Dividing an equilateral triangle into N equal (possibly non-connected) parts

Question

It’s easy to divide an equilateral triangle into $n^2$, $2n^2$, $3n^2$ or $6n^2$ equal triangles.

But can you divide an equilateral triangle into 5 congruent parts? Recently M. Patrakeev found an awesome way to do it — see the picture below (note that the parts are non-connected — but indeed are congruent, not merely having the same area). So an equilateral triangle can also be divided into $5n^2$ and $10n^2$ congruent parts.

Question. Are there any other ways to divide an equilateral triangle into congruent parts? (For example, can it be divided into 7 congruent parts?) Or in the opposite direction: can you prove that an equilateral triangle can’t be divided into $N$ congruent parts for some $N$?

                                           

(Naturally, I’ve tried to find something in the spirit of the example above for some time — but to no avail. Maybe someone can find an example using computer search?..)

I’d prefer to use finite unions of polygons as ‘parts’ and different parts are allowed to have common boundary points. But if you have an example with more general ‘parts’ — that also would be interesting.

Answer 1

In this MathOverflow thread, there are dissections with $5n^2$ pieces for all $n\ge 6$ where the pieces are simply connected quadrilaterals. The smallest such is pictured here:

                                                    

In the thread, it is mentioned that Michael Reid claims to have found a simply-connected dissection using $7n^2$ trapezoids for some $n$, but the result appears not to be published anywhere.

In the case where the tiles are triangles, the 1995 paper Tilings of Triangles by M. Laczkovich has many important results. In particular, it states that there is a dissection of an equilateral triangle into $2469600=2^5\cdot3^2\cdot5^2\cdot 7^3$ triangles with side lengths $7, 8,$ and $13$.

In general, Theorem 3.3 in the paper states

Let $x$ and $y$ be non-zero integers such that $x+2y\neq 0\neq y+2x$. Then there is a positive integer $k$ such that the equilateral triangle can be dissected into $n=|xy(x+2y)(y+2x)k^2|$ congruent triangles.

This yields dissections with a number of triangles whose squarefree part is any of $ 5, 6, 10, 13, 14, 15, 21, 30, 35, 39, 55, 65, 66, 70, 85, 95, 105, 119, 130,\ldots$