12.348. The side of the base of a regular triangular pyramid is equal to $a$, and the plane angle at the vertex of the pyramid is $\alpha$. Find the radius of the sphere inscribed in the pyramid.
Solution.

Let $EO_2$ be the height of the regular tetrahedron $EABC$ (Fig. 12.209), $ED$ the apothem of its lateral face $BEC$. Then $\angle BED = \frac{1}{2} \angle BEC = \frac{\alpha}{2}$. Since the pyramid is regular, the center $O_1$ of the inscribed sphere lies on the height $EO_2$. If $\angle EDO_2 = \gamma$, then $\angle O_1DO_2 = \frac{\gamma}{2}$, and from $\Delta O_1O_2D \left(\angle O_1O_2D = \frac{\pi}{2}\right)$ the radius of the inscribed sphere $r = O_2O_1 = O_2D \operatorname{tg} \angle O_1DO_2 = \frac{a \sqrt{3}}{6} \operatorname{tg} \frac{\gamma}{2}$.

In $\triangle EDB \left(\angle EDB = \frac{\pi}{2}\right): ED = BD \operatorname{ctg} \angle BED = \frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}$.

In $\triangle EOD \left(\angle EOD = \frac{\pi}{2}\right): ED = \frac{OD}{\cos \angle EDO} = \frac{a \sqrt{3}}{6 \cos \gamma} \Rightarrow \frac{a}{2} \operatorname{ctg} \frac{\alpha}{2} = \frac{a \sqrt{3}}{6 \cos \gamma} ; \cos \gamma = \frac{\sqrt{3}}{3} \operatorname{tg} \frac{\alpha}{2} = \operatorname{tg} \frac{\pi}{6} \operatorname{tg} \frac{\alpha}{2}$.

$\operatorname{tg} \frac{\gamma}{2} = \sqrt{\frac{1 - \cos \gamma}{1 + \cos \gamma}} = \sqrt{\frac{1 - \operatorname{tg} \frac{\pi}{6} \operatorname{tg} \frac{\alpha}{2}}{1 + \operatorname{tg} \frac{\pi}{6} \operatorname{tg} \frac{\alpha}{2}}} = \sqrt{\frac{\cos \left(\frac{\pi}{6} + \frac{\alpha}{2}\right)}{\cos \left(\frac{\pi}{6} - \frac{\alpha}{2}\right)}} = \sqrt{\frac{\sin \left(\frac{\pi}{3} - \frac{\alpha}{2}\right)}{\sin \left(\frac{\pi}{3} + \frac{\alpha}{2}\right)}} \Rightarrow$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1108.jpg?height=214&width=525&top_left_y=1498&top_left_x=206)

Answer: $\frac{a}{6} \sqrt{\frac{3 \sin \left(\frac{\pi}{3} - \frac{\alpha}{2}\right)}{\sin \left(\frac{\pi}{3} + \frac{\alpha}{2}\right)}}$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1109.jpg?height=631&width=571&top_left_y=64&top_left_x=94)

Fig. 12.210

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1109.jpg?height=695&width=618&top_left_y=14&top_left_x=658)

Fig. 12.211