Actually, I think you can read this straight off your Coxeter complex pictures. In particular, that the pictures for duals are always the same…

]]>I think I got that somewhat wrong.

It’s a “plane of symmetry” which connects each vertex with the midpoint of the opposite side (in the odd p case). Vertices are actually opposite other vertices…

]]>A bit sketchy, but…

From the Wikipedia article for “Quasiregular polyhedron”, these polyhedra (which must be edge-transitive) have a Schlafli symbol of {p over q}, with p=3, q=3,4,5 (octahedron (really “tetratetrahedron”), cuboctahedron and icosidodecahedron, respectively).

From Coxeter, “Regular Polytopes”, page 65 (Dover), for a quasi-regular polyhedron with a Schlafli symbol of {p over q} with odd p, each vertex must be opposite to the midpoint of an edge. Hence, you get a great circle. Which by edge-transitivity implies every edge is on a great circle. (Basically, the odd p thing excludes cases like the “zig-zag equator” of a cube with the “poles” at two opposite vertices.)

So the only thing remaining is to show that these “halfway through” polyhedra must be quasiregular. Or equivalently, edge-transitive.

And I think you can get there by noting that these polyhedra are each (by construction) the common core of a dual pair of regular polyhedra.

]]>Thanks! I’ll fix the typo… but more importantly, I think you’ve made some real progress here on Puzzle 1. Somehow this symmetry should arise from interpolating between a regular polyhedron and its dual. I don’t have time to think about this right now… maybe someone else can give it a try?

Another specially nice feature of these ‘halfway through’ polytopes is that ‘every edge looks alike’:

In other words, their symmetry group acts transitively on their set of edges. This is not true for any of other semiregular polyhedra in the tables above (except, of course, the regular ones):

You got it! The truncation process I’ve been talking about doesn’t get us the polyhedra with Coxeter diagrams like

•—4—o—3—•

or

•—4—•—3—•

(and the similar ones with the numbers 3 or 5 replacing the 4 here). So, that’s our next order of business: we want to meet those members of our three families!

The snub polyhedra will remain snubbed.

By the way, I guess you’re not related to Thorold Gosset, discoverer of the famous ‘Gosset polytopes’ in 6, 7 and 8 dimensions. But it would be cool if you were…

]]>More likely (and more generally), I suspect you’re looking for the cases where the two ends, or all three nodes of the Coxeter diagram are filled in. Those (I think) complete the list of all but the chiral Archimedian solids. (Which seemed to be snubbed in this scheme… ;-)

]]>Not sure this is what you’re looking for, but if you take the Archimedean solids, and exclude those with more than two types of polygon and those that are chiral, you get the solids in your charts, plus one: rhombicuboctahedron.

]]>Let me give an attempt at saying something about Puzzle 1. One thing I noticed about those ‘halfway through’ polytopes is their high degree of symmetry: fixing a vertex and rotating the whole thing around that vertex halfway around maps the polytope to itself. Therefore, for every edge hitting that vertex there is an opposite edge also hitting that vertex. Now we can consider that new edge and apply the same argument at its other vertex. In this way, we end up with a sequence of edges lying on a great circle.

So in order to solve the puzzle, I would try to find a good explanation for that symmetry.

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