21

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.

Alphonse
  • 6,462
  • 5
    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). – quarague Apr 07 '20 at 08:54

2 Answers2

36

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.

Image via: https://twitter.com/gregegansf/status/1003181379469758464

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

enter image description here

nickgard
  • 4,266
  • Wouah, it seems so cool! I love this beautiful example! I'll have a closer look and then accept your answer. – Alphonse Apr 06 '20 at 08:07
  • 1
    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. –  Apr 06 '20 at 17:06
  • 1
    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). – Ilmari Karonen Apr 07 '20 at 02:56
16

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

enter image description here

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.

orangeskid
  • 56,630
  • 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. – Alphonse Apr 06 '20 at 06:47
  • @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. – orangeskid Apr 06 '20 at 07:06
  • @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. – orangeskid Apr 06 '20 at 07:13
  • 2
    Your picture doesn't work: quadrilaterals can always tile the plane. – Alphonse Apr 06 '20 at 07:13
  • @Alphonse: That is interesting, was not aware of it. – orangeskid Apr 06 '20 at 07:15
  • 1
    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. – Jaap Scherphuis Apr 09 '20 at 12:00
  • @Jaap Scherphuis: Yes, avoiding a cut through the parallel sides is the key. If we have a pentagon such that no positive integer combination of angles equals $\pi$ or $2\pi$ then it cannot tile... that would bypass the classification. So perhaps a hexagon with explicit angles and the explicit cut. Although almost any choice will do it, of course. – orangeskid Apr 09 '20 at 21:49