It's also a fun exercise to take one of several images that show a football made out of hexagons and use the Euler characteristic to show that it is impossible to make such a football (no matter how much you bend or stretch your hexagons).
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.
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)?
Which I think means those points can't really lie on the edge of or interior of a hexagon. But I can't rule out that they might lie on a corner, it's hard to reason about.