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January 19

rotationally symmetric circle graphs equivalent to regular polyhedra vertices and edges.

let a circle graph be a set of points equally set in a circle with some number of points connected. The Tetrahedron is equivalent to a four point circle graph with all points connected. The Octahedron is a star of david set in a hexagon (everything connects except the ones that go through the center). The cube can be done with each connecting to +2 and -2 with and connected to +1 or -1 alternating (so a {8,2} star with 4 neighboring cell connectors) that only has 90 degree rotationally symmetrical...

What are the equivalents for the icosahedron (12 points and 30 edges) and dodecahedron (20 points and 30 edges)Naraht (talk) 01:27, 19 January 2022 (UTC)[reply]

I'm confused about what the rules are supposed to be. Are all the outside edges of the polygon supposed to be connected? How much symmetry are you requiring, or are you asking how much symmetry is possible? With a cube and 90 degree rotational symmetry, there is one possibility with the outer edges all connected, 2 with 4 of the outer edges connected and 1 with none of the outer edges connected; that's four total. There is no solution with 45 degree rotational symmetry. --RDBury (talk) 05:49, 19 January 2022 (UTC)[reply]

The circle is IMO a red herring. For a solid with n vertices, number its vertices in any way from 1 to n. Select n distinct points in the plane and also number them from 1 to n. They can be arranged in any way, in a circle or just scattered around, as long as no three are collinear. Then, for each edge of the solid, if it connects vertex A to vertex B, draw a line segment from point A to point B. Some arrangements give more symmetrical diagrams than others, but none will have the same symmetry group in the Euclidean plane as the original Platonic solid in Euclidean space. --Lambiam 09:21, 19 January 2022 (UTC)[reply]

Are you asking what arrangement of vertices on a circle gives the most symmetric diagram? —Tamfang (talk) 03:27, 24 January 2022 (UTC)[reply]