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Suppose a compact convex body $P \subset \Bbb R^3$ has only polygonal orthogonal projections onto a plane. Does this imply that $P$ is a convex polytope?

This question occurred to me when I was making exercises for my book. I figured this is probably easy and well known, but the literature hasn't been any help. One remark: if the number of sides of all polygons is bounded by $n$, the problem might be easier. Furthermore, if $P$ is assumed to be a convex polytope, this elegant paper by Chazelle-Edelsbrunner-Guibas (1989) gives a (perhaps, unexpectedly large) sharp $\exp O(n \log n)$ upper bound on the number of vertices of $P$ (ht Csaba Toth who generalized this to higher dimensions).

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Suppose not. By hypothesis, any finite intersection Q of supporting half-spaces of P strictly contains P. Because Q is a polytope, its orthogonal projections are polytopes. Let x be a point in Q\P. There is an orthogonal projection such that the image of x and that of P are disjoint. The corresponding polytope obtained by projecting Q strictly contains the projection of P. – Steve Huntsman Feb 18 '10 at 0:08
Let Q' be obtained by augmenting with supporting half-spaces such that the projections of Q and P coincide. By hypothesis, Q' still strictly contains P. By induction, we may therefore assume there is an infinite sequence of distinct points in Q that are not contained in P. Similarly, Q cannot be chosen so that all its orthogonal projections are the same as those of P. – Steve Huntsman Feb 18 '10 at 0:08
I am not sure I follow. You assume that $P$ is not a polytope and you conclude with a property implying that $P$ is really not a polytope. Now what? – Igor Pak Feb 18 '10 at 4:18
Let me make an easy general comment: any proof must use substantially the fact that this is a $3$-dim problem (it obviously fails in $\Bbb R^2$). Also, use the fact that $P$ is bounded, since all projections of a circular cone are polyhedral cones. – Igor Pak Feb 18 '10 at 4:19
There is a similar, but different, lemma in : If S is a subset of R^n such that the projection of S to every R^{d+1} is a union of finitely many d-dimensional polyhedra, then S is a union of finitely many d-dimensional polyhedra. – David Speyer Feb 19 '10 at 1:17
up vote 20 down vote accepted

Theorem 4.1 of this paper by Klee says yes. Moreover, the result generalizes to higher dimensions for projections of arbitrary dimension $\ge 2$.

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This is great! Many thanks. Let me read this paper carefully to understand what it really does and get back to you. – Igor Pak Feb 19 '10 at 20:03
After some thinking, it is now clear to me that what I wanted does easily follow from Klee. I am happy to accept this answer. Let me just mention that the key idea in Klee's paper is given in Mirkil's earlier paper, where the author proves that a closed convex cone is polyhedral if and only if all its 2-dim projections are closed and polyhedral (thus, circular cone doesn't work as it has a non-closed projection). Klee first refines and extends Mirkli's ideas, and then gives a "local" proof of the main result based on this. – Igor Pak Feb 21 '10 at 2:48

Here is a more direct proof of this statement.

A Short Proof of Klee's Theorem by John J. Zanazzi

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