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Question 1: Is there a polygon $P$ that

  1. cannot tile the plane

and

  1. tiles the plane when copies of $P$ and some other polygon(s) all similar in shape to $P$ but of different size(s) can be used?

Basically, with copies of $P$ alone we should be able to form a layout with gaps which are all similar in shape to P and of different size(s).

Motivation: In basic tiling, we are constrained to use congruent copies of a candidate polygonal region (congruent up to some isometry) to fill the plane without gaps. Here, we consider relaxing the constraint to allow scaled copies of the candidate region and try to see whether this relaxation can non-trivially improve the chance a candidate region has of being a tile.

Note: Two cases to this question – $P$ is convex and not necessarily so.

Question 2: Is there a $P$ such that it is not a rep-tile but a large copy of $P$ can be tiled by several units all similar to $P$?

Note: By rep-tile, we mean a polygon that can be cut into some finite number of equally scaled down copies of itself. So, the $P$ one is looking for cannot be tiled by any finite number of equally scaled down copies of itself but can be tiled with copies of itself which have been scaled down by factors that are different among themselves.

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  • $\begingroup$ If you say "it tiles the plane" do you also allow mirrored copies? Or would you consider a polygon, that does not tile the plane, but does so if we allow its mirror image, an example for Question 1? $\endgroup$
    – M. Winter
    Commented Jul 12, 2020 at 8:06
  • $\begingroup$ Mirrored copies are of same area - so one would want P and scaled copies of P to be needed in tiling the plane. Considering a mirrored copy of P as different from P could lead to another variant of the question though. $\endgroup$
    – Nandakumar R
    Commented Jul 12, 2020 at 13:11
  • $\begingroup$ So the answer to @MWinter is yes? $\endgroup$
    – Ville Salo
    Commented Jul 12, 2020 at 13:48
  • $\begingroup$ No. As far as the scope of question 1 as stated goes, we need a P which needs copies of itself and similar but scaled down copies of itself to tile the plane. $\endgroup$
    – Nandakumar R
    Commented Jul 12, 2020 at 18:03

6 Answers 6

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Non-convex solutions to Question 1

Consider the following polygon (the outward angle on the right is the same as the inward angle at the top)

a tile

Since I didn't know any better way to show it does not tile the plane, I brute-forced my way through some case distinctions.

The only non-convex corner of any tile must meet a corner of another tile. It can't be either of the two bottom corners or the top right corner (the "unoccupied angle" would be too small to fit another corner in). If it is the top left corner, then we end up in the situation sketched below in the left picture. If it is the rightmost corner, then the result is the right picture below.

top left corner does not work rightmost corner does not work

In both cases we clearly cannot complete the partial tiling to a tiling of the whole plane.

On the other hand, we can tile a strip in $\mathbb R^2$ with scaled copies of our polygon as follows.

tiling with two sizes

Edit: Here's another shape (essentially based on the same principle).

enter image description here

The proof that it doesn't tile the plane is similar to above, but we can get rid of most of the casework due to symmetry. Tiling with two different sizes is again possible.

enter image description here

Convex solution

As noted in the comments, cutting the "bowtie" tile along the central symmetry axis solves the convex version of Question 1. Also note that Rao's preprint shows that only pentagons from belonging to one of 15 families tile the plane, and we can choose the bowtie such that the resulting pentagon belongs to none of them.

Edit 2: I just found out that this convex solution is also presented in Figure 3 in this paper from 1982.

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  • $\begingroup$ Thanks for this very intuitive answer! And I would think the convex case answer you give works fine. The nice thing is you seem to have arrived at a convex case answer via the nonconvex answer. $\endgroup$
    – Nandakumar R
    Commented Jul 20, 2020 at 19:36
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    $\begingroup$ Your convex pentagon is indeed a solution, i.e., it does no tile space. At least we can choose an instance of that shape that does not. Your pentagon has a degree of freedom: its interior angles are $\pi/2$, $\alpha$ and $\beta:=\pi-\alpha/2$. We can choose $\alpha$ as an irrational multiple of $\pi$, and then the only combinations of these angles that give $2\pi$ are either $4\cdot \pi/2$, $2\cdot \pi/2 + \pi$ (if a vertex meets an edge) or $\alpha+2\beta$. It remains some straight-forward case analysis to show that these cannot be the only types of vertices of a tiling. $\endgroup$
    – M. Winter
    Commented Jul 21, 2020 at 9:06
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    $\begingroup$ @M.Winter Thanks for this argument. I actually found another proof based on a different degree of freedom: if we make the "long sides" long enough, then the side where we cut the bowtie becomes the shortest side, and from there it's not hard to show that this side must always map to another shortest side (hint: all angles are at least $\pi/2$). $\endgroup$
    – Florian Lehner
    Commented Jul 21, 2020 at 9:17
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Here is an answer to Question 2. The following shape (attributed to Karl Scherer on this website) tiles into similar shapes of different sizes.

