# Do plane projections determine a convex polytope?

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).

• 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 ams.org/mathscinet-getitem?mr=733052 : 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 E Speyer Feb 19 '10 at 1:17

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