11.130. Find the ratio of the volumes of a regular tetrahedron and an octahedron, given that their total surface areas are equal.
Solution.

$P O$ is the height of the regular tetrahedron $P M N K$ (Fig. 11.26). Let $M N=a$, then $M O=\frac{a \sqrt{3}}{3}, P O=\sqrt{M P^{2}-M O^{2}}=\sqrt{a^{2}-\frac{a^{2}}{3}}=\frac{a \sqrt{2}}{\sqrt{3}}$, the volume of the tetrahedron $V_{1}=\frac{1}{3} S_{\triangle M N K} \cdot P O=\frac{1}{3} \frac{a^{2} \sqrt{3}}{4} \frac{a \sqrt{2}}{\sqrt{3}}=\frac{a^{3} \sqrt{2}}{12}$, the total surface area of the tetrahedron $S_{1}=4 S_{\triangle M N K}=a^{2} \sqrt{3}$. The square $A B C D$ is the common base of two regular pyramids $E A B C D$ and $F A B C D$, which together form the given regular octahedron (Fig. 11.27), $O_{1}$ is the center of the square $A B C D$. Let $A B=b$. Triangles $A O_{1} E$ and $A O_{1} B$ are equal $\left(\angle A O_{1} E=\angle A O_{1} B=90^{\circ}, A O_{1}\right.$ is a common leg, $\left.A E=A B\right)$.

Therefore, $E O_{1}=A O_{1}=\frac{b \sqrt{2}}{2}$, and the volume of the octahedron $V_{2}=\frac{2}{3} S_{A B C D} \cdot E O_{1}=$ $=\frac{2}{3} b^{2} \cdot \frac{b \sqrt{2}}{2}=\frac{b^{3} \sqrt{2}}{3}$, and its total surface area $S_{2}=8 S_{\triangle A E B}=8 \cdot \frac{b^{2} \sqrt{3}}{4}=$ $=2 b^{2} \sqrt{3}$.

Since $S_{1}=S_{2}$, then $a^{2} \sqrt{3}=2 b^{2} \sqrt{3}, \frac{a}{b}=\sqrt{2}, \frac{a^{3}}{b^{3}}=2 \sqrt{2}$.

Then $\frac{V_{1}}{V_{2}}=\frac{a^{3} \sqrt{2}}{12}: \frac{b^{3} \sqrt{2}}{3}=\frac{a^{3}}{4 b^{3}}=\frac{1}{4} 2 \sqrt{2}=\frac{1}{\sqrt{2}}$.

Answer: $\frac{1}{\sqrt{2}}$.