I can bump your 12/31 value to 12/32. Start with your 6/12 octohedron; place it flat on a surface with two of the faces parallel to the surface. Copy the octohedron exactly one unit off the surface and add edges between the two pairs; we are now at 12 nodes and 24+6 or 30 edges, and the top octohedron has two degrees of freedom (fixing the bottom octohedron). Move the top octohedron so that one of its vertices forms the apex of a tetrahedron from the base triangle of the bottom octohedron; this lets you add two more edges from that vertex, and also two more edges from the corresponding upper plane, for a total of 32 edges. If this isn't clear I can draw a diagram. On Mon, Jun 20, 2016 at 7:52 AM, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
In 3D space, with n distinct points, what is the maximal number of unit distance lengths?
I'm trying to build up a sequence for OEIS, but I'm only positive I've got the right values on two or three of the entries. Here's what I have so far.
V -- E -- figure 4 -- 6 -- tetrahedron 5 -- 9 -- triangular bipyramid 6 -- 12 -- octahedron 7 -- 15 -- pentagonal bipyramid 8 -- 18 -- snub disphenoid or Raiskii spindle 9 -- 21 -- triaugmented triangular prism 10 -- 25 -- Nechustan spindle 11 -- 28 -- Augmented Nechustan spindle 12 -- 31 -- Double Pacman spindle 13 -- 36 -- Cuboctahedron + center 14 -- 40 -- Cuboctahedron + center + pyramid 15 -- 45 -- Icosahedron + 3 internal points 16 -- 50 -- Icosahedron + 4 internal points
http://math.stackexchange.com/questions/1830194/maximal-unit-lengths-in-3d-w... has a bit more info.
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