Seeing Garabed Gulbenkian's question (which was inspired by Joel Hamkins' question), reminds me of an analogous problem which I believe remains open, and which some might find intriguing. Define an equiprojective polyhedron $P \subset \mathbb{}R^3$ as one whose orthogonal projections—with the exception of projections in directions parallel to a face—are all $n$-gons for the same $n$. The definition is due to Shephard, 45 years ago. Thus, cubes are 6-equiprojective: their only projections to quadrilaterals are along directions parallel to faces. A triangular prism is 5-equiprojective. There are no 3- or 4-equiprojective polyhedra, as established in the paper that recently reopened this dormant subject:

Hasan, Masud, Mohammad Monoar Hossain, Alejandro López-Ortiz, Sabrina Nusrat, Saad Altaful Quader, and Nabila Rahman. "Some new equiprojective polyhedra." arXiv:1009.2252 (2010).

This paper establishes that both the equitruncated pyramid and the equitruncated triangular cupola are 10-equiprojective:

I believe this remains open:

Q1. Is there a $k$-equiprojective polyhedron for every $k \ge 5$? If not, for which $k$ does there exist $k$-equiprojective polyhedra?

Addendum. After seeing Ian's nice resolution of Q1, I went back to the cited paper and found this was already known: "(in fact, any $p$-gonal prism is $p + 2$-equiprojective)." My apologies! So I guess the real open problem here is:

Q2. Describe (or construct) all equiprojective polyhedra.

  • $\begingroup$ When I read the title I thought the problem was going to be: what can you say of a convex body in euclidean 3_space such that all of its orthogonal projections are projectively equivalent ? BTW, why are the bodies in your problem called "equiprojective"? $\endgroup$ – alvarezpaiva Jul 27 '13 at 9:11
  • $\begingroup$ "Equiprojective" because their projections are combinatorially equivalent. Your question is also nice! $\endgroup$ – Joseph O'Rourke Jul 27 '13 at 11:34

A $k$-drum (a product of a convex polygon and an interval, generalizing the prism and cube) is $k+2$-equiprojective for every $k$. The point is that one will see $m$ edges of the front face on the outer boundary, and $k-m$ edges of the lower face on the outer boundary for some $m$, and $2$ edges of two rectangular sides from any direction. Consider the projection of a polygon $\times \mathbb{R}$. This will project to an interval $\times \mathbb{R}$, where the endpoints of the interval are two extremal points of the polygon projection, cutting the polygon into $m$ and $k-m$ edges for some $m$. Now, cut this product at two levels to get a drum. The projection of the drum will be the convex hull of the projection of two faces, which will be two copies of the projection of the polygon, translated inside of the strip. So one sees $m$ edges from one face of the drum, $k-m$ from the other face, and $2$ edges from the projection of two edges corresponding to the extremal points $\times$ interval.

      (Wikipedia image of a 12-drum added by J.O'Rourke.)

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  • $\begingroup$ Very nice! I added an image of the 12-drum, whose projection is a 14-gon. $\endgroup$ – Joseph O'Rourke Jul 29 '13 at 15:45

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