[ Perpendicular Planes ] [Sphere Circumscribed around a Tetrahedron $]

In the triangular pyramid $S A B C$, the two equal lateral faces $A S B$ and $C S B$ are perpendicular to the plane of the base, and the face $A S C$ is inclined to the plane of the base at an angle $\beta$. Find the radius of the sphere circumscribed around the pyramid if the radius of the circle circumscribed around the base is $r$ and $\angle A B C=\alpha$.
The planes of the faces $ASB$ and $CSB$ are perpendicular to the plane of the base $ABC$ and intersect along the line $SB$. Therefore, the line $SB$ is perpendicular to the plane of the base $ABC$, i.e., $SB$ is the height of the pyramid $SABC$. From the equality of triangles $ASB$ and $CSB$, it follows that $AB = BC$. Therefore, triangle $ABC$ is isosceles. Let $K$ be the midpoint of $AC$. Then $BK$ is the bisector and the height of the isosceles triangle $ABC$. Therefore,

\[
\begin{gathered}
BK = BC \cos \angle KBC = BC \cos \frac{\frac{\alpha}{2}}{} = \\
= 2r \sin \angle BAC \cdot \cos \frac{\frac{\alpha}{2}}{2} = 2r \sin \left(90^{\circ} - \frac{\frac{\alpha}{2}}{2}\right) \cos \frac{\frac{\alpha}{2}}{2} = \\
= 2r \cos \frac{\frac{\alpha}{2}}{2} \cdot \cos \frac{\alpha}{2} = 2r \cos 2 \frac{\frac{\alpha}{2}}{2}
\end{gathered}
\]

Since $BK$ is the orthogonal projection of the inclined line $SK$ on the plane of the base $ABC$, by the theorem of three perpendiculars, $SK \perp AC$. Therefore, $BKS$ is the linear angle of the dihedral angle between the plane of the face $ASC$ and the plane of the base $ABC$. According to the problem, $\angle BKS = \beta$. From the right triangle $BKS$, we find that

\[
SB = BK \operatorname{tg} \angle BKS = 2r \cos 2 \frac{\frac{\alpha}{2}}{2} \operatorname{tg} \beta
\]

The center $O$ of the sphere circumscribed around the pyramid $SABC$ lies on the perpendicular to the plane of the base $ABC$ passing through the center $Q$ of the circle circumscribed around triangle $ABC$, and also in the plane perpendicular to the edge $SB$, passing through the midpoint $M$ of the segment $SB$. Let $R$ be the radius of this sphere. The lines $OQ$ and $SB$ are perpendicular to the same plane $ABC$, so $QD \parallel SB$. In the rectangle $OQBM$, it is known that

\[
OQ = MB = \frac{1}{2} SB = r \cos 2 \frac{\frac{\alpha}{2}}{2} \operatorname{tg} \beta, \quad QB = r.
\]

Therefore,

\[
R = OB = \sqrt{OQ^2 + QB^2} = \sqrt{r^2 \left(\cos^2 \frac{\alpha}{2} \operatorname{tg}^2 \beta + 1\right)} = r \sqrt{1 + \cos^2 \frac{\alpha}{2} \operatorname{tg}^2 \beta}
\]

## Answer

$\sqrt{1 + \cos^2 \frac{\alpha}{2} \operatorname{tg}^2 \beta}$