A planar set is said to be a rep-tile if it can be tiled by congruent shapes, each similar to the original. If there are $k$ such shapes, each scaled down by a factor of $\sqrt{k}$, it is said to be rep-$k$. Rep-tiles exist for $k=2,3,4,5$:
More examples (not all triangular) are at Wikipedia.
From Michael Beeson's paper here, we know that triangular rep-$k$ tiles exist if and only if $k$ is a square, a sum of two squares, or three times a square. (This yields $k=2,3,4,5,8,9,10,12,13,16,17,18,20,25,26,27,\ldots$)
The Gosper curve is a rep-tile of order $7$.
I have found what looks like an instance of a rep-$6$ tile here:
However, this is a pretty atrocious shape. Are there any polygonal rep-$6$ tiles? I'd settle for simply connected, in a pinch.
Best Answer
If I understood the problem correctly, a possible (though boring) solution would be to take a rectangle of width $\sqrt{3}$ and height $\sqrt{2}$ and divide it like this:
This way we get six smaller rectangles of height $\sqrt{2}/2 = \sqrt{3}/\sqrt{6}$ and width $\sqrt{3}/3 = \sqrt{2}/\sqrt{6}$, which are obviously congruent to the original rectangle, thus forming a polygonal rep-$6$ tile.
More generally, for any pair of integers $m,n$ the rectangle with sides of length $\sqrt{m}, \sqrt{n}$ is a rep-$(m\cdot n)$ tile. In particular, taking $m=k, n=1$ we obtain rep-tiles for any positive $k$.