Re: [math-fun] Non-standard "polyhedra" ...
Combinatorily a polyhedron can be just thought of as a decomposition of the sphere into geodesic polygons. If you expand the definition to include other compact surfaces besides the sphere, there are many lovely examples of regular polyhedra. For instance, the 3-holed torus can be tiled in a regular way with 8 hexagon, with each vertex surrounded by 4 of them. Resulting in a symmetry group of size 96. (Determined by: one hexagon going to any of the 8, in any of 12 ways.) But if you try to tile the sphere with hexagons, you find it won't work. Suppose there are N hexagons averaging q >= 3 around each vertex. The number of vertices must be V = 6N/q. Then the number of edges must be E = 6N/2 = 3N. The number of faces is F = N. Since the Euler characteristic X(S^2) = 2, this means 2 = X(S^2) = V - E + F = 6N/q - 3N + N = N(6/q - 2). But since q >= 3, this is impossible. —Dan Colin writ: ----- Forgive me if this is obvious and well-known. Because of Euler we know that for polyhedra without holes we have V+F=E+2. As a result we can't have a polyhedron made entirely of hexagons, and need at least 12 pentagons or an equivalent set of shapes. Etc. But things change if we allow intersecting faces. Clearly we have a lot of scope for variations, but we can always retain limits to make things reasonable. But ... Can we make a "polyhedron" consisting only of hexagonal "faces"? -----
On 2020-05-06 22:27, Dan Asimov wrote:
Combinatorily a polyhedron can be just thought of as a decomposition of the sphere into geodesic polygons.
Well, yes and no. More about that in a minute.
If you expand the definition to include other compact surfaces besides the sphere, there are many lovely examples of regular polyhedra.
For instance, the 3-holed torus can be tiled in a regular way with 8 hexagon, with each vertex surrounded by 4 of them. Resulting in a symmetry group of size 96. (Determined by: one hexagon going to any of the 8, in any of 12 ways.)
That is nice ... thank you.
But if you try to tile the sphere with hexagons, you find it won't work.
Yes ... my comment tried to make it clear that I knew that. In particular, I wrote:
Because of Euler we know that for polyhedra without holes we have V+F=E+2. As a result we can't have a polyhedron made entirely of hexagons, and need at least 12 pentagons or an equivalent set of shapes. Etc.
I thought that would be clear, but thanks for including the entire proof. However ... Consider the stellated dodecahedron. In the usual, obvious, trivial sense that has 60 faces. But in a more relaxed sense we could consider it to have 12 mutually intersecting faces, each of which is a pentagram. So there are different things that can be relaxed. We can work on surfaces with holes, as you did earlier, and as is the case with the Szilassi polyhedron. But we could choose to relax a different condition, and have the "faces" penetrate/intersect. If we do that, can we get "polyhedra" with only hexagons? Cheers, Colin -- I've worked out why language is weird. It's because it's invented by, and used by people. And people are weird.
Possibilities: The Tetrahedron of David, with 8 intersecting faces, each a regular 6/2 hexagonal star. It seems allowable as a regular solid, depending on how the rules are bent. Dan Asimov also mentioned something, many years past, based on heptagonal stars, I think 7/2. I may have this confused with a different idea, using regular heptagons, joining four or five around each vertex. Another possibility is based on 8/3 octagonal stars extended from an interior central cube, but I think this needs some triangular faces to close properly. I've been curious how the star notion works out in higher dimensions: There are six regular hyper-solids in 4-space, but for 5+ dimensions, the only fully compliant objects are the generalized cube, tetrahedron, and octahedron. What happens if we allow stars? Rich --- Quoting Colin Wright <maths@solipsys.co.uk>:
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On 2020-05-06 22:27, Dan Asimov wrote:
Combinatorily a polyhedron can be just thought of as a decomposition of the sphere into geodesic polygons.
Well, yes and no. More about that in a minute.
If you expand the definition to include other compact surfaces besides the sphere, there are many lovely examples of regular polyhedra.
For instance, the 3-holed torus can be tiled in a regular way with 8 hexagon, with each vertex surrounded by 4 of them. Resulting in a symmetry group of size 96. (Determined by: one hexagon going to any of the 8, in any of 12 ways.)
That is nice ... thank you.
But if you try to tile the sphere with hexagons, you find it won't work.
Yes ... my comment tried to make it clear that I knew that. In particular, I wrote:
Because of Euler we know that for polyhedra without holes we have V+F=E+2. As a result we can't have a polyhedron made entirely of hexagons, and need at least 12 pentagons or an equivalent set of shapes. Etc.
I thought that would be clear, but thanks for including the entire proof.
However ...
Consider the stellated dodecahedron. In the usual, obvious, trivial sense that has 60 faces. But in a more relaxed sense we could consider it to have 12 mutually intersecting faces, each of which is a pentagram.
So there are different things that can be relaxed. We can work on surfaces with holes, as you did earlier, and as is the case with the Szilassi polyhedron. But we could choose to relax a different condition, and have the "faces" penetrate/intersect.
If we do that, can we get "polyhedra" with only hexagons?
Cheers,
Colin -- I've worked out why language is weird. It's because it's invented by, and used by people. And people are weird.
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On Wed, May 6, 2020 at 3:56 PM Colin Wright <maths@solipsys.co.uk> wrote:
Consider the stellated dodecahedron. In the usual, obvious, trivial sense that has 60 faces. But in a more relaxed sense we could consider it to have 12 mutually intersecting faces, each of which is a pentagram.
Using V - E + F = 2 - 2g, the small stellated dodecahedron has genus 4, so it's doing both. This link seems relevant: https://en.wikipedia.org/wiki/Isotoxal_figure -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
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