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…

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.

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):

or

(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…

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.