1989 AHSME Problems/Problem 26

Problem

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

$\mathrm{(A) \frac{\sqrt{3}}{12} } \qquad \mathrm{(B) \frac{\sqrt{6}}{16} } \qquad \mathrm{(C) \frac{1}{6} } \qquad \mathrm{(D) \frac{\sqrt{2}}{8} } \qquad \mathrm{(E) \frac{1}{4} }$

Solution

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…

$\frac{\frac{x^3}{6}}{x^3} \rightarrow \frac{1}{6}$ or $\fbox{C}$

Solution 2

Split the cube in half along a cross-section parallel to two of the bases. Because of symmetry, we only need to compare the remaining square pyramid and the octahedron. Because the area of the base of the pyramid is half the area of the square base of the pyramid, and we divide by three for the pyramid's volume, our final answer is $\frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$ or $\fbox{C}$ .

1989 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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1989 AHSME Problems/Problem 26

Contents

Problem

Solution

Solution 2

See also