# Why are there only 13 Archimedean solids and not 14?

I just finished a project on solids for geometry and I couldn’t help but wonder out of curiosity why are there only 13 Archimedean solids and not 14 or more?

• Do you know why there are only 5 Platonic solids? Mar 26 '19 at 22:47
• Ummm... why did you write "not 14" rather than "not 12"? ....or not "73"? Mar 26 '19 at 22:48
• It's analogous to, but a bit grungier than, the enumeration of Platonic solids: Any such polyhedron is a triangulation of $S^2$ and thus has $V - E + F = 2$, and the sum of face angles at a given vertex is at most $2\pi$. Now regularity enforces a couple of inequalities that only have a few solutions in positive integers. Mar 26 '19 at 22:53
• David G. Stork because I want to know why there aren’t 14 or more? I know it’s nothing less than 13 because if you take the 5 Platonic solids truncate them you get 7 Archimedean solids then take those and you get 4 more then you take the Platonic solid and you can snub it so you get the last two all together you get 13 so you can’t have less. That’s why I want to know why it’s not 14 OR more (that includes your “73” because it’s more than 14....) Mar 26 '19 at 22:54
• Checking the Wikipedia page - not only are there two infinite families that would fit the definition if they weren't specifically excluded, there's also a fourteenth that fits a slight variation on the definition (local symmetry instead of global symmetry). It's all a matter of exactly what rules we apply. Mar 26 '19 at 22:55

This question has been answered in this publication : Walsh, T. R. S. Characterizing the Vertex Neighbourhoods of Semi-Regular Polyhedra.'' Geometriae Dedicata 1, 117-123, 1972. link