Yes, 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.)

It 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 :-) )

An issue of Journal of the ACM featured all six.

From Utah.