It is widely known that we can square a square, which is defined as follows:

"Squaring a square" is tiling a square with integer side lengths with smaller squares with integer side lengths.

In reality it is very simple; we can for example tile a $2\times2$ square with four unit squares. Thus, people've thought of the restriction that all tiles must have different sizes, which they've named perfect. This turns out to be possible, with the smallest perfectly squared square being a $112\times112$ square:

Squared square

My question is simple: can we do the same with equilateral triangles? Meaning, can we tile an equilateral triangle with integer side lengths with smaller equilateral triangles with integer (but all different) side lengths? If so, what would be the smallest perfectly triangled triangle?


So I've been doing some thinking and I think I'm getting closer.

Let's have a look at the smallest triangle tile (we'll call it $\tau$). We'll split cases.

Case 1: $\tau$ is in a vertex. If the little triangle is a tile in one of the three vertices of the parent triangle, then it has one edge left; only triangles smaller than or as small as $\tau$ can fill that edge, but $\tau$ was the smallest triangle. This is impossible.

Case 2: $\tau$ is on the border. If $\tau$ is on the border of the parent triangle, but not in the corner, then either side of $\tau$ is touching that of a bigger triangle, which makes those two neighboring tiles overlap. This is also impossible.

Assume now $\tau$ is not on the border of the parent triangle.

Case 3: An edge of $\tau$ is touching a vertex. If an edge is touching a vertex, then since the neighboring tiles are bigger, we have the following (the orange tile being $\tau$, ignore that the blue triangle is the same size, it's beside the point):

Case 3 visual

Which makes this case equivalent to case 2.

And then the last case is when none of the above hold, but I'm quite stuck on that.


I don't think it is possible with a finite number of equilateral triangles.

First I realized that one can't tile a non-equilateral triangle with equilateral triangle (because the vertices of the non-equilateral triangle have an angle $\theta \neq 60$ so you can't stick an equilateral triangle there). So we can't divide and conquer

Then I thought of Sierpinski's triangle; however, there will be an awful number of repeated triangles that we will need to subsequently divide in a different way. To do that we need to figure out the different ways we have to partition one equilateral triangle with equilateral triangles (back to square one).

Now lets consider an equilateral triangle (ET for short) with side $n=q!$ where q is a really big number (so that we can have lots of divisors).

The only way we can partition the thing is like this: Tiling of ET

Lets consider an initial tile of size $m<n$, that means that we automatically get another triangle of size $n-m$. Lets suppose $m>n-m$. That means that we have a row of ETs with side $n-m$. If we try to divide the ET of side $m$ we have the same problem again, we have another row with a lot of small triangles of the same size(as shown in the figure).

So we have to focus on the triangles with side $n-m$, we need to further divide them in a different way, so we need to figure out the different ways to do that (back to square one). Each time you divide a triangle in a given way, you get even more ETs that you need to take care of.


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