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- Statement
Identify or construct all k-equiprojective polyhedra. A polyhedron P is

*k-equiprojective*if its orthogonal projection to a plane is a k-gon in every direction not parallel to a face of P. Thus a cube is 6-equiprojective.- Origin
Geoffrey Shephard in [She68].

- Status/Conjectures
Open.

- Partial and Related Results
A characterization is detailed in [HL08]: “A polyhedron is equiprojective iff its set of edge-face pairs can be partitioned into compensating pairs.” For term definitions, see the original paper. Building on this work, a recent paper [HHLO+10] establishes that any equiprojective polyhedron has at least one pair of parallel faces, that there is no 3- or 4-equiprojective polyhedron, and the triangular prism is the only 5-equiprojective polyhedron.

- Related Open Problems
A generalization of the problem was posted on MathOverflow, 11Feb11: [O'R11]

- Appearances
Also in [CFG90], Problem B10.

- Categories
polyhedra

- Entry Revision History
J. O’Rourke, 31 Dec. 2010; 11 Feb 2011.

- [O'R11]
Joseph O’Rourke. What is determined by the combinatorics of the shadows of a convex polyhedron? http://mathoverflow.net/questions/55124/, February 2011.

- [CFG90]
H. P. Croft, K. J. Falconer, and R. K. Guy.

*Unsolved Problems in Geometry*. Springer-Verlag, 1990.- [HHLO+10]
Masud Hasan, Mohammad Houssain, Alejandro Lopez-Oritz, Sabrina Nusrat, Saad Quader, and Nabila Rahman. Some new equiprojective polyhedra. http://arxiv.org/abs/1009.2252, 2010.

- [HL08]
Masud Hasan and Anna Lubiw. Equiprojective polyhedra.

*Comput. Geom. Th. Appl.*, 40(2):148–155, 2008.- [She68]
Geoffrey C. Shephard. Twenty problems on convex polyhedra—II.

*Math. Gaz.*, 52:359–367, 1968.