11.007. Find the ratio of the volume of a cube to the volume of a regular tetrahedron, the edge of which is equal to the diagonal of the face of the cube.
## Solution.

For a tetrahedron where all edges are equal, denote the edges as $a$. The radius of the circumscribed circle around the base of the tetrahedron is $R=\frac{a \sqrt{3}}{3}$. The foot of the height $H$ is the center of the circumscribed circle around $\triangle A B C$ (see Fig. 11.5). Therefore, $H=\sqrt{a^{2}-R^{2}}=\frac{\sqrt{2} a}{\sqrt{3}}$ and the volume $V_{1}=\frac{1}{3} S_{\text {base }} \cdot H$. Since $\triangle A B C$ is equilateral, $S_{\text {base }}=\frac{a^{2} \sqrt{3}}{4}$. Thus, $V_{1}=\frac{a^{3} \sqrt{2}}{12}$. Let the edge of the cube be $b$. The diagonal of the cube is $b \sqrt{2}$. It is equal to the edge of the tetrahedron $b \sqrt{2}=a$, hence $b=\frac{a}{\sqrt{2}}$. The volume of the cube $V_{2}=b^{3}=\frac{a^{3}}{2 \sqrt{2}}$. The ratio $\frac{V_{2}}{V_{1}}=\frac{a^{3} \cdot 12}{2 \sqrt{2} \cdot a^{3} \sqrt{2}}=3$.

Answer: 3.