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We add a bit to Which polygons tessellate the hyperbolic plane?.

Question: Are there hyperbolic quadrilaterals with all angles different (not necessarily irrational fractions of π) that tile the hyperbolic plane? What about quads with 3 of the angles equal and 1 different? If the answer to either question is "yes" one could ask for conditions under which tiling happens.

YCor
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Nandakumar R
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Here's a simple solution: start with the equilateral hyperbolic triangle with angles $2\pi/7$, which tiles the hyperbolic plane. Cut it into three congruent quadrilaterals with angles $(2\pi/7, \pi/2+\epsilon, 2\pi/3, \pi/2-\epsilon)$, meeting at the center of the triangle.

Timothy Budd
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    I think one of the quadrilateral angles should be $2 \pi / 3$, not $\pi / 3$. – Zerox Apr 07 '23 at 20:24
  • Yes, thanks Zerox. – Timothy Budd Apr 07 '23 at 20:38
  • Thanks for this very simple answer. The same approach - equipartitioning a regular equilateral with a 3-fan - appears to work for the elliptical plane as well. It also seems that if all angles of a quad are irrational in units of pi, then it cannot be a tile. – Nandakumar R Apr 08 '23 at 05:38
  • Since not all convex hyperbolic quads can tile, it might be of some interest to characterize those that do tile - similar to finding all convex pentagons that can tile the euclidean plane. – Nandakumar R Apr 08 '23 at 06:18
  • If it can be shown (say) that "for large enough n, any n-gon that tiles the hyperbolic plane has necessarily to have all angles except at most, a constant number (or very small fraction of n) of angles equal" that could be hopefully interesting. – Nandakumar R Apr 08 '23 at 16:09