# 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.

• maybe triangular numbers have some part to play? Commented Jun 24, 2016 at 21:33
• A reference for the information you provided would make it easier to understand and possibly answer your question Commented Jun 24, 2016 at 22:29
• @John Of course I've tried! But more you look at the solution for $n=5$ more miraculous it seems: yellow, orange and green parts differ by a rotation, blue and pink differ by a reflection — and then these two groups magically fit together… After a couple of months I still have no idea how to generalise this — but maybe someone can find other examples using brute-force computer search, for example… Commented Jun 26, 2016 at 18:15
• How did you manage to split the triangle into $10n^2$ congruent parts? Commented Mar 2, 2021 at 16:20
• @Vosatorp it's not quite obvious, but it can be done: dev.mccme.ru/~merzon/visual/pic_triangle10.png Commented Mar 3, 2021 at 18:29

Recently Pavel Guzenko found a way to divide an equilateral triangle into 15 congruent parts (and also into 30 congruent parts).

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 $$2\pi/3$$).

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 $$5, 6, 10, 13, 14, 15, 21, 30, 35, 39, 55, 65, 66, 70, 85, 95, 105, 119, 130,\ldots$$

In a rather different vein, a properly sized spherical equilateral triangle can be divided into 14 congruent, connected parts.

Begin with the seven-way division of an equilateral triangle shown below.

The segments connecting the original vertices of $$\triangle ABC$$ with those of $$\triangle A'B'C'$$ are angle quadrisectors at the former vertices, omitting the bisectors.

In Euclidean geometry this division gives only those congruences admitted by the $$D_3$$ symmetry. But in spherical geometry $$\triangle ABC$$ can be sized to render its vertex angles $$\psi=144°$$. With this sizing all seven pieces become equal in area, and the equilateral $$\triangle A'B'C'$$ becomes congruent with its edge-sharing neighbors. The outer pieces lying along the edges of $$\triangle ABC$$ are not yet congruent with the others, but these are also isosceles. So all seven pieces are isosceles, and next we divide each of them in half by drawing the bisector of an apex angle.

This gives the claimed 14 congruent pieces on the sphere, all of them being spherical triangles with vertex angles measuring $$36°,72°,90°$$.

The existence of this 14-way division is connected with the fact that certain generalized regular polyhedra (Kepler-Poinsot polyhedra) will cover the sphere exactly seven times. See this question and in particular Dr. Richard Klitzing's answer.