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Fix a finite subset $S$ of the natural numbers $\mathbb{N}$, each element $\ge 3$.

Is there a convex polyhedron $P$ that has among its shadows regular $n$-gons for each $n \in S$? Does such a $P$ exist for every $S$?

By shadow I mean the orthogonal projection of $P$ onto a 2D plane, for example, the shadow on the $xy$-plane, with $P$ above ($z>0$) that plane and the light at $L=(0,0,+\infty)$. By "among its shadows" I mean that you can rotate $P$ as you wish, and the goal is to find orientations of $P$ that cast the requisite shadows. What shadows are cast at other orientations is not relevant.

I believe the answer to my question is No. But the answer is Yes for a (seemingly minor) variation of the question, exactly the same except that the light is not infinitely far away, but rather say it is at $L=(0,0,h)$, so that the projection is central projection from $L$ rather than orthogonal projection, and one can translate and rotate $P$ (or equivalently, move $L$ and the projection plane around). Then one could build a $P$ for a given $S$ as follows. Let me use $S=\{5,7,11\}$.

Form $P$ as the convex hull of a regular 5-gon, a regular 7-gon, and a regular 11-gon, stacked in parallel planes, each scaled so that the hull includes all the vertices and edges of each regular $n$-gon:
Polyhedron 5,7,11
Then placing $L$ just above the center of the pentagon produces a regular pentagon shadow, moving $L$ straight up enough yields a regular heptagon shadow, and raising $L$ high enough vertically, a regular hendecagon shadow. It seems not difficult to extend the construction to any $S$.

But I cannot see how to accomplish this with the light at $L=(0,0,+\infty)$. My first idea was to arrange the regular $n$-gons roughly as inscribed in longitudinal great circles of a sphere, so that rotating $P$ on a North-South "spit" will project each in turn.
                    Longitude Lines
But I think this approach succeeds only with $S$ of small cardinality.

[This question is a small bite out of an earlier MO question, "What is determined by the combinatorics of the shadows of a convex polyhedron?", and it can be viewed as a lower-dimensional version of another MO question, "Are the Platonic solids shadows of 4-polytopes?".]

Either a construction of a $P$ for any $S$, with $L=(0,0,+\infty)$, or an argument why this is possible only for some $S$, would be much appreciated. In fact, a single counterexample $S$ would enlighten. Thanks for ideas!

Addendum. Here is an illustration of the elegant idea of Andreas Blass, for regular 4-, 8-, and 12-gons (blue, green, and red respectively):
Polyhedron 4,8,12
His construction raises interesting questions. Can any three regular polygons be achieved as shadows (without the (mod 4) restriction)? Is there another construction that achieves more than three regular shadows, for perhaps similarly restricted $n$-gons?

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If one allows only rotations along a given axis, then the answer seems to be no, even when members of S are odd numbers. Arrange the desired polygons so that they share a common base line and a common opposite vertex. Cut vertical slits in them and slide them together so that they share a common altitude. Fan the pieces out and hang them from a string. It should be clear that the triangle shadow implies the other pieces should be fanned at a large angle away from the triangle to not contribute to the shadow. Gerhard "Ask Me About System Design" Pasman, 2011.08.22 –  Gerhard Paseman Aug 23 '11 at 0:34
    
@Gerhard: I agree that rotations along a given axis (a "spit" in barbecuing terminology) is insufficient! –  Joseph O'Rourke Aug 23 '11 at 1:32
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The point is that, between any two shadows is some rotation. If a triangle is required to be one of the shadows, that puts restrictions on the size and shape of any shadow within rho degrees rotation, where rho is significantly large. For any two numbers m,n, there should be a calculable angle rho_{m,n} which is rewuired to rotate between shadows. Gerhard "Ask Me About System Design" Paseman, 2011.08.22 –  Gerhard Paseman Aug 23 '11 at 2:49
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2 Answers

