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Since Q is formed by a Cartesian product, the vertices V (Q) of Q can be partitioned d 2d into 2 disjoint sets, say V (Q) = Si=1 Qi. Where the convex hull of the vertices in Qi is combinatorially equivalent to a copy of the original polytope P for i = 1,..., 2d. By the pigeonhole principle, of the n +2d vertices selected to form the set S, there is at least one d set of the vertex partition of V (Q), say Qk, that contains at least (n +2d)/2  vertices of S. Let H be a supporting hyperplane for Q at Qk which contains no other vertices of Q. This is also a supporting hyperplane for S, so the convex hull of the vertices in S ∩ Qk forms a face of Q∗, which is at most dimension d′. d ′ By hypothesis each vertex of P is incident with at most (n +2d)/2  − d facets. Any ′ d ′ ′ d vertex of P is in at least d facets, so (n +2d)/2  − d ≥ d . This implies (n +2d)/2  ≥ ′ ′ ′ 2d , so S ∩ Qk contains at least 2d vertices. We claim at most d − 1 of these vertices are from U. To see this, note that since S ∩ Qk is the set of vertices of a face of the convex hull of S, it consists of some (possibly empty) set of vertices from T , and some (possibly empty) set of vertices from U. These vertices from U lie in general position to one another and are additionally in general position with respect to T . Thus the fact that the convex hull of ′ ′ S ∩ Qk has dimension at most d yields that |U ∩ Qk| ≤ d + 1. ′ ′ ′ If |U ∩ Qk| = d + 1 then |T ∩ Qk| ≥ d − 1 ≥ 1, hence S ∩ Qk contains d + 2 vertices in general position, so the convex hull has dimension greater then d′, a contradiction. Similarly ′ ′ ′ if |U ∩Qk| = d then |T ∩Qk| ≥ d ≥ 2, and once again S ∩Qk would contain d +2 vertices in ′ d ′ general position. Therefore |U ∩ Qk| ≤ d − 1, hence |T ∩ Qk| ≥ (n +2d)/2  − d + 1. Since these vertices are in T and in a face of the convex hull of S, the convex hull of T ∩Qk forms a ∗ ∗ d ′ proper face of P . Therefore P has a facet with at least (n +2d)/2  − d + 1 vertices, and thus P has a vertex incident with the same number of facets, contrary to assumption. Hence there is no such subset S of the vertices of Q such that the convex hull of S is combinatorially equivalent to Q∗.

Corollary 1. Let P be an n-gon with n ≥ 3 · 2d − 2d +1. Let Q be the simple polytope formed by taking the Cartesian product of P with the d-dimensional cube. Then there is no subset S of the vertices of Q where the convex hull of S has the same combinatorial type as the dual polytope Q∗.

Proof. Observe that P satisfies the hypothesis of Theorem 1 as d′ = 2 and n ≥ 3 · 2d − 2d +1 d ′ implies (n +2d)/2  − d ≥ 2. Each vertex of P is incident with at most 2 facets, so indeed by Theorem 1 there is no subset S of the vertices of Q where the convex hull of S has the same combinatorial type as the dual polytope Q∗.

Conjecture 2. The truncated d-simplex, the truncated d-cube and the truncated d-cross- polytope for d ≥ 4 are families of where Conjecture 1 holds.

2 Acknowledgements

The author thanks Richard Ehrenborg and Margaret Readdy for inspiring conversations and comments on an earlier draft, and G´abor Hetyei for comments.

References

[1] Gunter¨ M. Ziegler, “Lectures on polytopes,” Graduate Texts in Mathematics, 152, Springer-Verlag, New York.1995

University of Kentucky, Department of Mathematics, Lexington, KY 40506. [email protected].

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