[math-fun] Can a polyhedron

with its vertices on a sphere and its edge lengths all equal have more than 92 faces? I'm looking at one (a doenneacontahedron?) with 12 regular pentagons (pentaga?), 60 equilateral triangles, and 20 non-regular (but equilateral) hexagons. There are 90 vertices, all tetravalent, so maybe there is a C_90 fullerane (as opposed to fullerene). I also made a pentacontahedral candidate for C_48 fullerane with 30 squares, 8 equilateral triangles, and twelve rhombi. --rwg