Please see the original problem specification (which Joseph O'Rourke was responding to in his answer) below.
I'm interested in a particular case of the problem where one wants to pack equiradius spheres around a central sphere s.t. every sphere has the same kissing number (i.e. contacts with its nearest-neighbors) and its nearest-neighbors are arranged in a rotationally symmetric fashion. In the problem specification (below), the centerpoint of each sphere represents a vertex, and edges represent kissing contacts between nearest-neighbors.
It troubles me that I have no intuitive understanding for why the set of solutions to this problem should be restricted to the platonic solids, and I'd really like a better appreciation for why this is so. Intuitively it feels like there should be some set of solutions involving more vertices/spheres than the dodedecahedron or icosahedron allows.
I'd like to decompose a sphere (say, or radius $R_s$) into a fully connected graph where:
(1) - The degree of each vertex is fixed at $N$,
(2) - Each edge in the decomposition graph has a constant bending angle $\theta$ and fixed length $L$,
(3) - The $N$ edges around each vertex, $v_i$, are arranged in a rotationally symmetric pattern,
(4) - The vertices $v_i$ lie on the sphere one is attempting to approximate.
As a function of the bending angle/curvature of each edge $\theta$, the uniform length of each edge $L$, and the number of edges around each vertex $N$, when can I generate a fully-connected graph that approximates the closed surface of the sphere, and how many vertices will it contain?
Again, any and all feedback is greatly appreciated!
Note - This is similar to my earlier question (which disallowed curved edges): "Constructing a graph that approximates a sphere using rotationally symmetric building blocks with equal numbers of edges". I now realize (thanks to André Henriques) that the requirement of having a vertex-transitive/isohedral graph, with fixed-length straight edges and rotational symmetry around vertices, should restrict one to the five platonic solids (though I'm not entirely sure if the rotational symmetry constraint matters in this case).
Update - I have now removed the constraint that the graph is vertex-transitive.