A rep-tile is a shape that can tile larger copies of the same shape.

Question 1: Are there any convex pentagons that are also rep-tiles?

Remarks: 15 convex pentagonal tiles of the plane are known and none of them appears to be a rep-tile. Assuming this observation is right, one can invoke a proof given in 2017 by Michel Rao - that these 15 are the only convex pentagonal tiles possible - to answer our question in the negative. However, I don't know if Rao's proof has been validated and if there is a simpler (elementary) proof that there are say no convex pentagonal rep-tiles. So, here one is actually asking if there is a simpler proof for a weaker claim.

Definition: Let us say a multi-way rep-tile is a polygon P with the property: if P1 and P2 are magnified copies of P and P1 can be tiled with m units of P and P2 can be tiled with n units of P with n> m, then, a layout of n units can form P2 without m of the units in the layout together forming a P1. As shown on this page: https://en.wikipedia.org/wiki/Rep-tile, there are isosceles trapeziums with multi-way property (with m= 4 and n = 9). On the other hand, the square is obviously a rep-tile but not multi-way.

Question 2: Are there other convex polygons with this multi-way rep-tile property?

  • $\begingroup$ IIRC the proof of there being exactly 15 classes of tiling pentagons is specifically for periodic tilings, in which case a different pentagon that tiles via rep-tiles (i.e., aperiodically) wouldn't necessarily fall under that heading (but would be shocking in its own right as a convex aperiodic-only tiling!). I feel like one way of attacking this would be by looking at the various five-element planar graphs to see which might be realizable as adjacency graphs of a pentagonal rep-tile. $\endgroup$ Jul 6 '21 at 20:51
  • $\begingroup$ @StevenStadnicki Why would rep-tile tilings be aperiodic? I mean, I suppose they could be, but (non-fractal) rep-tile tilings are typically periodic. $\endgroup$
    – Buzz
    Jul 6 '21 at 22:12
  • $\begingroup$ @Buzz There's a tiling directly induced by the rep-tile (replace the original by its internal copies, scale up and possibly rotate so a subtile is in the same orientation and place as the original tile, iterate and take the limit, which is a tiling) and AFAIK that one's usually aperiodic. If a convex pentagon that wasn't on the list of the 15 families that tile the plane periodically turned out to be a rep-tile, then that induced tiling would per force have to be aperiodic. $\endgroup$ Jul 6 '21 at 22:30
  • $\begingroup$ If a pentagon rep-tile exists, then the supertile would form a planar graph with all faces having at least 5 edges. The Euler formula then gives a lower bound for the number of pentagons in the supertile (12 I believe). This should give that one vertex must have "high degree", meaning one angle must be very small. I wonder if angle chasing can show then that this is not possible. $\endgroup$
    – Nick S
    Jul 7 '21 at 2:52
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    $\begingroup$ @StevenStadnicki: I believe the classification of monohedral pentagonal tilings does consider aperiodic tiles, and finds none - see e.g. this Quanta article for a description of the classification strategy, which proceeds by trying local patches until a contradiction is reached rather than assuming periodicity. $\endgroup$ Jul 7 '21 at 20:07

Question 2 can be answered in the affirmative, at least: there are many triangles with the multi-way rep-tile property.

Every triangle has a simple tiling of itself with $k^2$ copies (just take the affine image of the standard equilateral tiling); these are called quadratic tilings by Michael Beeson in his 2010 paper on tiling triangles by congruent triangular tiles. So for a triangle to have this property, it suffices to have any non-quadratic tiling of size $m$, since we can choose a larger quadratic tiling with some $n>m$ tiles: the quadratic tiling has no subtriangles that are not themselves tiled quadratically, as should be obvious from looking at which straight lines can be formed. (This condition is of course also necessary.)

So we can then consult the linked paper for a variety of triangles with non-quadratic tilings: this is true of the $(30^\circ,60^\circ,90^\circ)$ triangle and any right triangle whose legs are in a rational ratio, which I think is exhaustive though I haven't read the full paper in a while to confirm this (it focuses mostly on the realizable numbers of tiles, which is related but not quite identical to our question here).

If you want to revise your definition to the stricter condition that there are two self-tilings, neither of which contains any other nontrivial self-tiling, there are still triangular examples: consider the unique $3$-tiling of the $(30^\circ,60^\circ,90^\circ)$ triangle by itself and its $k=2$ quadratic tiling.

As an addendum, the question of convex hexagonal rep-tiles ought to be much easier to resolve, since there are only three classes of convex hexagon which can tile the plane, and I believe none of them can even tile a half-plane (which is a prerequisite to being a rep-tile, though the proof involves a somewhat messy compactness argument). You can at least use the fact that any rep-tile must have an angle of at most $90^\circ$ to narrow the scope of possible tiles a little. Combined with a negative answer to Question 1, this would effectively reduce the classification problem of Question 2 to the convex quadrilaterals, which seems like it might prove rather difficult - it might end up being easiest to just classify all convex quadrilateral rep-tiles and then work out which ones are multi-way.

  • $\begingroup$ Thanks! The paper you linked shows (fig 1) that three 30-60-90 triangles can tile a bigger 30-60-90 triangle and this big triangle will not appear as a 'sub-layout' of any quadratic tiling with the unit triangle. A nice and simple example. And thanks for the thoughts placing question 1 in a wider context. $\endgroup$ Jul 8 '21 at 20:10

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