12.212. The base of the pyramid is an isosceles triangle with equal sides of length $b$; the corresponding lateral faces are perpendicular to the base plane and form an angle $\alpha$ with each other. The angle between the third lateral face and the base plane is also $\alpha$. Find the radius of the sphere inscribed in the pyramid.
## Solution.

Let the lateral faces $SAB$ and $SAC$ of the pyramid $SABC$ (Fig. 12.79) be perpendicular to the plane $ABC$ of its base. Then their common edge $SA$ is the height of the pyramid. In $\triangle ABC$, $AB = AC = b$, $\angle BAC = \alpha$.

If $E$ is the midpoint of $BC$, then $\angle BAE = \frac{\alpha}{2}$, $AE \perp BC \Rightarrow SE \perp BC$ and $\angle SEA$ is the angle of inclination of the face $BSC$ to the plane of the base, $\angle SEA = \alpha$.

Let $O$ be the center of the sphere inscribed in the given pyramid. Since the sphere touches the faces of the dihedral angle with edge $BC$, $O$ lies on the bisector of the linear angle $SEA$ of this dihedral angle.

Let $M$ and $K$ be the points of tangency of the sphere with the faces $ABC$ and $SAB$ of the pyramid, respectively. Thus, $OM \perp ABC$, $OK \perp SAB$, $OM = OK = r$, where $r$ is the desired radius of the sphere.

In the plane $SAB$, draw the perpendicular $KD$ to the line $AB$. Since the planes $SAB$ and $ABC$ are perpendicular, $KD \perp ABC \Rightarrow KD \perp MD$. In the quadrilateral $OKDM$, the angles at vertices $K$, $D$, and $M$ are right angles, so it is a rectangle, and since $OK = OM$, $OKDM$ is a square, $DM = OM = r$.

In $\triangle AEB \left(\angle AEB = 90^\circ\right): AE = AB \cos \angle BAE = b \cos \frac{\alpha}{2}$.

In $\triangle ADM \left(\angle ADM = 90^\circ\right): AM = \frac{DM}{\sin \angle BAE} = \frac{r}{\sin \frac{\alpha}{2}}$.

In $\triangle OME \left(\angle OME = 90^\circ\right): ME = OM \cot \angle OEM = r \cot \frac{\alpha}{2}$.

$$
AM + ME = AE \Rightarrow \frac{r}{\sin \frac{\alpha}{2}} + \frac{r \cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}} = b \cos \frac{\alpha}{2} \Leftrightarrow r = \frac{b \cos \frac{\alpha}{2} \sin \frac{\alpha}{2}}{1 + \cos \frac{\alpha}{2}} ; r = \frac{b \sin \alpha}{4 \cos^2 \frac{\alpha}{4}}
$$

Answer: $\frac{b \sin \alpha}{}$

$4 \cos^2 \frac{\alpha}{4}$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0961.jpg?height=787&width=634&top_left_y=81&top_left_x=81)

Fig. 12.80