Three regular solids: tetrahedron, hexahedron, and octahedron have equal surface areas. What is the ratio of their volumes?
I. solution: Let the edges of the bodies be denoted by $a, b, c$, their surface areas by $F_{1}, F_{2}, F_{3}$, and their volumes by $k_{1}, k_{2}, k_{3}$.

$$
F_{1}=4 \frac{a^{2}}{4} \sqrt{3}=a^{2} \sqrt{3}, \quad F_{2}=6 b^{2}, \quad F_{3}=8 \frac{c^{2}}{4} \sqrt{3}=2 c^{2} \sqrt{3}
$$

According to the problem, $F_{1}=F_{2}=F_{3}$, that is,

$$
\begin{array}{lll}
a^{2} \sqrt{3}=6 b^{2}, & \text { from which } & a=\sqrt{2 \sqrt{3} b^{2}}=b \sqrt[4]{12} \\
2 c^{2} \sqrt{3}=6 b^{2}, & \text { from which } & c=\sqrt{\sqrt{3} b^{2}}=b \sqrt[4]{3}
\end{array}
$$

The volumes of the bodies:

$$
\begin{gathered}
k_{1}=\frac{1}{3} \frac{a^{2}}{4} \sqrt{3} \cdot \sqrt{a^{2}-\left(\frac{2}{3} \cdot \frac{a^{2}}{2} \sqrt{3}\right)^{2}}=\frac{a^{2}}{12} \sqrt{3} \cdot \sqrt{a^{2}-\frac{a^{2}}{3}}=\frac{a^{2}}{12} \sqrt{3} \cdot a \frac{\sqrt{2}}{\sqrt{3}}=\frac{a^{3} \sqrt{2}}{12} \\
k_{1}=b^{3}, \\
k_{3}=2 \cdot \frac{1}{3} c^{2} \cdot \frac{c \sqrt{2}}{2}=\frac{c^{3} \sqrt{2}}{3}
\end{gathered}
$$

Substituting the values of $a$ and $c$

$$
\begin{gathered}
k_{1}=\frac{(b \sqrt[4]{12})^{3} \sqrt{2}}{12}=\frac{b^{3} \sqrt[4]{12^{3}} \sqrt[4]{4}}{12}=\frac{b^{3} \sqrt[4]{3^{3} 4^{3} 4}}{12}=\frac{b^{3} \sqrt[4]{3^{3}}}{3} \\
k_{3}=\frac{(b \sqrt[4]{3})^{3} \sqrt{2}}{3}=\frac{b^{3} \sqrt[4]{3^{3} \sqrt[4]{4}}}{3}=\frac{b^{3} \sqrt[4]{3^{3} 4}}{3}
\end{gathered}
$$

Thus,

$$
k_{1}: k_{2}: k_{3}=\frac{\sqrt[4]{3^{3}}}{3}: 1: \frac{\sqrt[4]{3^{3} 4}}{3}=1: \sqrt[4]{3}: \sqrt{2}=\sqrt[4]{1}: \sqrt[4]{3}: \sqrt[4]{4} \approx 1: 1.316: 1.414
$$

Géza Biczó (Bp., II., Rákóczi g. II. o. t.)

II. solution: Let $F$ be the common surface area.
a) $\frac{F}{4}=\frac{a^{2} \sqrt{3}}{4}, \quad$ from which $\quad a=\sqrt{\frac{F \sqrt{3}}{3}}$,

and thus

$$
k_{1}=a^{3} \frac{\sqrt{2}}{12}=\frac{F \sqrt{3}}{3} \cdot \sqrt{\frac{F \sqrt{3}}{3}} \cdot \frac{\sqrt{2}}{12}=\frac{F \sqrt{F}}{36} \sqrt[4]{12}
$$

b) $\frac{F}{6}=b^{2}, \quad$ from which $\quad b=\sqrt{\frac{F}{6}}$, and thus

$$
k_{2}=b^{3}=\frac{F}{6} \sqrt{\frac{F}{6}}=\frac{F \sqrt{F}}{6 \sqrt{6}}=\frac{F \sqrt{F}}{36} \sqrt{6}
$$

c) $\frac{F}{8}=\frac{c^{2} \sqrt{3}}{4}, \quad$ from which $\quad c=\sqrt{\frac{F \sqrt{3}}{6}}$,

and thus

$$
k_{3}=c^{3} \frac{\sqrt{2}}{3}=\frac{F \sqrt{3}}{6} \sqrt{\frac{F \sqrt{3}}{6}} \cdot \frac{\sqrt{2}}{3}=\frac{F \sqrt{F}}{18} \sqrt[4]{3}=\frac{F \sqrt{F}}{36} \sqrt[4]{48}
$$

Thus,

$$
k_{1}: k_{2}: k_{3}=\sqrt[4]{12}: \sqrt[4]{36}: \sqrt[4]{48}=\sqrt[4]{1}: \sqrt[4]{3}: \sqrt[4]{4}
$$