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0 pointsjacobolus2y ago0 comments

It's possible if you attach 6 pairs of your hexagons along two adjacent sides. https://archive.bridgesmathart.org/2017/bridges2017-237.html

For example, you can cover an octahedron with 4 regular hexagons, with 2 of the hexagons coming together at each octahedron vertex. Or you can divide these hexagons each into any Löschian number (integers of the form a² + b² + ab).

You might protest that these "hexagons" are now not connected like a honeycomb anymore, and now you have 12 "pentagons" mixed in, and that's true. But this idea turns out to still be very practically useful for making grids on a sphere, etc.

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lupire2y ago· 2 in thread

How do you attach a pair along 2 adjacent sides? By folding it?

jacobolusOP2y ago

You can cover e.g. an octahedron, a triangular prism, or a hexagonal dihedron, each of which has 6 corners where there is only 240° of surface meeting at the corner, exactly the amount covered by 2 corners from adjacent regular hexagons. So you end up attaching those hexagons along each of their two adjacent sides meeting at that corner. This makes them metrically hexagonal, but topologically pentagonal (if you consider the two distinct edges between adjacent faces to merge together).

lupire2y ago

Oh so it's like taking two hexagons attached at a common side and then stretching and pinching together another pair of sides.

In a computation sense, ignoring rigid geometry, you have 2 hexagons with sides 12 sides A1 to A6 and B1 to B6, and you are identifying A1 = B1 to make an edge AB1 and A2 = B2 to make a edge AB2, and then fusing AB1 + AB2 = AB12.

What's not obvious to me is why it's useful to fuse edges like that. I guess that it's only possible when the vertex between the edges has only 2 edges, with the same faces, and in discrete geometry that pre-fusion vertex isn't doing any work. But then why not fuse all possible edges, leaving you with a unigonal dihedron (double-sided circular disc)?

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