How to draw a hyperbolic triangle?

Question

Any ideas about how to draw this with TikZ? I tried several things, but nothing satisfactory. What I tried was to doing in the simplest way, that is,

\begin{tikzpicture}
    \draw (point) arc (angle contitions);
\end{tikzpicture}

What I need is something like the image:

I tried to use this code also (from Use tikz (for example) to draw pictures in hyperbolic geometry)

    \begin{document}

\newcommand{\hgline}[2]{
\pgfmathsetmacro{\thetaone}{#1}
\pgfmathsetmacro{\thetatwo}{#2}
\pgfmathsetmacro{\theta}{(\thetaone+\thetatwo)/2}
\pgfmathsetmacro{\phi}{abs(\thetaone-\thetatwo)/2}
\pgfmathsetmacro{\close}{less(abs(\phi-90),0.0001)}
\ifdim \close pt = 1pt
    \draw[blue] (\theta+180:1) -- (\theta:1);
\else
    \pgfmathsetmacro{\R}{tan(\phi)}
    \pgfmathsetmacro{\distance}{sqrt(1+\R^2)}
    \draw[blue] (\theta:\distance) circle (\R);
\fi
}

\begin{tikzpicture}
\draw (0,0) circle (1);
\clip (0,0) circle (1);
\hgline{30}{-30}
\hgline{180}{270}
\hgline{30}{120}
\hgline{0}{180}

\end{tikzpicture}

But this allow me to draw lines embedded on the disk, and when I try to remove the circle, the code is unuseful.

Thank you.

A hyperbolic triangle.
The first I tried was to use the most simple thing using \draw (point) arc (angle). However, I could not control the lines so I searched here:

https://tex.stackexchange.com/questions/16617/use-tikz-for-example-to-draw-pictures-in-hyperbolic-geometry

I tried everything they put, but their solutions were about drawing hyperbolic lines on the disk, and when I tried to remove the disk the code began to do strange things. — Rub, Feb 18 '19 at 16:58
I did it, and it was too much complicated for what I need, I need just a code for drawing a simple hyperbolic triangle... — Rub, Feb 18 '19 at 17:00
@RubénFernándezFuertes We need to know what your simple hyperbolic triangle looks like. Please make your question in some formats like this: I want to draw this Here is what I have tried . — , Feb 18 '19 at 17:02
In the simplest possible case, you only have to draw three curves joining three points. — , Feb 18 '19 at 17:03
I edit the question giving more details. I thought such a simple question would not need it. Sorry for that. — Rub, Feb 18 '19 at 17:14

score 1 · Accepted Answer · 2019-02-18T18:02:07.013

1

Something like this?

\documentclass[tikz,border=3.14mm]{standalone}
\begin{document}
\begin{tikzpicture}[font=\sffamily]
 \path (0,0) coordinate (A) (4,1) coordinate (B) (2,-2) coordinate (C);
 \draw[thick,path picture={
 \foreach \X in {A,B,C}
 {\draw[line width=0.4pt] (\X) circle (1);}}] (A) node[left]{$O$} to[bend right=12] 
 (B) node[above right]{$g_2^{-1}\cdot O$} to[bend right=15] 
 (C) node[below]{$g_1^{-1}\cdot O$} to[bend right=20] cycle;
 \node at (barycentric cs:A=1,B=1,C=1) {$<180^\circ$};
\end{tikzpicture}
\end{document}

EDIT: With explicitly constructed circular arcs.

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[font=\sffamily]
 \path (0,0) coordinate (A) (4,1) coordinate (B) (2,-2) coordinate (C);
 \draw[thick,path picture={
 \foreach \X in {A,B,C}
 {\draw[line width=0.4pt] (\X) circle (1);}}]
 let \p1=($(B)-(A)$),\p2=($(C)-(B)$),\p3=($(C)-(A)$),
 \n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)},\n3={atan2(\y3,\x3)},
 \n4={veclen(\y1,\x1)},\n5={veclen(\y2,\x2)},\n6={veclen(\y3,\x3)} in
 (A) node[left]{$O$}  arc(-90-15+\n1:-90+15+\n1:{\n4/(2*sin(15))})
 --(B) node[above right]{$g_2^{-1}\cdot O$} 
  arc(-90-15+\n2:-90+15+\n2:{\n5/(2*sin(15))})
 --(C) node[below]{$g_1^{-1}\cdot O$} 
 arc(90-15+\n3:90+15+\n3:{\n6/(2*sin(15))}) -- cycle;
 \node at (barycentric cs:A=1,B=1,C=1) {$<180^\circ$};
\end{tikzpicture}
\end{document}

edited Feb 18 '19 at 18:02

answered Feb 18 '19 at 17:17

I have a small request: How can I add the angles? – Feb 18 '19 at 17:18
1

@JouleV Added. (With path picture it is very easy.) – Feb 18 '19 at 17:22
1

It is perfect now :) Unfortunately I have reached my 40 votes limit, so I can't vote yet. – Feb 18 '19 at 17:28
@JouleV No worries! (Even if I cared, I am reputation capped today.... ;-) – Feb 18 '19 at 17:29
The thing is that the sides must be arcs of circles, and I'm not sure they are exactly like that, aren't they? (the labels are not neccesary). Many thanks. – Rub Feb 18 '19 at 17:51
@RubénFernándezFuertes I think they are. Anyway, I added a version that explicitly constructs these arcs as circular arcs. – Feb 18 '19 at 18:02
1

I did it but I don't know why it did not appear! Sorry, I did it again, and now it seems to be okay. Many thanks again. – Rub Feb 18 '19 at 22:58

How to draw a hyperbolic triangle?

1 Answers1