The planar graph associated with the octahedron --- see eg. first diagram at http://mathworld.wolfram.com/OctahedralGraph.html has symmetry group of size 6,3,2,1 according to the lengths assigned to its edges. Replacing edge-adjacent triangular faces by tangent circles excludes 3-cyclic symmetry; hence the numbers in my rider. WFL On 10/19/15, William R. Somsky <wrsomsky@gmail.com> wrote:
I *think* (to the extent that my insomniated brain functions at this hour) that a necessary (though not necessarily sufficient) condition for an N-fold rotationally symmetric solution is that cos[pi/N] be rational, which (Niven's theorem?) only occurs for N=1,2,3. N=1 is essentially no symmetry (which is a whole 'nother ball-of-wax), so it'll be two "orbits" of two or of three for symmetric cases. On Oct 19, 2015 05:58, "Fred Lunnon" <fred.lunnon@gmail.com> wrote:
This problem impinged during my customary brainstorming session in search of convincing reasons not to get out of bed. I have not examined it, so have no idea whether it is well-known, easy, hard or impossible.
Deventer's "Gear Shift" puzzle http://www.jaapsch.net/puzzles/gearshift.htm is topologically equivalent to an octahedron with a gear on each face, meshing with three gears adjacent at edges.
Is it possible to morph this concept into its planar map: a flat train comprising one ring enclosing two 3-planet tiers around one sun?
Furthermore, do such trains exist with 6-fold, 2-fold, and no symmetry?
Fred Lunnon