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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.


          enter image description here
          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 $\mathbb{Z}^d$.


See also: Tiling the plane with incongruent isosceles triangles.

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    $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Jul 29 '18 at 22:04
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    $\begingroup$ @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. $\endgroup$ – Gerry Myerson Jul 29 '18 at 22:48
  • $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Jul 29 '18 at 23:17
  • $\begingroup$ Why does "displacing corners" lead to incommensurate tilings? Based on the picture, it appears that every edge is shared by two triangles, so... $\endgroup$ – Victor Protsak Jul 29 '18 at 23:19
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    $\begingroup$ @JoelDavidHamkins: Finally fixed the coloring when this was bumped to the front page. $\endgroup$ – Joseph O'Rourke Jan 12 at 19:31

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