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The set of convex polygons which tile the plane is, as of $2017$, known: it consists of all triangles, all quadrilaterals, $15$ families of pentagons, and three families of hexagons. Euler's formula rules out strictly convex $n$-gons with $n\ge 7$. (The pentagonal case is by far the most difficult one.)

I am interested in pairs of convex polygons that can collectively tile the plane. Specifically, I am curious how many sides a polygon in such a pair can have.

Here are some conditions to impose on such a tiling, from weakest to strongest:

  • There is at least one copy of each tile. (Without this condition, one can trivially take a pair consisting of a tiling polygon and any other convex polygon, and just never use the latter shape.)

  • There are at least $k$ copies of each tile.

  • There are infinitely many of each tile.

  • Every tile borders a tile of the other type.

  • The tiling is $2$-isohedral, i.e., every tile can be carried to any other tile of the same shape by a symmetry of the tiling.

Each of these conditions implies those above it.

In the weakest case, the number of sides is unbounded, as exhibited by the following example:

enter image description here

(The tiling is constructed by decomposing "wedges" of central angle $2\pi/N$ into congruent isosceles triangles, and then combining the central triangles to yield an $N$-gon in the center.)

Requiring at least $k$ of each tile still yields arbitrarily high numbers of sides, by taking the above construction for $N=M\cdot k$ and subdividing the $N$-gon into $k$ "wedges" which are $(M+2)$-gonal.

On the other end of the spectrum, I have found a $2$-isohedral tiling using regular $18$-gons, shown below:

enter image description here

After consulting this paper, it seems that the tiling pictured above is of type $4_2 18_{12}-1\text{a}\ \text{MN}\ \text{p}6\text{m}$ in their classification scheme (shown at the bottom of page 109); there are no $2$-isohedral tilings which allow for any higher number of contacts between different shapes, although type $3_1 18_{12}-1\text{a}\ \text{MN}\ \text{p}6\text{m}$ also works (and can be obtained from the above construction by cutting each kite-shaped tile in two). Thus, it is maximal among $2$-isohedral tilings.

What are the maximal tilings under weaker conditions? The maximal number of sides under each successively stronger restriction is a weakly decreasing sequence which goes $\infty, \infty, ?, ?, 18$. So far, I have no bounds on the missing two terms except that they are each at least $18$.

Some notes on this problem:

  • It is not necessarily the case that one of the tiles may tile the plane on its own; see this math.SE question for a counterexample.

  • If convexity is relaxed for either piece, the number of sides is unbounded even in the $2$-isohedral case (in fact, both pieces can simultaneously have arbitrarily many sides).

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  • $\begingroup$ Originally from Math StackExchange here, with no progress made on the question. $\endgroup$ Commented Jan 27, 2021 at 16:23
  • $\begingroup$ I wonder if you may be able to divide a square into copies of a convex $n$-gon and a convex $m$-gon for $n$ large.If you can do that, since the square tiles the plane periodically, you would get a tiling satisfying all conditions. $\endgroup$
    – Nick S
    Commented Feb 6, 2021 at 20:36
  • $\begingroup$ @NickS: This cannot be done with just one copy each, since their union would be a convex polygon with at least $n+m-4$ sides. With multiple copies, this would at least attain the "infinitely many tiles" condition, but I do not know if it is possible (my intuition leans against it). $\endgroup$ Commented Feb 6, 2021 at 21:29
  • $\begingroup$ I was thinking of finitely many copies. It is very easy to achieve if one polygon is allowed to be non-convex. $\endgroup$
    – Nick S
    Commented Feb 6, 2021 at 22:28
  • $\begingroup$ To make clear what I had in mind (maybe someone can fix this idea), consider the following: Lets say that the square has vertices $(2,0), (0,2),(-2,0), (0,-2)$. One piece is a regular $4n$-gon with 4 of the vertices at $(1,0), (0,1), (-1,0), (0,-1)$. Next, cut the remaining piece in 4 identical pieces by the segments $(1,0)-(2,0); (0,1)-(0,2); (-1,0)-(2,0)$ and $(0-1,0-2)$... The only issue is that the 4 leftover pieces are not convex. $\endgroup$
    – Nick S
    Commented Feb 6, 2021 at 22:31

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