Hyperbolic plane has a beautiful triangulation by congruent hyperbolic triangles where all the vertices of the triangulation have degree 7, this is of course not possible in the euclidean plane, even more, I think it is not possible to triangulate the euclidean plane with triangles that are $C$-bilipschitz with a unit triangulation for some constant $C$, because that would imply there is a bilipschitz map from $\mathbb{H^2}$ to the euclidean $\mathbb{R^2}$ and this is not possible, is that correct?


So my question is: is it possible to triangulate the euclidean plane with any kind of euclidean triangles such that at every vertex has degree 7. I've been trying to do it by hand but always a lot of little triangles start appearing and I'm not certain if I can continue triangulating in such a way that the whole plane is covered.



Euclidean triangulation of the plane with degree 7 at each vertex.
Unless I am missing something: start with the image of a degree-seven triangulation of the hyperbolic plane (coming from the (2, 3, 7) triangle group -- that is the triangulation both you and @Will seem to be alluding to) in the Beltrami-Klein model (this will be a triangulation of the disk with Euclidean straight edges). Then lift the vertices of which are distance $r$ from the origin by some monotoically increasing convex function of $r,$ such that $f(1) < \infty, f(0) = 0$ This will move your triangulation to a straight-line triangulation of a convex "cup." Now centrally project this from the point $(0, 0, f(1)).$ The image will be a combinatorially equivalent triangulation of the plane.


