# 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?

• First reaction, written without much thought: what's the polar dual question? Is that easier? – Alexander Woo Oct 31 '11 at 4:26
• @Alexander Woo, I have to think a bit more to understand what the equivalent formulation would be on the dual of a polyhedron, but I have no reason to believe this would simplify the problem. – UltraBlue06 Oct 31 '11 at 4:43
• For my comfort, I restrict discussion to subsets of R^3. Since each pair of edges forms a (possibly degenerate) tetrahedron after taking its convex hull, the question may boil down to an (overlapping) decomposition of such a polyhedron into tetrahedra which have at least two edges in common with the polyhedron. I imagine Bill Thurston or Joseph O'Rourke will have something further to say. My sense is that all regular polyhedra do, and some (Csakar?) polyhedra will not, and it will be a combinatorial result. Gerhard "Ask Me About System Design" Paseman, 2011.10.30 – Gerhard Paseman Oct 31 '11 at 4:58
• @Gerhard Paseman, I very much agree with your analysis, though I was hoping for an easier method of tackling the problem. – UltraBlue06 Oct 31 '11 at 5:05
• I am also perfectly fine restricting treatment of this question to R^3. – UltraBlue06 Oct 31 '11 at 5:06

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 $d\ge 4$.
• So the opposite graph $−G$ is constructed by mapping the vertices of $G$ to antipodal positions on the unit sphere (centered at $v$) and then connecting the vertices via the opposite or alternative edge path around the sphere. As such, $−G$ should be combinatorially isomorphic to $G$. Furthermore, when $G$ and $−G$ intersect, which is guaranteed if $v$ is internal to $P$, we know there exists a chord through that point of intersection which also intersects $v$ at the center of the sphere. – UltraBlue06 Nov 1 '11 at 20:54