# Convex polygons that do not tile the plane individually, but together they do

I am looking for two convex polygons $$P,Q \subset \Bbb R^2$$ such that $$P$$ does not tile the plane, $$Q$$ does not tile the plane, but if we allowed to use $$P,Q$$ together, then we can tile the plane.

Here I do not require the tilings to be lattice tilings, or even periodic tilings. I allow tilings by congruent copies of $$P$$ and/or of $$Q$$, i.e. I am allowing rotations and reflections!

I haven't found any example, and maybe there could be none.

• Did you know that the equivalent question in $\mathbb{R}^3$ has a very simple elegant answer? Regular tetrahedrons and regular octahedrons do not tile space individually but combined they do (assuming the same edge length). Apr 7, 2020 at 8:54

There is a tiling of the plane made from regular heptagons and irregular pentagons.

We know that regular heptagons cannot tile the plane.

The irregular pentagon has four equal sides and one shorter side. A tiling of the plane by these pentagons would require two pentagons to share the short side (as they do in the image), but the resulting angle cannot then be tiled by other pentagons, so this irregular pentagon does not tile the plane.

I think the reference is to this paper: https://erikdemaine.org/papers/Sliceform_Symmetry/paper.pdf

• Wouah, it seems so cool! I love this beautiful example! I'll have a closer look and then accept your answer. Apr 6, 2020 at 8:07
• The hexagon-pentagon tiling shown in the paper (p. 207) has a more interesting symmetry, it's not as simple, regular and not obviously periodic. He calls it 'almost-regular'. Wonder why there are different arrangements. Maybe one is a near miss.
– user762332
Apr 6, 2020 at 17:06
• One detail perhaps worth pointing out, though it's kind of obvious in hindsight, is that the argument that any "tiling of the plane by these pentagons would require two pentagons to share the short side" involves the fact that all corners of the pentagon are obtuse, and thus it's not possible for the corners of two (or more) pentagons to meet in the middle of an edge of a third pentagon (as e.g. in this tiling). Apr 7, 2020 at 2:56

HINT:

Consider a convex hexagon that can tile the plane. There are three types of tiling hexagons, we take one of type 1, which has two opposite sides parallel and equal

Cut it into two pentagons. There are $$15$$ types of pentagons that tile the plane, see link. We can arrange the cut so that the obtained pieces do not tile the plane.

• Oups, I forgot to say that I allow tilings by congruent copies of $P$ and/or of $Q$, i.e. I am allowing rotations and reflections! So I think in your case the two convex polygons we obtain still tile the plane individually. Apr 6, 2020 at 6:47
• @Alphonse: Yes, I got that. Added a picture, perhaps it makes it clearer. But I have to think whether either of them tiles the plane individually, not sure. Apr 6, 2020 at 7:06
• @Alphonse: However, we can do the cutting procedure with any polygon that tiles. I have a feeling that at some point both pieces will not tile. Apr 6, 2020 at 7:13
• Your picture doesn't work: quadrilaterals can always tile the plane. Apr 6, 2020 at 7:13
• It's nice that you have been able to fix your argument. As long as the cut does not go through the parallel opposite sides (else you get type 1 tileable pentagons) but through another opposite pair, the resulting pentagons most likely won't be tileable. Choosing the cut so that each pentagon has distinct side lengths rules out almost all tilable pentagon types immediately. Apr 9, 2020 at 12:00