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Any member of a nontrivial family like this has to be a rep-tile; looking among those will give you many examples. Some specific examples: The family of all rectifiable polyominoes, which includes al …

If you impose no other restrictions, something like this works just fine: For a slightly more complicated variant incorporating both tile shapes:

Every such $S$ has a periodic tiling, in which finitely many disjoint copies form a set with one representative for each translate of some discrete lattice $L$ - see Bhattacharya 2016 or Greenfeld and … For an example of a disconnected $S$ that cannot form a lattice tiling, consider $\{(0,0),(2,0)\}$; it requires two copies to form a patch that tiles by lattice translation. …

tiling triangles by congruent triangular tiles. … the unique $3$-tiling of the $(30^\circ,60^\circ,90^\circ)$ triangle by itself and its $k=2$ quadratic tiling. …

Yes, it is possible; in fact, we can do it entirely with $5-12-13$ right triangles at different scales. First, note that we can three triangles at scales in the ratio $5:12:13$ to form a $5\times 12$ …

This MSE question exhibits two non-regular tetrahedra which can be decomposed into 8 smaller copies congruent to themselves; this yields $8^n$ for any $n$.

Yes. View the FCC packing as a series of stacked square packings, with spheres of unit radii centered at the points $(2a,2b,2\sqrt{2}c)$ and $(2a+1,2b+1,(2c+1)\sqrt{2})$ for all $a,b,c,\in\mathbb Z$: …

It produces every Penrose tiling using a 5-tuple of real numbers in $[0,1]$ (up to a measure-0 set of invalid choices), so it is very easy to generate samples from. … (From there, of course, it's easy to transform into a kite-and-dart tiling.) By normalizing and rotating, you can ensure your tiling is drawn from $\mathcal{P}$. …

I think there's no obstruction to just greedily choosing a working color by going outwards in a spiral: Each new tile never borders more than three pre-colored neighbors.

But by extending our tiling outwards in the strip between these two lines (at every point allowing for arbitrarily far-reaching extensions of the tiling that avoid $v$), we can fill up the entire strip … So that's the (rather tedious) proof - we try making a simple compactness argument for tiling the plane, unless it breaks, in which case we have a similar compactness argument for tiling a half-plane, …

It is not the case that all strongly obtuse subdivisions with a threshold over $120^\circ$ are possible through vertex-to-vertex connections. Consider the following dissection of a pentagon into trian …

Here is a solution with $24$ pieces: In general proving impossibility results will be extremely hard, as it is with almost all tiling problems of this form. …