Nonlinear Tessellations

Background

Hyperbolic distance and triangle congruency

Something funny must happen with hyperbolic distance if a region of infinite hyperbolic extent can be contained within a bounded region of the Euclidean plane.

Roughly, the closer we are to the circle C bounding the Poincare disc, the more our Euclidean eyes will see the hyperblic ruler shrinking.

Vertical segments indicate equal hyperbolic lengths

Here is a sample calculation of hyperbolic distance.

Computing distances in hyperbolic geometry is difficult, so constructing tilings may appear particularly challenging. Two hyperbolic triangles that look different may be congruent, but this will be a bother if determining congruence requires establishing that corresponding sides have the same lengths, and that must be done by integration.

However, in some ways hyperbolic geometry is much more rigid than Euclidean geometry, because

in hyperbolic geometry, similarity implies congruence.

That is, to show hyperbolic triangles are congruent, it suffices to examine their angles. If corresponding angles are equal, the triangles are congruent.

The proof of this is not too difficult, but uses Saccheri quadrilaterals. This is a construction with which Saccheri attempted to prove the parallel postulate was a consequence of the other four. In fact, he was proving theorems in hyperbolic geometry, though he did not know this.

Using similarity implies congruence, establishing that our constructions are tilings is relatively simple.

Looking at this diagram, we see many hyperbolic triangles. Corresponding angles in all these triangles are equal and so all the triangles are congruent.

There are also congruent hyperbolic hexagons in this tiling. Can you find them?

Return to hyperbolic geometry.