Let $P$ be a polyhedron in $\mathbb{R}^3$. Say that $P$ combinatorially shadows a sequence of natural numbers $S$ if there is a continuous rotation of $P$ such that its orthogonal-projection shadows are polygons whose number of sides coincide with the elements of $S$ in order. For example, $S$ might be the odd primes: $S=(3,5,7,11,13,\ldots)$, and we want the shadows to be a triangle, then a pentagon, then a septagon, then a hendecagon, etc.

Q0. Let $S$ be an increasing sequence of natural numbers, whose first element is $\ge 3$. Does there exist a polyhedron $P$ that combinatorially shadows $S$?

I think the answer here is Yes, illustrated just for $S=(3,5)$ with this example:
       Polyhedron Shadows
The generalization is that the needed increase above the previous element in the sequence is achieved by bumping out near the centroid of an appropriate face (the centroid $c$ of face $(1,2,3)$ in the above example is bumped out to $\lbrace a, b \rbrace$), shallow enough to be hidden for the previous element (middle image), but sufficent so that a rotation will simultaneously expose the additional vertices (right image). So if I am correct here, there is a prime polyhedron that realizes the odd primes—either up to any given prime, or all odd primes if an infinite number of faces are countenanced.

Correction (9Dec12): I now think the above sketch fails to allow many vertices to appear in the shadow simultaneously. Better is to split existing vertices into two ... [remaining bad idea deleted]. 23Dec12: Now I believe the construction posted in a separate answer settles Q0 (positively).

My question concerns arbitrary—not necessarily increasing—sequences:

Q1. Let $S$ be an arbitrary sequence of natural numbers, each $\ge 3$. Does there exist a polyhedron $P$ that combinatorially shadows $S$?

Ideas, even half-baked, or pointers to relevant literature welcomed! Thanks!

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    $\begingroup$ Try flowering. Perhaps you can make the word picture work or say why it fails. From one edge of the polyhedron, draw a semicircle. Divide that semicircle into n_0 pieces, and use that for your first face. For each of those n_0 edges, draw another semicircle (tilted away from the shadow projection, and perhaps translated so that each edge gets a rectangle followed by a semicircle) and repeat,dividing each semicircle into as many edges as necessary. There are problems with this to be solved, but I do not see 2 as an obstacle. Gerhard "After Two Comes Very Many" Paseman, 2012.12.13 $\endgroup$ Dec 14, 2012 at 6:14
  • $\begingroup$ @Gerhard: I have come to the same conclusion: doubling is not an obstacle. I also have been exploring an avenue akin to your idea. If/when time permits, I hope to eventually post an example. As always, thanks for your insights! $\endgroup$ Dec 14, 2012 at 12:23

1 Answer 1


This answers my Q0. The polyhedron is the convex hull of these faces:
           Prime Polyhedron 17

When rotated, the shadow has $3, 5, 7, 11, 13, 17$ vertices:
           Prime Shadows
Clearly the same design (not dissimilar from Gerhard's suggestion) suffices to capture any sequence with $s_{i+1}-s_i \ge 2$. This suffices for a prime polyhedron, my original (recreational) goal.
           Prime Polyhedron 23

It is not difficult to capture sequences that also sometimes increase by just $1$, by arranging for rotation to expose a (thin) triangle face. Thus the answer to Q0 is Yes.


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