Tiling with similar tiles Question 1: Is there a polygon $P$ that

*

*cannot tile the plane

and


*tiles the plane when copies of $P$ and some other polygon(s) all similar in shape to $P$ but of different size(s) can be used?

Basically, with copies of $P$ alone we should be able to form a layout with gaps which are all similar in shape to P and of different size(s).
Motivation: In basic tiling, we are constrained to use congruent copies of a candidate polygonal region (congruent up to some isometry) to fill the plane without gaps. Here, we consider relaxing the constraint to allow scaled copies of the candidate region and try to see whether this relaxation can non-trivially improve the chance a candidate region has of being a tile.
Note: Two cases to this question – $P$ is convex and not necessarily so.
Question 2: Is there a $P$ such that it is not a rep-tile but a large copy of $P$ can be tiled by several units all similar to $P$?
Note: By rep-tile, we mean a polygon that can be cut into some finite number of equally scaled down copies of itself. So, the $P$ one is looking for cannot be tiled by any finite number of equally scaled down copies of itself but can be tiled with copies of itself which have been scaled down by factors that are different among themselves.
 A: Here is an answer to Question 2. The following shape (attributed to Karl Scherer on this website) tiles into similar shapes of different sizes.

Convincing myself that it is not a rep-tile took me several case distinctions - I found it easiest to start with one of the right angles and construct the tiling from there until deriving a contradiction (angles of $\pi/3$ can only be "filled" in one way, right angles and angles of $2\pi/3$ can be "filled" in two different ways).
A: 
This tile $P$ tiles an open half-plane in a hyperbolic fashion. It's "tiled" in the sense that it's a union of copies of tiles similar to $P$ with disjoint interiors. So you can sort of tile the plane with tiles similar to it, you miss just one line. More interesting (and probably what you meant) is what happens if you restrict to finitely many similar copies of $P$, since I guess any polygon's similar copies tile a full measure subset of the plane.
I should probably also add Paint as an answer to
Time-saving (technology) tricks for writing papers .
A: It seems you did not restrict to finitely many scaled down copies of the tile. So here is one interesting tile for which you need infinitely many scaled down copies in order to tile the plane. On the left is the tile itself, and on the right is how to use scaled copies to tile an L tromino, after which it is trivial to tile the plane.

A: Are you willing to allow tiles with fractal boundary?  (I can see that you write "polygon" throughout, so maybe not?).
If so, then another example is the so-called "Koch snowflake".  See https://en.wikipedia.org/wiki/Koch_snowflake#Tessellation_of_the_plane .  If you allow such tiles, then this is also a positive answer to your Question 2.
A: Another family of fractal examples is provided by Thurston's famous unpublished notes:
http://timo.jolivet.free.fr/docs/ThurstonLectNotes.pdf .
Look at Figure 9.5.
A: 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.
