# For which sidelengths are there polyominos composed of three squares that tile the plane?

Given three naturals $a<b<c$. We consider polyominos, connected or not, which are composed of three squares of sides $a,b,c$.

How can one characterize all triples $a,b,c$ for which such a polyomino exists which tiles the plane?

Translations and rotations are allowed (and, why not, reflections, but I don't know if they make a big difference).

It is easy to find such polyominos if either $b=2a$ or $c=\lambda a+\mu b$ with $(\lambda,\mu)\in\{ (1,1), (1,2),(2,1),(2,2),(0,2),(-1,2),(-2,2)\}$. In each case, we can build a fundamental domain by taking a certain such polyomino plus a copy rotated by 180°. Some examples are displayed here.

I wonder if this condition is necessary (and thus yields the answer). It seems quite obvious that for triples $a,b,c$ not satisfying it, there cannot be a simple fundamental domain as above. But it might be possible (though I don't think so) that there exist some 'sporadic' constructions involving 90° rotations and/or reflections. Is there a way to rule those out?

• Consider the adjacency relationships: How many squares share part of a side with a given square. I think you will find that there will be tight restrictions on b and c given that you can't have too many smaller squares surrounding a square of size c. – The Masked Avenger Mar 25 '15 at 18:23
• @TheMaskedAvenger Yes that may be a good start. But there can always be parts of size c around, so I wonder how this idea might be formalized. Certainly that needs some case analysis. – Wolfgang Mar 25 '15 at 21:02
• What's the smallest triple that doesn't satisfy any of those equations? – Gerry Myerson Mar 25 '15 at 22:31
• @GerryMyerson after adding one case I had missed, the smallest unknown is 1,3,9. – Wolfgang Mar 26 '15 at 7:46
• I know :) and the question is if there are polyominos as above that tile the plane without involving generalized hexagons. – Wolfgang Mar 26 '15 at 16:03