Find the volume of a triangular pyramid, five edges of which are equal to 2, and the sixth is equal to $\sqrt{6}$.

#
Let $A B C D$ be a triangular pyramid, in which

$$
A D=B D=C D=A B=A C=2, B C=\sqrt{6} \text {. }
$$

If $D O$ is the height of the pyramid, then $O$ is the center of the circumcircle of triangle $A B C$ (since $A D=B D=C D$). Let $R$ be the radius of this circle, $M$ be the midpoint of $B C$, and $\angle A B C=\alpha$. Then

$$
\begin{gathered}
A M=\sqrt{A B^{2}-B M^{2}}=\sqrt{2^{2}-\left(\frac{\sqrt{6}}{2}\right)^{2}}=\sqrt{4-\frac{3}{2}}=\frac{\sqrt{5}}{\sqrt{2}} \\
\sin \alpha=\frac{A M}{A B}=\frac{\sqrt{5}}{2 \sqrt{2}}, R=\frac{A C}{2 \sin \alpha}=\frac{2 \sqrt{2}}{\sqrt{5}} \\
D O=\sqrt{A D^{2}-R^{2}}=\sqrt{4-\frac{8}{5}}=\sqrt{\frac{12}{5}} \\
V=\frac{1}{3} S_{\triangle A B C} \cdot D O=\frac{1}{3} \cdot \frac{1}{2} B C \cdot A M \cdot D O=\frac{1}{6} \cdot \sqrt{6} \cdot \frac{\sqrt{5}}{\sqrt{2}} \cdot \sqrt{\frac{12}{5}}=1
\end{gathered}
$$

Find the volume of a regular triangular pyramid with side length $a$ and radius $R$ of the circumscribed sphere.

## Answer

$\frac{a^{2}\left(R \sqrt{3}+\sqrt{3 R^{2}-a^{2}}\right)}{12}$.