqrt{3}}{6 \cos 2 \beta}=\frac{7 a \sqrt{3}}{6}$.
In $\triangle E D C\left(\angle E D C=90^{\circ}\right):$

$$
\operatorname{tg} \frac{\alpha}{2}=\frac{C D}{E D}=\frac{a}{2}: \frac{7 a \sqrt{3}}{6}=\frac{\sqrt{3}}{7} \Rightarrow \cos \alpha=\frac{1-\operatorname{tg}^{2} \frac{\alpha}{2}}{1+\operatorname{tg}^{2} \frac{\alpha}{2}}=\frac{1-\frac{3}{49}}{1+\frac{3}{49}}=\frac{23}{26}
$$

Answer: $\frac{23}{26}$.