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Consider the 3-simplex, or tetrahedron, in 3-space. Regardless of the positions of the vertices, every point in the volume defined by the triangular faces of the polytope's skeleton graph can lie along a chord between two non-adjacent edges. Or, equivalently, every interior point can lie 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?

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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
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2 Answers

up vote 22 down vote accepted

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 $d\ge 4$.

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Nice answer. =) –  UltraBlue06 Oct 31 '11 at 5:32
    
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
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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).

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