12.350. The lateral edges and two sides of the base of a triangular pyramid have the same length $a$, and the angle between the equal sides of the base is $\alpha$. Find the radius of the circumscribed sphere.
## Solution.

Let $DO$ be the height of the pyramid $DABC$, with $DA = DB = DC = BA = BC = a$ (Fig. 12.211). Then point $O$ is the center of the circle circumscribed around the base of the pyramid, and the points on the line $DO$ are equidistant from the vertices of $\triangle ABC$. The radius of this circle is

$$
\begin{aligned}
& \begin{array}{l}
\text { of this circle } OA = R_1 = \frac{BC}{2 \sin \angle BAC} = \frac{a}{2 \sin \left(\frac{\pi}{2} - \frac{\alpha}{2}\right)} = \frac{a}{2 \cos \frac{\alpha}{2}}. \\
\text { In } \triangle AOD \left(\angle AOD = \frac{\pi}{2}\right):
\end{array} \\
& DO = \sqrt{DA^2 - OA^2} = \sqrt{a^2 - \frac{a^2}{4 \cos^2 \frac{\alpha}{2}}} = \frac{a}{\cos \frac{\alpha}{2}} \sqrt{\cos^2 \frac{\alpha}{2} - \frac{1}{4}} = \\
& = \frac{a}{\cos \frac{\alpha}{2}} \sqrt{\cos^2 \frac{\alpha}{2} - \cos^2 \frac{\pi}{3}} = \frac{a}{\cos \frac{\alpha}{2}} \sqrt{\frac{1 + \cos \alpha}{2} - \frac{1 + \cos \frac{2\pi}{3}}{2}} = \\
& = \frac{a}{\cos \frac{\alpha}{2}} \sqrt{\frac{\cos \alpha - \cos \frac{2\pi}{3}}{2}} = \frac{a \sqrt{\sin \left(\frac{\pi}{3} + \frac{\pi}{2}\right) \sin \left(\frac{\pi}{3} - \frac{\alpha}{2}\right)}}{\cos \frac{\alpha}{2}}
\end{aligned}
$$

The center $E$ of the sphere circumscribed around the pyramid $DABC$ lies on the ray $DO$.

If $F$ is the midpoint of $DA$, then $EF \perp DA$ and $\triangle EFD \sim \triangle AOD$, $\frac{DO}{AD} = \frac{FD}{DE}$.

The radius of the circumscribed sphere

$$
\begin{aligned}
& \text { The radius of the circumscribed sphere } \\
& R = DE = \frac{AD \cdot FD}{DO} = \frac{\frac{a}{2} \cdot a}{\frac{a}{\cos \frac{\alpha}{2}} \sqrt{\sin \left(\frac{\pi}{3} + \frac{\alpha}{2}\right) \sin \left(\frac{\pi}{3} - \frac{\alpha}{2}\right)}} = \\
& = \frac{a \cos \frac{\alpha}{2}}{2 \sqrt{\sin \left(\frac{\pi}{3} + \frac{\alpha}{2}\right) \sin \left(\frac{\pi}{3} - \frac{\alpha}{2}\right)}}.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1111.jpg?height=800&width=561&top_left_y=50&top_left_x=93)

Fig. 12.212

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1111.jpg?height=466&width=622&top_left_y=385&top_left_x=642)

Fig. 12.213

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