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The centers of the faces of a cube are also the vertices of polyhedron. How to Compute the ratio of the volume of the polyhedron to that of the cube containing it?

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Hint: do you know what kind of polyhedron it is? Can you express the length of its side in terms of the length of the cube? You can decompose the polyhedron into two pieces that you probably know how to find the volume of.

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The dual to the cube is the octahedron, which is made up of two pyramids glued together. Draw a picture of the appropriate cross sections to derive the properties of these pyramids. If the side length of the cube is $2$ then the lengths of the edges of the octahedron are $\sqrt{2}$, so the area of the base of the pyramids is $2$ and the height is $1$ (half the cube). So the volume of the two pyramids is $$\frac{2}{3}(\text{area of base}) \cdot (\text{height})= \frac{4}{3} .$$ The volume of the cube is $8$ so that the ratio is $\frac{1}{6}$.

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