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Apr 4, 2021 at 22:55 comment added Yaakov Baruch I think that the idea of commensurability together with each tile having to be a triangle (see my partial answer) could go a long way to attack the problem in a combinatorial sort of way, not too ramified. For a example assuming that there is a simple intersection somewhere (only 2 lines) it may not be too hard to show that the tiling can be continued in only 2 ways, leading to patterns (4) and (6) in the question.
Mar 23, 2021 at 13:06 history edited Will Brian CC BY-SA 4.0
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Mar 23, 2021 at 12:59 comment added Will Brian @WlodekKuperberg: I don't see how. Actually, I would think the reverse could be more tractable: prove that every periodic fair cutting is extra-fair. But I don't see how to do that either.
Mar 22, 2021 at 23:21 comment added Wlodek Kuperberg Nice observation. Do you think it would be not too hard to prove that every [extra-] fair cutting is periodic?
Mar 22, 2021 at 14:27 history answered Will Brian CC BY-SA 4.0