4 ms·
By that method, there's eight. Also included would be the constructions of regular triangles meeting six at a vertex, and regular squares meeting four at a ver
by T-hawk 13y ago
By that method, there's eight. Also included would be the constructions of regular triangles meeting six at a vertex, and regular squares meeting four at a vertex.
- jonsen 13y agoYes, and the triangular infinihedron forms a dual pair with the hexagonal infinihedron. The quadratic infinihedron is its own dual, as the tetrahedron is. (Polyhedronal duality: Surface midpoints are vertices of the dual. Ex: The six surface midpoints of a cube are the six vertices of a regular octahedron; and vice versa.)
- nickzoic 13y agoIt occurs to me that (per the second argument) these have Schläfli symbol {s,m} = {6,3}, {3,6} and {4,4} ... for each of these, working out 1/s + 1/m - 1/2 = 1/E gives 1/E = 0 ... which makes sense (kind of) since those three cases are infinite regular tilings instead of regular polyhedra ... (EDIT: Just noticed jonsen's comment here too, "infinihedron" suddenly makes sense to me :-) )