When does every point in a polytope lie along a chord between its edges? Consider the 3-simplex, or tetrahedron, in 3-space.  Regardless of the positions of the vertices, every point in the simplex lies on a chord between two non-adjacent edges of the simplex.  Or, equivalently, every interior point lies along a straight line segment which intersects two non-adjacent edges.
When is this property true of other convex (or non-convex) polyhedra?  How does this property extend to the general $N$-simplex?
 A: I think that the proposed solution is slightly incomplete since the original question asks for nonadjacent edges of a 3-polytope. It can be easily fixed by studying the intersection of the graphs G and -G.  
Regarding possible n-dimensional versions of the problem, the following result can be proved. Let P be an n-dimensional polytope and k and m positive integers such that k + m = n + 1. For any point x in P there are two faces, F and G, of P such that dim F \le k - 1, 
dim G \le m - 1, and x is in conv(F U G). 
A: The question asks whether every point $v$ in the interior of a 3-polytope  $P \ $ is on an interval between two edge-points.  This is easy.  Project the edges of $P$ onto a unit sphere centered at $v$.  Call the resulting graph $G$ blue.  Take the opposite $-G$ and call this graph red.  Clearly red and blue graphs intersect, since otherwise one must lie in the face of another, which is impossible since $v$ is interior.  Thus the line through the intersection point and $v$ is as desired.  
As for higher dimensions, this is clearly not possible already for dim-reasons.  We are talking about 2-parametric family of intervals, which cannot possibly cover the interior of a $d$-polytope, for $d\ge 4$.  
