The $L$-tromino does not tile a $3\times 3$ square. However, it can tile $3\times3$ squares glued together in various ways:
I am interested in the polyominoes $P$ for which the $L$-tromino can tile a copy of $P$ scaled up by a factor of $3$ along both axes.
It is easy to see that some $P$ will fail, such as any $1\times k$ rectangle for odd $k$. However, it appears that every polyomino $P$ with an even number of cells can be tiled by $L$ shapes when it is scaled up by $3$. This holds for all polyominoes $P$ with $2, 4, 6, \ldots, 12$ cells.
One might be tempted to argue from induction: given a polyomino $P$ of size $2k$, subtract a $3\times 6$ domino from $P$ (which can be tiled), and tile the remaining size-$(2k-2)$ polyomino with $L$s by the inductive hypothesis. The problem with this is that there are arbitrarily large polyominoes for which it is not possible to remove a domino and leave a single connected polyomino behind; in fact, there are polyominoes that cannot be split into any collection of smaller polyominoes of even size. (This is one example.)
A variant of this inductive argument that might work is to show that from any polyomino $P$, one can remove either a domino, a $T$-tetromino, or two $L$ trominoes, to leave a smaller connected polyomino. If this were true, it would suffice for the induction, but I don't see a great way to prove it. (It may well have a counterexample!)
Can every polyomino with an even number of squares be tiled by L shapes of one-third the side length?
Best Answer
Yes, I think they can. This is an outline of a proof; there may be problems in the details that I missed.
Any even polyomino with a $2\times 2$ sub-square can be cut into two pieces through the square in at least two different ways, one that leaves two odd pieces and one that leaves two even pieces. Therefore it is possible to always cut such an even polyomino into two even polyominoes.
We can continue this until have all pieces dominoes, which are tileable, or thin polyominoes, polyominoes with no such $2 \times 2$ subsquares.
If a cell in a thin polyomino has exactly two neighbors, we can again make a cut in two ways to leave either two odd or two even polyominoes, so the same trick work as before. We end up with dominoes, or polyominoes with cells that have 1, 3, or 4 neighbors (and no subsquares).
We call the ones with more than one neighbor junctions. If a polyomino has more than one junction, we can make cuts at junctions so that we break the polyomino into smaller polyominoes with more than one cell and at most one junction. The arms of these junctions (or what used to be a junction) can only be one cell (otherwise we would have a cell with two neighbors).
These polyominoes are the right tromino, T-tetromino, (and maybe not strictly necessary) the skew-tetromino and the X-pentomino, all of which are tileable by right trominoes (when scaled).
I think a similar approach can work for analyzing odd polyominoes, although it is more difficult. In addition to the scaled $1 \times k$ rectangles (for odd $k$), we can also not tile certain thin L-shapes (such as the V-pentomino scaled) and certain thin T-shapes of (such as the T-pentomino scaled) -- I have not worked through a full proof but the reason is pretty much the same as for why scaled $1 \times k$ rectangles are not tileable. My conjecture is that an odd polyomino is tileable unless it is thin, and it cannot be broken into smaller polyominoes by cuts without leaving a $1\times k$ rectangle with $k$ odd.