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@Wolfgang Right. They only have tilings if they also have a parallelogram tiling, but some of those tilings don't look at all like any parallelogram tiling. They can have some tiles that go across the edge where two (or even three) parallelograms would meet.
@Wolfgang I've tried a few different shapes made out of triangles of side $5$ and in every case I've found that they have a tiling iff they also have a tiling by that parallelogram. It feels like this should be telling me something but I don't know what it is. (Obviously if we could prove this then we'd be done, since that parallelogram can't tile a triangle (for colouring reasons)).
@Vincent That claim comes from this paper (Conway and Lagarias. Tiling with polyominoes and combinatorial group theory. 1990.) I think the argument is that colouring arguments can only be used to disprove the existence of signed tilings in which some copies of an "anti-tile" are allowed to appear. The tile homotopy group always detects whether or not a signed tiling exists and can sometimes tell you more. I think all this is best explained in the Michael Reid paper which I linked to above.
