Track / Stage 4 / 41 of 340 #301 of 1964

Problem 301

AMC 12 late, AIME early

Geometry Difficulty 4.5 Find the answer

Among the eight vertices of a cube, four happen to be the vertices of a regular tetrahedron. The ratio of the surface area of the cube to the surface area of the regular tetrahedron is
(A) $\sqrt{2}$
(B) $\sqrt{3}$
(C) $\frac{\sqrt{6}}{2}$
(D) $\frac{\sqrt{6}}{3}$

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Official solution

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.