enter image description here

Convincing myself that it is not a rep-tile took me several case distinctions - I found it easiest to start with one of the right angles and construct the tiling from there until deriving a contradiction (angles of $\pi/3$ can only be "filled" in one way, right angles and angles of $2\pi/3$ can be "filled" in two different ways).

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  • $\begingroup$ That is a nice and simple example which has mostly settled the question except for the non-convex non- polyomino non-fractal case.. With reference to both questions, one could go farther and ask if there are other types of quadrilaterals (quads with different angle sets) which have this 'similar tiling' property, whether they have any shared property (properties) like say, being cyclic and so forth... And pentagons with the property …. $\endgroup$
    – Nandakumar R
    Commented Jul 21, 2020 at 17:16
  • $\begingroup$ The website behind the link in the answer has many more examples of polygons that can be decomposed into similar tiles of different sizes (including several non-convex ones); I wouldn't be surprised if some of them were non rep-tiles as well. $\endgroup$
    – Florian Lehner
    Commented Jul 21, 2020 at 17:50
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enter image description here

This tile $P$ tiles an open half-plane in a hyperbolic fashion. It's "tiled" in the sense that it's a union of copies of tiles similar to $P$ with disjoint interiors. So you can sort of tile the plane with tiles similar to it, you miss just one line. More interesting (and probably what you meant) is what happens if you restrict to finitely many similar copies of $P$, since I guess any polygon's similar copies tile a full measure subset of the plane.

I should probably also add Paint as an answer to Time-saving (technology) tricks for writing papers .

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  • $\begingroup$ In case of tiling problems it is implicitly assumed because of such examples that you cannot have infinitely many copies in a bounded region. But I agree that it would be useful to have more definitions in the question. $\endgroup$
    – domotorp
    Commented Jul 12, 2020 at 16:19
  • $\begingroup$ In the hyperbolic metric you do have finitely many tiles in a bounded region. Also, the open half plane is not tiled in the sense I define, by for instance a typical pentagon. (I think.) $\endgroup$
    – Ville Salo
    Commented Jul 12, 2020 at 16:36
  • $\begingroup$ Actually, I'm not sure about that comment or my answer. Does this tile a half plane in a more interesting sense than any other polygon? $\endgroup$
    – Ville Salo
    Commented Jul 12, 2020 at 16:39
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    $\begingroup$ Yes, the cover is locally finite. $\endgroup$
    – Ville Salo
    Commented Jul 12, 2020 at 16:45
  • $\begingroup$ @VilleSalo in the hyperbolic metric all tiles are congruent $\endgroup$
    – John Dvorak
    Commented Jul 21, 2020 at 17:24
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It seems you did not restrict to finitely many scaled down copies of the tile. So here is one interesting tile for which you need infinitely many scaled down copies in order to tile the plane. On the left is the tile itself, and on the right is how to use scaled copies to tile an L tromino, after which it is trivial to tile the plane.

The tile looks like an L tromino but with an inner quarter gone. That inner quarter can be recursively tiled by copies scaled down to 1/2, 1/4, 1/8 and so on.

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  • $\begingroup$ I had not thought of infinitesimal tiles. But what you point out seems quite interesting! $\endgroup$
    – Nandakumar R
    Commented Jul 20, 2020 at 19:38
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Are you willing to allow tiles with fractal boundary? (I can see that you write "polygon" throughout, so maybe not?).

If so, then another example is the so-called "Koch snowflake". See https://en.wikipedia.org/wiki/Koch_snowflake#Tessellation_of_the_plane . If you allow such tiles, then this is also a positive answer to your Question 2.

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  • $\begingroup$ I hadn't thought of fractals. But what you point out is certainly of interest. $\endgroup$
    – Nandakumar R
    Commented Jul 22, 2020 at 6:48
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Another family of fractal examples is provided by Thurston's famous unpublished notes [Groups, tilings, and finite state automata: Summer 1989 AMS Colloquium lectures]:

http://timo.jolivet.free.fr/docs/ThurstonLectNotes.pdf .

Look at Figure 9.6.

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