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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.
enter image description 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?

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  • $\begingroup$ 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. $\endgroup$
    – The Masked Avenger
    Commented Mar 25, 2015 at 18:23
  • $\begingroup$ @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. $\endgroup$
    – Wolfgang
    Commented Mar 25, 2015 at 21:02
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    $\begingroup$ What's the smallest triple that doesn't satisfy any of those equations? $\endgroup$
    – Gerry Myerson
    Commented Mar 25, 2015 at 22:31
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    $\begingroup$ @GerryMyerson after adding one case I had missed, the smallest unknown is 1,3,9. $\endgroup$
    – Wolfgang
    Commented Mar 26, 2015 at 7:46
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    $\begingroup$ I know :) and the question is if there are polyominos as above that tile the plane without involving generalized hexagons. $\endgroup$
    – Wolfgang
    Commented Mar 26, 2015 at 16:03

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