Cube in an octohedron in a cube Start with a cube.  Inscribe an octohedron in the cube by connecting the centers of the cube's faces.  Then inscribe another cube inside the octohedron by connecting the centers of the octohedron's faces.  What is the ratio of the side lengths of the inner and outer cube?
I have a solution to this, but I want to see if anyone can come up with a more elegant one.  (In particular, an "elegant" solution would be one that generalizes immediately to the n-dimensional analog of this problem.  Hint:  In dimension $n$, the ratio of the outer cube side to inner cube side is $n$.)
 A: The idea is to note that in $\mathbb R^n$, a vertex of a $n$-cube of side length $2$ and centered at the origin is given by the vector $\boldsymbol x = (x_1, \ldots, x_n) \in \{(\pm 1, \ldots, \pm 1)\}$, where  $x_i \in \{-1, +1\}$ for each $i = 1, \ldots, n$.  Without loss of generality, choose the $n$ faces incident to the vertex $\boldsymbol v = (1, 1, \ldots, 1)$.  The midpoints of these faces are $\boldsymbol m_j = (0, \ldots, 0, 1, 0, \ldots, 0)$ where the $j^{\rm th}$ coordinate is unity, for $j = 1, \ldots, n$.  These describe the $n$-octahedron face, the center of which is simply $\boldsymbol c = \frac{1}{n} \sum_{i=1}^n \boldsymbol m_j =  \frac{1}{n} \boldsymbol v$.  It follows that the ratio of the edge lengths of the outer to inner $n$-cube is $n$.
The above could probably be phrased more elegantly, but the idea is fairly elementary and straightforward.
A: Let $a$ the side of the first cube.
Then $a$ is also equal to the distance between two opposite vertex of the octahedron.
The second cube has side equal to $\frac{a}{3}$.
Thus the ratio of the side lengths of the inner and outer cube is equal to $\frac{1}{3}$.
