# Apparent existence of a semi-regular polyhedron, but that I cannot find in any table.

I propose the existence of a semi-regular polyhedron with one square, one hexagon and two triangles at each vertex. The sum of angles at reach vertex is $$330°$$ and therefore the external angle is $$30°$$, which divides $$720°$$. That would imply $$\frac{720}{30} = 24$$ vertices, from which can be easily calculated that there are $$48$$ sides and $$26$$ faces ($$4$$ hexagons, $$6$$ squares and $$16$$ triangles). That´s fine, but I cannot see any sign of this supposed polyhedron in lists of the $$13$$ Archimedean solids, etc. What condition does this polyhedron violate? Why does it not exist?

• I am a bit confused. If you go around a vertex, in what order do the faces appear in your suggested construction? Dec 1, 2018 at 1:38
• In other words, can you draw the planar graph corresponding to your suggestion? I tried and I couldn't make all vertices look the same (remember that semi-regularity requires a group of isometries transitive on the vertices, not just the same unordered collections of faces at each vertex) but, perhaps, I'm misunderstanding your suggestion. Dec 1, 2018 at 1:49
• Have you tried making one (or part of one) out of card? Does it seem to fit together? Dec 1, 2018 at 1:57