discrete geometrypolygonstiling
I am looking for two convex polygons $P,Q \subset \Bbb R^2$ such that
$P$ does not tile the plane, $Q$ does not tile the plane, but if we allowed to use $P,Q$ together, then we can tile the plane.
Here I do not require the tilings to be lattice tilings, or even periodic tilings. I allow tilings by congruent copies of $P$ and/or of $Q$, i.e. I am allowing rotations and reflections!
I haven't found any example, and maybe there could be none.
You are asking several questions, I understand only the first one, the rest will require some major clarification before they become answerable:
Question 1. Let $M$ is a Riemannian surface homeomorphic to the plane. Does $M$ admit a tiling? Here a tiling means a partition of $M$ into pairwise isometric relatively compact regions with piecewise-smooth boundary, such that two distinct tiles intersect along at most one boundary curve.
This question has a very easy an negative answer. For instance, start with the Euclidean plane $E^2$ and modify its flat metric on an open ball $B$, so that the new metric has nonzero (at some point) curvature in $B$ and remains flat (i.e., of zero curvature) outside of $B$. (This modification can be even made so that the surface $M$ is isometrically embedded in the Euclidean 3-space $E^3$: start with the flat plane in $E^3$ and make a little bump on it.) The resulting manifold admits no tiling, since all but finitely many tiles would be disjoint from $B$ and, hence, have zero curvature metric. At the same time, at least one tile $T_1$ will contain points where the curvature is not zero. Therefore, $T_1$ cannot be isometric to the tiles which are disjoint from $B$.
The same (or similar) construction can be used for any (connected, noncompact) smooth 2-dimensional manifold $S$, no matter what its topology is. This manifold always admits a metric $g_0$ of constant curvature $-1$. Now, modify the metric $g_0$ on a small ball $B\subset S$ making it into a metric of nonconstant curvature in $B$. Since $S$ is non-compact, the same argument as above shows that the new Riemannian manifold admits no tiling. If $S$ is compact, then, of course, each metric on $S$ admits a tiling, consisting of a single tile. If you require at least two tiles, one can always construct metrics which do not admit tilings, but this, requires more work.
However, if your manifold $M$ is, say, a flat 2-torus, i.e., genus 1 closed surface equipped with a zero curvature metric, then such $M$ always admits a tiling consisting of more than one tile. Let me know if you want to see a proof (it is quite easy).
We can achieve this using two polyominoes with one having double the dimensions of the other and the second copy rotated $90^\circ$ and reflected.
The basic element showing the two polyominoes is below. Its bounding box is $165\times98$ and the key to the tiling is that the longer vertical edges are both $82$ units long while the shorter vertical edges are both $16$ units long. It's clear that one polyomino can't tile the plane on its own; the second polyomino is needed for them to hook together.
And here's the tiling (click to enlarge):
The general approach is to hook two polyominoes together in the pattern below. Imagine this initially as a series of rectangles from bottom-left to top-right: $2\times1$ at bottom-left, then $2\times4$ just above that, then $8\times4$ and finally $8\times16$ in the large rectangle at top-right, with the other dimensions being fully determined from this.
A Python program was used to vary the dimensions of the smallest rectangle and redraw the resulting polyominoes. The data from the program was used to find the correct size of rectangle to generate the required matching edges. In the final polyomino the small rectangle is $2\times16$.
Best Answer
There is a tiling of the plane made from regular heptagons and irregular pentagons.
We know that regular heptagons cannot tile the plane.
The irregular pentagon has four equal sides and one shorter side. A tiling of the plane by these pentagons would require two pentagons to share the short side (as they do in the image), but the resulting angle cannot then be tiled by other pentagons, so this irregular pentagon does not tile the plane.
Image via: https://twitter.com/gregegansf/status/1003181379469758464
I think the reference is to this paper: https://erikdemaine.org/papers/Sliceform_Symmetry/paper.pdf