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I've been working on the following problem, and although I've made some progress, I don't know how to finish. Throughout the exercise, we are working in the Poincaré disc.

Consider the points $p=(-1/3,0)$ and $q==(1/3,2/3)$, and let $T$ be the hyperbolic translation such that $T(p)=q$. Let $R$ be the hyperbolic rotation around $p$ in $\pi/2$. Determine if $TR$ is a hyperbolic translation, rotation or parallel displacement.

Progress so far: The translation $T$ takes a point and sends it to another point that lies on the Euclidean circle that passes through the point and the points $(\frac{-12\sqrt{13}-15}{61},\frac{-10\sqrt{13}+18}{61})$ and $(\frac{12\sqrt{13}-15}{61},\frac{10\sqrt{13}+18}{61})$. I did this by a direct computation, by finding the hyperbolic line that goes through $p$ and $q$ and seeing where it intersects the unit circle. Next, the rotation around $p$ in $\pi/2$ is the composition of two reflections, one with respect to the line $y=0$ and the other one with respect to the circle $(x+5/3)^2+(y-4/3)^2=32/9$. I also did this by direct computation.

Now, in order to see what $TR$ is, I believe I would want to be looking for fixed points, for example. However, the only way I can think of doing this is by explicitly writing out the formula for $TR$. This doesn't seem elegant or desirable. Can anyone help me?

$^*$I recently posted this accidentally using someone else's account that also uses this computer; I deleted that question and am asking it again.

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    $\begingroup$ I assume the coordinates/equations in your post referring to points inside the unit disc model of the hyperbolic plane? $\endgroup$ Commented Aug 1, 2021 at 16:52
  • $\begingroup$ Yes! I'll edit the question. $\endgroup$ Commented Aug 1, 2021 at 16:59
  • $\begingroup$ This can be proved in a purely "synthetic" fashion, using neutral geometry axioms/theorems (meaning axioms/theorems that are satisfied by both Euclidean and hyperbolic geometry). Would an answer like that be satisfactory to you? $\endgroup$ Commented Aug 1, 2021 at 17:01
  • $\begingroup$ @LeeMosher I would most certainly be interested in an answer like that. Any help would be appreciated. $\endgroup$ Commented Aug 1, 2021 at 17:07
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    $\begingroup$ Defining $T$ as "the hyperbolic translation such that $T(p)=q$" isn't unique. My guess would be that this is intended to be the translation that fixes the line (i.e. geodesic) connecting $p$ and $q$, but the wording doesn't say so. There is a family of translations matching the wording, parameterized by two real degrees of freedom, e.g. representing the ideal fixed points of the translation. $\endgroup$ Commented Aug 3, 2021 at 18:11

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Ok, I've been able to work it out. Here's the answer (I've omitted many of the explicit calculations since they are fairly easy and a bit tedious to write out):

The hyperbolic line that goes through $p$ and $q$ has equation $N:(x+5/3)^2+(y-2)^2=\frac{52}{9}$. The hyperbolic line that goes through $p$ and is orthogonal to this circle has equation $M:(x+5/3)^2+(y+8/9)^2=208/81$, and by an explicit calculation we find that the translation $T$ can be obtained by reflecting around the circle $L:(x+5)^2+(y+14/3)^2=25+14^2/9-1$ and then reflecting around $M$.

Now to calculate the rotation around $p$, let $K$ be the circle $(x+5/3)^2+(y+4/15)^2=25/9+16/15^2-1$. This circle passes through $p$ and intersects $M$ in an angle of $\pi/4$. Therefore, rotation around $p$ in $\pi/4$ can be calculated by reflecting first around $K$ and then $M$. Therefore, $TR$ can be obtained by reflecting first around $K$, then $M$, then $M$ and finally $L$. In particular, this is equal to reflecting around $K$ and then $L$. Since $K$ and $L$ are disjoint, we obtain that $TR$ is a translation.

I hope this makes sense.

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