Does there exist infinite 3-valent pentagon-faced polytope

[ubersketch]

Basically, I'm asking whether you can have a 3 to a corner pentagon tiling that goes on infinitely, like the Euclidean plane, even as an abstract polytope.
Asking for a friend.

[Klitzing]

An infinite tiling without gaps and overlaps, even allowing for various arbitrary shaped finite pentagons, which are asked to tile the plane only three-valent, does not work. This is because you always get a medium angular defficiency of 360 - 3 x 108 = 360 - 324 = 36 degrees.

But whenever you provide any additional freedom to the above restrictions, either by allowing additional four-valent vertices, or some additional polygons with more vertices, or even just infinite pentagons, then you'll get back into play.

--- rk

[ubersketch]

What about an abstract case where there are no restrictions based on geometry?

[Klitzing]

For an abstract three-valent pentagonic-faced polyhedron you clearly have

Code: Select all

10N |   3 |  3
----+-----+---
  2 | 15N |  2
----+-----+---
  5 |   5 | 6N


and here any N would do.

That is: not only N=1 (the elliptical space case with identified antipodes of the sphere)
and N=2 (the spherical space dodecahedron),
but esp. too the limit for N towards infinity, yielding a truely infinite tiling.

So, having no idea for a corresponding realisation space geometry for the latter case ...

--- rk

[wendy]

Having three pentagons at a corner would lead to just two variations, the {5,3} of spherical and elliptic space.

You could play with a {5,6} by identifying opposite pentagons at each vertex as the same pentagon. In this case, a pentagon would represent the yickle formed in this way, But i believe it reduces to a mapping of {5,3} as a subset group of {5,6}.