Similar Triangles in Tiling a Plane When tiling the infinite plane with triangles, is it necessary for two of the triangles to be similar?
I've tried different methods to construct, but none work. My idea was to use right triangles to show that it's not necessary, but it doesn't work. 
 A: Hint Tile the plane with squares (or rectangles).
Then for each square pick a point inside. Connect it to the four vertices. Show that you can pick the points in such a way that the triangles are not similar...
A: Note that if you pick any set of isolated points in the plane that is unbounded in every direction you can join them up into a tiling of the plane with triangles. Now intuitively there are just way more places for points to be than there are possible pairs of triangles to be similar, so you should be able to nudge points around by small amounts to disrupt any similarities that may have arisen.
You may be able to formalise this argument by enumerating the (possibly infinitely many) pairs of similar triangles, and then arguing that in any such pair there is at least one vertex that belongs to one triangle and not the other, and you can move that vertex so that the similarity is disrupted and no new similarities are created – you have uncountably many places to move it to and only countably many triangles you might accidentally match with, so you'll be able to miss them all.
Then you simply need to argue that doing this for every pair in your enumeration only needs to move each vertex finitely many times, so there is an eventual "final resting place" for each vertex, and take all those resting places as your final plane-tiling. (This will be much simpler if you started with a tiling of the plane that had some reasonable-sounding restrictions on it, like each vertex only neighbouring finitely many triangles.)
I mean. The construction in the other answer is perhaps simpler. But I'm hoping this answer will illustrate the following stronger claims:


*

*not only is this possible, but it's possible using arbitrarily small modifications to any existing triangular tiling (we haven't shown exactly that, but something close)

*not only is this possible, it's possible without coming up with any clever ideas or intricate constructions, instead by just doing the thing.

