$23 \cdot 24$ A cube has eight vertices, four of which are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.
(A) $\sqrt{2}$.
(B) $\sqrt{3}$.
(C) $\sqrt{\frac{3}{2}}$.
(D) $\frac{2}{\sqrt{3}}$.
(E) 2 .
(31st American High School Mathematics Examination, 1980)
[Solution] The edges of the tetrahedron are the face diagonals of the cube, so if $s$ is the edge length of the cube, the area of each face of the tetrahedron is
$$
\begin{aligned}
& \frac{1}{2} \cdot \frac{\sqrt{3}}{2}(s \sqrt{2}) \cdot s \sqrt{2} \\
= & \frac{(s \sqrt{2})^{2} \sqrt{3}}{4} \\
= & \frac{s^{2} \sqrt{3}}{2},
\end{aligned}
$$

The required ratio is $\frac{6 \cdot s^{2}}{4 \cdot\left(\frac{s^{2} \sqrt{3}}{2}\right)}=\sqrt{3}$.
Therefore, the answer is $(B)$.