12.255. Two faces of a triangular pyramid are equal right-angled triangles with a common leg equal to $d$. The angle between these faces is $\alpha$. Two other faces of the pyramid form a dihedral angle $\beta$. Find the radius of the sphere circumscribed about the pyramid.
## Solution.

Let the faces $SAB$ and $SAC$ of the pyramid $SABC$ (Fig. 12.121) be equal, $\angle SAB = \angle SAC = 90^\circ, SA = d$. Then the segment $SA$ is perpendicular to the plane of the base of the pyramid and is the height of the pyramid, $\triangle BAC$ is isosceles. If $D$ is the midpoint of $BC$, then $AD \perp BC, DS \perp BC$, $\angle SDA$ is the linear angle of the dihedral angle formed by the faces $BSC$ and $BAC$ of the pyramid, $\angle SDA = \beta. BA \perp SA, CA \perp SA \Rightarrow \angle BAC$ is the angle between the faces $SAB$ and $SAC, \angle BAC = \alpha$.

$O_1$ is the center of the circle circumscribed around $\triangle ABC, E$ is the midpoint of $SA$, $EA = \frac{d}{2}$.

The center $O$ of the sphere circumscribed around the pyramid is the point of intersection of the perpendicular dropped from the plane $ABC$ at point $O_1$ and the perpendicular bisector of the segment $SA$ lying in the plane $SAD, OA$ is the radius of the sphere.

In the quadrilateral $AEOO_1$, the angles at vertices $A, E, O_1$ are right angles, so $AEOO_1$ is a rectangle and $OA = EO_1$.

In $\triangle SAD (\angle SAD = 90^\circ): AD = SA \operatorname{ctg} \angle SDA = d \operatorname{ctg} \beta$.

In $\triangle ADB (\angle ADB = 90^\circ): BD = AD \operatorname{tg} \angle BAD = d \operatorname{ctg} \beta \operatorname{tg} \frac{\alpha}{2}$.

$$
BC = 2BD = 2d \operatorname{ctg} \beta \operatorname{tg} \frac{\alpha}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1008.jpg?height=773&width=747&top_left_y=78&top_left_x=305)

Fig. 12.122

$O_1A$ is the radius of the circle circumscribed around $\triangle ABC$. Then

$$
O_1A = \frac{BC}{2 \sin \angle BAC} = \frac{2d \operatorname{ctg} \beta \operatorname{tg} \frac{\alpha}{2}}{2 \sin \alpha} = \frac{d \operatorname{ctg} \beta \operatorname{tg} \frac{\alpha}{2}}{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = \frac{d \operatorname{ctg} \beta}{2 \cos^2 \frac{\alpha}{2}}
$$

In $\triangle EAO_1 (\angle EAO_1 = 90^\circ):$

$$
\begin{gathered}
EO_1 = \sqrt{EA^2 + AO_1^2} = \sqrt{\frac{d^2}{4} + \frac{d^2 \operatorname{ctg} \beta}{4 \cos^4 \frac{\alpha}{2}}} = \frac{d}{2 \cos^2 \frac{\alpha}{2}} \sqrt{\cos^4 \frac{\alpha}{2} + \operatorname{ctg}^2 \beta} \\
\text{Answer: } \frac{d}{2 \cos^2 \frac{\alpha}{2}} \sqrt{\cos^4 \frac{\alpha}{2} + \operatorname{ctg}^2 \beta}
\end{gathered}
$$