Do you mean you rescale the triangle on each iteration, so that its circum-circle/in-circle has radius 1?

Could you please clarify the question? (I suppose that the last sentence of your post is in doubt and you would like to know if it's a true statement?)

Nichts ueber dein Deutsch zu sagen, aber each triangle which maps onto (0,1,\infty) in the upper half-plane thus becomes an ideal hyperbolic triangle. Actually, a hyperbolic triangle is made like follows: you take three circles in the upper half-plane {(x,y)|y>0} such that all of them are orthogonal to {y=0}. Then their respective segments form the sides of a triangle. This is a hyperbolic triangle. However, each hyperbolic triangle is uniquely determined by its internal angles or side lengths, so you may be looking at a more general object which you name a circular triangle.

@AlexB.: I thought that the centre of $GL(3,\mathbb Z)$ was $\mathbb Z_2$.

@sasquires: all right, I see what you mean. I just picked up the above two matrices virtually at random (I mean I did several tries, but without too much consideration). Your example is much more clever in this regard and finally dots the "i".

If Ax=ax and Bx=bx, in'nit ABx= A(bx) = b Ax = ba x = ab x = a Bx = B(ax) = BAx, for every matrices A, B, and eigenvector x with eigenvalues a, b with respect to these matrices? Then, a, b, should not be necessarily 1.