# Can one always tile the plane with a regular polygon + a single other shape?

We know that regular triangles, squares and hexagons can tile the plane without leaving any "hole".

However, I've noticed that many regular polygons can tile the plane if we allow for a single type of "hole" (i.e., another shape) to be present.

The following image contains an example with pentagons and rhombi:

What we obtain in this case is not a periodic tiling, but rather an aperiodic one: Still, we are able to tile the plane with these two shapes.

Another example, this time with decagons + "concave hexagons":

My question is: Can we always tile the plane by combining a regular polygon and a single other shape?

I'm also interested in the extension to star polygons.

Edit

As pointed out in the comments, the way I asked the question was imprecise. I guess that we should add the additional constraint that no "hole" can be in contact with another "hole", otherwise there will be trivial solutions to the problem.

• Take any square grid and place a "small" regular polygon at each "corners". Jul 1, 2021 at 14:41
• You can put the regular polygons in a rectangular bounding box, tile the rectangles, and call the space outside the regular polygons "holes". Jul 1, 2021 at 14:42
• @peterwhy (and also achille hui) Ok, the way I asked the question was not precise. I guess that what I want is the additional constraint that each polygon must have at least one side in common with another? I don't know whether this is enough to obtain what I have in mind though. Jul 1, 2021 at 14:45
• @valerio Then you can still put two regular polygons together on one common side, put that $(2n-2)$-gon in a rectangle (possibly slightly larger than a bounding rectangle, to merge holes), and tile the rectangles. Jul 1, 2021 at 15:01
• @valerio you still can, the paired up regular polygons have some parallel sides, so when bounding the paired polygons in rectangles, make sure the parallel sides are on the bounding rectangles. Then tile the rectangles by translation, so that each regular polygon also touches another regular polygon from an outside rectangle. Jul 1, 2021 at 15:56

Yes, it is always possible.

Suppose $$n>3$$ is odd. Then we arrange the polygons as shown (examples below for $$n=5,7,9$$):

Note also that $$n=3,4$$ are trivially possible.

Now suppose $$n>4$$ is even. Then we take the construction from $$n/2$$ (which we may assume exists by induction) and simply truncate the corners of the $$n/2$$-gons to produce $$n$$-gons which share the same adjacency graph, while causing equal perturbations to the "holes" so that they remain congruent to one another.

• Ok, this is true. I formulated the question in an imprecise way. I think (but I am not sure) that what I want is the additional constraint that each polygon has at least 1 side in common with another identical polygon. Jul 1, 2021 at 14:51
• Thanks for the clarification - I've edited the post to answer the updated question. Jul 1, 2021 at 15:40
• Thank you for your updated answer! I'm thinking now that the actual condition I wanted is that each polygon has two sides in common with other identical polygons. I have to think a bit about your answer, but it seems quite convincing. PS: I noticed that in all the examples you brought the "trick" is to arrange the polygons in an hexagonal configuration. I wonder if one can reformulate the question based on this observation. Jul 1, 2021 at 15:56
• I believe all of the secondary polygons in these construction will have symmetry under horizontal and vertical flips, so you can cut them into quarters to obtain smaller pieces which touch two others of the same shape. Jul 1, 2021 at 19:27
• No, this construction works for any odd $n$ (in fact, it also works for even $n>8$). See here for an $n=11$ example - the general construction is just placing translated pairs of vertically reflected $n$-gons so they align on sides with a certain slope. We actually have some flexibility in which side to join the polygons - the only requirements are that it be rotated more than $90^\circ$ from the horizontal side, and that the two diagonally-adjacent $n$-gons don't overlap. Here is another way to place $11$-gons. Jul 2, 2021 at 16:18