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7
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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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22
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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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2
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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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