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

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…
$\frac{\frac{x^3}{6}}{x^3} \rightarrow \boxed{\frac{1}{6}}$
\frac{1}{6}
Level 5
Geometry
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:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the volume ratio of the octahedron to the cube, we first need to determine the volumes of both shapes. 

Let the side length of the cube be denoted as \( s \). The volume of the cube is given by the formula:

\[
V_{\text{cube}} = s^3
\]

Next, we need to find the volume of the octahedron formed by joining the centers of the faces of the cube. The centers of the faces of the cube can be represented as points in a three-dimensional coordinate system. For a cube centered at the origin with vertices at \( (\pm \frac{s}{2}, \pm \frac{s}{2}, \pm \frac{s}{2}) \), the centers of the faces are located at:

\[
\left( \frac{s}{2}, 0, 0 \right), \left( -\frac{s}{2}, 0, 0 \right), \left( 0, \frac{s}{2}, 0 \right), \left( 0, -\frac{s}{2}, 0 \right), \left( 0, 0, \frac{s}{2} \right), \left( 0, 0, -\frac{s}{2} \right)
\]

The octahedron has a volume formula given by:

\[
V_{\text{octahedron}} = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]

However, a more straightforward approach is to use the known formula for the volume of a regular octahedron, which is:

\[
V_{\text{octahedron}} = \frac{\sqrt{2}}{3} a^3
\]

where \( a \) is the length of the edge of the octahedron. The edge length \( a \) can be determined by calculating the distance between two adjacent face centers. For example, the distance between \( \left( \frac{s}{2}, 0, 0 \right) \) and \( \left( 0, \frac{s}{2}, 0 \right) \) is:

\[
a = \sqrt{\left( \frac{s}{2} - 0 \right)^2 + \left( 0 - \frac{s}{2} \right)^2 + (0 - 0)^2} = \sqrt{\left( \frac{s}{2} \right)^2 + \left( \frac{s}{2} \right)^2} = \sqrt{\frac{s^2}{4} + \frac{s^2}{4}} = \sqrt{\frac{s^2}{2}} = \frac{s}{\sqrt{2}}
\]

Substituting \( a \) into the volume formula for the octahedron gives:

\[
V_{\text{octahedron}} = \frac{\sqrt{2}}{3} \left( \frac{s}{\sqrt{2}} \right)^3 = \frac{\sqrt{2}}{3} \cdot \frac{s^3}{2\sqrt{2}} = \frac{s^3}{6}
\]

Now, we can find the ratio of the volumes:

\[
\text{Volume Ratio} = \frac{V_{\text{octahedron}}}{V_{\text{cube}}} = \frac{\frac{s^3}{6}}{s^3} = \frac{1}{6}
\]

Thus, the final answer is \( \frac{1}{6} \).<|END_OF_LATENT|> The final answer is $\frac{1}{6}$.