Building equilateral triangles by reflecting tokens

Three tokens are placed at the vertices of an equilateral triangle with side length 1.
A move is to reflect a token at any other token. After several moves the tokens build again an equilateral triangle.

Which side lengths are possible for this new triangle?

The possible side lengths are

just 1 unit.

And here's why:

Imagine a triangular grid:

No matter how many times you reflect, your tokens will always be on the vertices of this grid. So you can't make any smaller equilateral triangles.

But what about bigger ones? Well, all moves are reversible. So if you made a bigger equilateral triangle, you could just undo those moves to shrink your triangle back to one unit. And if that were possible, you could shrink your size-1 triangle by performing the same moves!
So you can't shrink or grow your triangle; the only possible equilateral triangle has side length 1.

• Nice and clean proof. I like the reversibility argument. Good one! Nov 19, 2021 at 7:01
• I don't get it. If I "reflect" B on A, but leave C where it is, won't I get a triangle with sides 1,1, and (2 times equilateral height)? Nov 19, 2021 at 14:22
• @GeorgeMenoutis: It's not "again an equilateral triangle". Nov 19, 2021 at 16:26
• Not to be a party pooper, but are you sure the symmetry argument works? Assume that getting from a side-1 triangle to a side-2 triangle is possible by using reflections on the side-1 grid. By symmetry, getting from a side-2 triangle to a side-1 triangle is possible by using reflections on the side-1 grid. By similarity, this means getting from a side-1 triangle to a side-0.5 triangle is possible by using reflections on a side-0.5 grid. Does that necessarily contradict the fact that it's not possible by using reflections on the side-1 grid? Nov 22, 2021 at 5:03
• @Randal'Thor: The reflections aren't relative to any particular grid. The grid arises from the properties of the reflections. Drawing a different grid doesn't change the possible reflections. Nov 22, 2021 at 6:19