Dividing an equilateral triangle into N equal (possibly non-connected) parts 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.
 A: Recently Pavel Guzenko found a way to divide an equilateral triangle into 15 congruent parts (and also into 30 congruent parts).

A: In a recent preprint https://arxiv.org/abs/1812.07014 M.Beeson shows how to  divide an equilateral triangle into $15×3^6=10935$ equal triangles (with sides 3, 5, 7 and one angle equal to $2\pi/3$).

A: 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Reid claims to have found a simply-connected dissection using $7n^2$ trapezoids for some $n$, but the result appears not to be published anywhere.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$
A: As the equilateral triangle has three axes of symmetry, congruent equilateral triangles can be split into a square number of shapes, multiplied by $1, 2,$ or $3$, so $6$ also works.
Any equilateral triangle can be split into $n^2$ sub-equilateral triangles, 
First, we need to find a square number that is divisible by $n$. As seen in the picture you posted above, each row of the triangle adds the next odd number of triangles: $1,3,5,7,9...$ This is the pattern used by square numbers:
$1=1||1+3=4||4+5=9||9+7=16||16+9=25...$ For this reason, any equilateral triangle can be split into $n^2$ equal shapes.
Due to the three axes of symmetry, any equilateral triangle can be divided into $2,3,$ or $6$ shapes, as seen in the picture below.

If we combine these two possibilities, we can get an equilateral triangle into $n^2,2n^2,3n^2,$ or $6n^2$ shapes. No other shapes are possible because of these two laws.
So, we can only split an equilateral triangle into $n=2,3,4,6,8,9,12,16,18,20,24,25,etc...$ congruent shapes. However, like in the example you gave, splitting a triangle into separate congruent parts is indeed possible.
A: I suppose there must be some way to do this with infinitesimals. Basically, think of this as a Riemann's sum. The equilateral triangle starts with the bottom left corner at the origin as shown.

I then make a right hand riemann's sum with "a" subdivisions.
Then I make Nth subdivision a part of each of the N non-connecting pieces of the triange.
Below is an example for "a"= 10 and N = 2

As you can see, the more you increase "a", the closer orange looks to blue.
So if you find the limit as "a" approaches infinity, you get the perfect shape of the triangle, and, for N = 2, the two pieces (orange and blue) are congruent.
This works for N any real number, you could do every 3rd column for N = 3 and so on. As long as "a" is infinity and not 10.

So basically in the end, you end up with N congruent "densities", not really N congruent shapes. But it's the best I can come up with :)
