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:
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!
