1989 AHSME Problems/Problem 26

A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is

$\mathrm{(A) \frac{\sqrt{3}}{12} } \qquad \mathrm{(B) \frac{\sqrt{6}}{16} } \qquad \mathrm{(C) \frac{1}{6} } \qquad \mathrm{(D) \frac{\sqrt{2}}{8} } \qquad \mathrm{(E) \frac{1}{4} }$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Call the length of a side of the cube x. Thus, the volume of the cube is $x^3$ . We can then find that a side of this regular octahedron is the square root of $(\frac{x}{2})^2$ + $(\frac{x}{2})^2$ which is equivalent to $\frac{x\sqrt{2}}{2}$ . Using our general formula for the volume of a regular octahedron of side length a, which is $\frac{a^3\sqrt2}{3}$ , we get that the volume of this octahedron is...

$(\frac{x\sqrt{2}}{2})^3 \rightarrow \frac{x^3\sqrt{2}}{4} \rightarrow \frac{x^3\sqrt{2}}{4}*\frac{\sqrt{2}}{3} \rightarrow \frac{2x^3}{12}=\frac{x^3}{6}$

Comparing the ratio of the volume of the octahedron to the cube is…

$\frac{\frac{x^3}{6}}{x^3} \rightarrow \framebox[1.1\width]{(C) \frac{1}{6} }$ (Error compiling LaTeX. Unknown error_msg)

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1989 AHSME Problems/Problem 26