# Tiling with incommensurate triangles

Say that two triangles are incommensurate if they do not share an edge length or a vertex angle, and their areas differ. Suppose you'd like to tile the plane with pairwise incommensurate triangles. I can think of at least one strategy.

Spiral out from an initial triangle in the pattern depicted below, with the red extensions chosen to avoid length/angle/area coincidences with all previously constructed triangles.

The central $$\triangle$$ is slightly non-equilateral. All $$\triangle$$s incommensurate.
It seems clear that this approach could work, although it might not be straightforward to formalize to guarantee incommensurate triangles. Which brings me to my question:

Q. What is a scheme that details a lattice tiling—all vertices at points of $$\mathbb{Z}^2$$—composed of pairwise incommensurate triangles?

This requires a more explicit design that effectively describes the triangle corner coordinates in a way that makes it evident that no lengths/angles/areas are duplicated. Without such a clear description, it is not even immediately evident (to me) that it is possible.

The same question may be asked for incommensurate simplex tilings with vertices in $$\mathbb{Z}^d$$.

• Do four zigzags (one for each quadrant, (1,0) to (0,1) to (x,0) to (0,y) and so on), where if need be record all lengths involved, and choose the next x and y to leave room for lengths for the other zigzags. (You may need to divide areas or follow pattern 2 discretely, but a zigzag should work.) Busy packing, I'll let you fill in. Gerhard "Going Down Under, American Style" Paseman, 2018.07.29. – Gerhard Paseman Jul 29 '18 at 22:04
• @Gerhard, if I understand your description, you have two triangles sharing the edge joining $(1,0)$ and $(0,1)$, so the triangles don't meet Joseph's definition of incommensurate. – Gerry Myerson Jul 29 '18 at 22:48
• Right. Neither does Joseph's first method. However, with an outward triangular spiral, there is still a chance. Gerhard "Maybe One Guess Will Work" Paseman, 2018.07.29. – Gerhard Paseman Jul 29 '18 at 23:17
• Why does "displacing corners" lead to incommensurate tilings? Based on the picture, it appears that every edge is shared by two triangles, so... – Victor Protsak Jul 29 '18 at 23:19
• @JoelDavidHamkins: Finally fixed the coloring when this was bumped to the front page. – Joseph O'Rourke Jan 12 at 19:31

Edit: This is a solution for non-commensurate side lengths only. The question also asks for non-commensurate angles, which this example does not provide.

I follow your approach. I start with a triangle with points in $$\mathbb{Z}^2$$ and add three incommensurate triangles, which leads to a new large triangle that is congruent to the first one and has points in $$\mathbb{Z}^2$$. We can repeat this process to tile the plane.

The first points are $$A(0,0), B(2,0), C(2,1)$$ as in the following image.

We now add three new points $$A'(-4,0), B'(2,-14), C'(24,12)$$ on the rays. We note:

(1) All new side lengths are strictly larger than the side lengths of the triangle $$A,B,C$$.

(2) All new side lengths are distinct (exercise).

(3) The new triangle $$\Delta C'A'B'$$ is congruent to $$\Delta ABC$$, in fact there is a right angle at $$A'$$ and the side $$A'C'$$ is double as large than $$A'B'$$.

The new arrangement of the four triangles is incommensurate by (1) and (2).

We can now repeat this process. Since (3) holds, we can use the congruency from $$\Delta ABC$$ to $$\Delta A'B'C'$$ to get the next larger triangle. By (1), all new lengths are larger than all old lengths and the new lengths are also distinct by (2). Moreover all points lie on $$\mathbb{Z}^2$$.

We can fill up all of $$\mathbb{R}^2$$ with incommensurate triangles.