I'm answering only an absurdly small piece of the question, in the hope that this $\varepsilon$ isn't exactly 0. I claim that if $S$ consists of just three numbers, each divisible by 4, then there is a convex polytope in $\mathbb R^3$ whose orthogonal projections to the three coordinate planes are regular polygons (centered at the origin) with the prescribed numbers of sides. Instead of describing the desired polytope $P$ directly, I'll describe its polar polytope, $Q$, the set of points $q$ in the dual vector space such that $(q,p)\leq 1$ for all $p\in P$. The orthogonal projections of $P$ to the coordinate hyperplanes have as their polar polygons the corresponding intersections of $Q$ with the coordinate hyperplanes. So I want those intersections to be regular polygons, centered at the origin, with the prescribed numbers of sides. There is still some freedom about the size and orientation of these polygons, but here I'll make very simple choices. In each of the three coordinate hyperplanes, inscribe in the unit circle a polygon of the desired number of sides, with a vertex at the intersection of the circle with each coordinate axis. (This uses that the number of sides is divisible by 4.) Let $Q$ be the convex hull of the set of all vertices of these three polygons. What needs to be checked is that the intersection of $Q$ with any coordinate hyperplane, say the $x$-$y$-plane, is the polygon that we put there, not some superset resulting from the convex hull of what we put into the other two coordinate hyperplanes. Fortunately, this follows immediately from the following observation: If you take the convex hull of the union of the unit circles of the $x$-$z$ and $y$-$z$-planes and intersect that convex hull with the $x$-$y$-plane, what you get is the square whose vertices are the intersection points of the $x$ and $y$ axes with the unit circle. (Proof of observation: The whole convex hull of the union of the two circles, if projected into the $x$-$y$-plane, gives the convex hull of the union of the projections of the two circles. Those projections are the segments, from $-1$ to $1$, of the $x$ and $y$ axes.) Since the polygon we want in the $x$-$y$-plane includes the square, that polygon is the intersection of the $x$-$y$-plane and $Q$, as required.

Note that, since I took the three polygons that generate $Q$ to have the same circumradius, their polars, the projections of $P$, will have the same inradius.

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Beautiful! I added an illustration as an addendum to my question. Thanks! –  Joseph O'Rourke Aug 23 '11 at 20:41
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It may be possible to replace one of the polygons by one with a different number of sides mod 4 and still have an orthogonal projection. However, you will need to find four points on the perimeter of the new polygon that not only form a square, but such that the opposite vertices "are seen from both sides", meaning a certain bounding hyperplane of the polygon contains a vertex of the square and is parallel to a diagonal of the square. I suspect any other solutions will be not orthogonal projections or necessarily off center or both. Gerhard "Ask Me About System Design" Paseman, 2011.08.23 –  Gerhard Paseman Aug 23 '11 at 22:15
    
Also, I was inspired by the 'trip-lets' on the cover of Hofstadters GEB book. I used a bandsaw and other tools to make custom versions for my parents and for a few others. I suspect that one can use a slight distortion of similarly sized shapes for the three orthogonal projections. Using exact shapes will often entail an off-center placement or a slight skew from orthogonal arrangement. Gerhard "Ask Me About Wood Shaping" Paseman, 2011.08.23 –  Gerhard Paseman Aug 24 '11 at 3:24
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I've decided to expand my comments into a challenge. Hopefully those with the skills and interest will take it up and provide some definite answers for Joseph.

Take two polygons (with two distinct numbers of sides) T and S. For me it is helpful to think of T as a square and S as a triangle, but you are welcome to choose your two favorite polygons. Consider normal vectors in R^3 to each of these objects, and set the normal of S to be part of the z-axis. Compute the set of all vectors v in P(S,T) such that if the normal of T has the same direction as v, the intersection of the two cylinders induced by S and T and the normals is a set that has the desired shadows as indicated by the normals.

I claim (by intuition and without proof) that the set P(S,T) is (after a suitable normalization) bounded and bounded away from the normal of S. Even the generalization where the normals do not share a basepoint and all orientations of T are considered, even that set is bounded and bounded away from the normal of S.

The idea now is that one can define a disk on a sphere through whose center passes the normal of S, and serves as a guarantee that the shadow of T will not be found by orienting the normal of T to pass through the disk.

My belief is that these disks are large enough to allow an estimate that looks like "If S (where now S is the finite set of integers in Joseph's post) has n+3 numbers less than n^2 , The polygons of S cannot all be realized by a single connected convex polytope in R^3".

The challenge is to compute the size of these disks for many pairs S and T.

Gerhard "Ask Me About System Design" Paseman, 2011.08.23

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"For me it is helpful to think of $T$ as a square and $S$ as a triangle." For me it is helpful to think of $S$ as a _S_quare and $T$ as a _T_riangle! Sorry, Gerhard, I couldn't resist! :-) Your intuition coincides with mine. I do believe your sketch can be expanded to a proof. Great insight! –  Joseph O'Rourke Aug 24 '11 at 0:40
    
Good. It's nice to know someone takes care in reading my posts. An additional intuition is that the disk will be much larger for consecutive integers than for integers of the same parity. I hope you or someone else will take up some form of this challenge. Gerhard "Dyslexics of the World, Untie!" Paseman, 2011.08.23 –  Gerhard Paseman Aug 24 '11 at 2:04
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