3 ms·
You could say the sixth is an infinitely large one made of regular hexagons.
by jonsen 13y ago
You could say the sixth is an infinitely large one made of regular hexagons.
- T-hawk 13y agoBy 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 :-) )
- kazagistar 13y agoThat does not work, the sum of infinite zeroes is still zero.
- samatman 13y agoIt doesn't have to be infinitely large. It could simply be on a bounded manifold. And as another commenter noted, there are eight by this criterion.
- danbee 13y agoThe sixth is a teapot.
- ctdonath 13y agoAn issue of Journal of the ACM featured all six.
- lifeformed 13y agoFrom Utah.