![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png) simont |
Wed 2003-12-24 04:36 |
| I think there's a standard result that means a solid of this type needs twelve 5-edge vertices if the rest are all going to be 6-edge, otherwise you get a flat triangular lattice rather than a closed solid. So yes, there are twelve "pentagons" visible, and I think this would continue to be the case on any solid of this type as you increase the number of points, no matter how many 6-edge vertices go in between. In the 60-point case: if you remove each 5-edge vertex and replace the five faces surrounding it with a pentagon, I don't think you get anything particularly interesting. If you look at the visible 5-edge vertices: 
you see that actually they aren't even consistently separated. The two "pentagons" I've marked in the upper left end up contiguous; the central one is separated from both of them by a single layer of triangles (not enough to form hexagons); and the central and lower ones touch at a vertex rather than an edge. This is unquestionably an irregular polyhedron, not a trivial knockoff of buckminsterfullerene. |
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