Construction of equilateral triangle in Poincare disc model

Question

Points A and B are given in Poincare disc model. Construct equilateral triangle ABC. Any kind of help is welcome.

Answer 1

This very much depends on the geometric primitives you have at hand. If you know how to construct a circle given its midpoint and a point on the circle, then you can simply intersect the circle around $A$ through $B$ with the circle around $B$ through $A$. A circle in the Poincaré disc is an Euclidean circle, but the hyperbolic center doesn't usually coincide with the Euclidean one.

If you had coordinates, you could compute the coordinates of the Euclidean center, but as you asked about a construction, I'll not take this computational approach.

You would be better off if you were to construct a circle through three given points, as in that case the Euclidean and the Poincaré interpretation don't differ. Let's concentrate on the circle around $A$ through $B$ for now. If you know how to perform a point reflection in hyperbolic geometry, you can use the mirror image of $B$ in $A$ as the second point. But you'll still need a third point, which has to lie on some other line. I'd suggest you draw a different hyperbolic line thorugh $A$, and then transfer the length $\lvert A,B\rvert$ to that line. You can define lengths in the disc model using cross ratios, and you can transfer cross ratios from one line to another using a sequence of perspectivities.

The following illustration shows one possible way to construct a $B''$ such that $\lvert A,B\rvert=\lvert A,B''\rvert$. The second hyperbolic line (i.e. Euclidean circle) through $A$ was chosen arbitrarily. The rest of the construction is fixed. $A_{\text{inv}}$ is the image of $A$ under inversion in the unit circle. The fact that the line $AB'$ was chosen to go through $A_{\text{inv}}$ apparently ensures that $Q'$, $Q''$ and $C_2$ are collinear, so that this all works out. I only know this fact from experimentation, I don't know the name of the theorem behind it.

There might be a shorter route, but this is the first thing that came to my mind.

Answer 2

There is indeed a shorter approach. Euclid's Proposition I holds in Hyperbolic Geometry just as well as it holds in Euclidean Geometry.

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI1.html