Covering all the edges of a hypercube? Consider an arbitrary $n$- dimensional hypercube:
If we select $n - 1$ corners of that hypercube and highlight all $(n - 2)$ dimensional elements that originate from each of the corners is it possible to cover all the $(n - 2)$ dimensional faces of the cube? Also, is it ever possible to cover all the faces originating from a corner besides one of the $n - 1$ originally selected?
Some insight:
I failed to cover all the edges of a cube using 2 points and the set of edges extended from them. The same applies for the case squares and lower. 

The $n$-hypercube has $2n(n-1)\ \ \ \ (n-2)$-faces. Each corner touches $2n-3\ \ \ \ (n-2)$-faces. At most $(2n-3)(n-1)$ faces are covered, and at least $3(n-1)$ remain uncovered.
