# Dissecting an equilateral triangle into equilateral triangles of pairwise different sizes

It is know that a square can be dissected into other square such that no two of the squares have the same size. This is the simplest dissection of that kind: Is it also possible to dissect an equilateral triangle into equilateral triangles such that no two of them have the same size?

I think the following argument holds:

Suppose that $ABC$ is the great equilateral triangle with $A$ pointing upwards and that there exists a finite partition of $ABC$ into equliateral triangles such that every two triangles have different size lengths.

First note that in the vertex $A$ the angle must be filled with a small equilateral triangle $AXY$. There are a number of equilateral triangles $T_1,..,T_k$ in the partition with one side contained in $XY$ and the vertices pointing downwards. Among these there is one with the smallest side length: $D_1E_1F_1$ (with $D_1,E_1$ on $XY$).

The angles formed by the triangles $(T_j)$ in $D_1,E_1$ must be filled near their vertices with two equilateral triangles, one of which has smaller side length than $D_1E_1$. Denote with $A_1$ this vertex and $A_1B_1C_1$ this smaller triangle.

I think that now we can continue recursively to find always a smaller triangle:

• the triangle $A_1B_1C_1$ behaves just like the initial triangle $AXY$ and we will find one smallest triangle pointing downwards with side contained in $B_1C_1$

• this triangle has two neighbor triangles and one of them is smaller, which we denote by $A_2B_2C_2$, and so on.

We can continue this procedure indefinitely because all the triangle sides are different. • opinions? is this right? – Beni Bogosel Mar 10 '13 at 20:55
• To me the proof seems perfectly valid. And I think its a lot more elegant than the other one, too. – Dominik Mar 11 '13 at 15:26
• @Dominik: The article proves a more general thing, I guess, that's why it is more complicated. – Beni Bogosel Mar 11 '13 at 19:15

This paper claims it is impossible.

• Is my proof valid? – Beni Bogosel Mar 11 '13 at 6:32

Although as Ross Millikan points out, such a dissection is impossible, it is of interest to note that, provided we count upwardly oriented triangles as different from downwardly oriented, there is a dissection into 15 triangles. That is, none of the upwardly oriented triangles are same sized, nor any of the downwardly oriented.

http://arxiv.org/abs/0910.5199

[Note that site allows a free PDF download.]