8,9

Calculate the volume of a regular tetrahedron if the radius of the circle circumscribed around its face is $R$.

#
Let the edge of a regular tetrahedron \(ABCD\) be \(a\). Since the radius of the circle circumscribed around an equilateral triangle with side \(a\) is \(\frac{a \sqrt{3}}{3}\), we have the equation \(\frac{a \sqrt{3}}{3} = R\), from which \(a = R \sqrt{3}\). Let \(DM\) be the height of the regular tetrahedron \(ABCD\). From the right triangle \(AMD\), by the Pythagorean theorem, we find that

\[
DM = \sqrt{AD^2 - AM^2} = \sqrt{a^2 - R^2} = \sqrt{3R^2 - R^2} = R \sqrt{2}
\]

Therefore,

\[
V_{ABCD} = \frac{1}{3} S_{\triangle ABC} \cdot DM = \frac{1}{3} \cdot \frac{a^2 \sqrt{3}}{4} \cdot R \sqrt{2} = \frac{1}{3} \cdot \frac{3R^2 \sqrt{3}}{4} \cdot R \sqrt{2} = \frac{R^3 \sqrt{6}}{4}.
\]