7
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @JoelDavidHamkins: Finally fixed the coloring when this was bumped to the front page. $\endgroup$ – Joseph O'Rourke Jan 12 at 19:31

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.