## 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)

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.

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.

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

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.

## 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.

scaled copies of Pto 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. $\endgroup$