Creating a Tessellated Hyperbolic Disk with Tikz

Question

I'm looking to create this using Tikz:

I've thought of trying to use conformal mappings although I highly doubt that's the correct way of doing it. Any help or guidance would be greatly appreciated!

Source of the image: https://www.math.stonybrook.edu/~bishop/lectures/FWCG10.pdf

Have you looked at https://tex.stackexchange.com/q/16617/86 ? There's a few different approaches to drawing hyperbolic diagrams there. — Andrew Stacey, Jan 26 '21 at 07:08
for a reference, you can see and modify this one https://tex.stackexchange.com/a/452022/140722 — Black Mild, Jan 26 '21 at 16:27

Juan Castaño · Accepted Answer · 2021-01-26T10:42:31.363

A small example with tikz. It should be possible to simplify more, but I do not know the guideline to place the smaller circles (centers, radii, angles, etc.).

\documentclass[border=2mm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
  \draw[very thick] (0,0) circle (4);

  \begin{scope}
    \clip (0,0) circle (4);
    \foreach\i in {0,72,...,360}
    {
      \draw (\i:5)       circle (4);
      \draw (\i+12:5)    circle (2);
      \draw (\i-12:4.5)  circle (1.5);
      \draw (\i+36:4.25) circle (1);
    }
  \end{scope}
\end{tikzpicture}
\end{document}

EDIT: Another code, following Andrew Stacey suggestion. Now the circles meet each one perpendicularly, I think.

\documentclass[border=2mm]{standalone}
\usepackage{tikz}
\def\R{4}
\newcommand{\mycircle}[2] % radius, angle
{
  \pgfmathsetmacro\d{sqrt(\R\R+#1#1)};
  \draw(#2:\d) circle (#1);
}
\begin{document}
\begin{tikzpicture}
  \draw[very thick] (0,0) circle (\R);

  \begin{scope}
    \clip (0,0) circle (\R);
    \foreach\i in {0,72,...,360}
    {
      \mycircle{5}{\i}
      \mycircle{1.55}{\i+12};
      \mycircle{1.55}{\i-12};
      \mycircle{0.8}{\i+36};
      \mycircle{0.8}{\i-36};
    }
  \end{scope}
\end{tikzpicture}
\end{document}

Just FYI, in a hyperbolic diagram then the inner circles are meant to meet the outer one perpendicularly. So the centres aren't located on the outer circle but at the intersection of certain tangents to the outer circle. — Andrew Stacey, Jan 26 '21 at 07:10
@AndrewStacey, Thank you!!! I guess I should know that... :( — Juan Castaño, Jan 26 '21 at 07:15

score 3 · Answer 2 · answered Jan 26 '21 at 06:03

3

A template in PSTricks only for emergency, fun, and comparison purposes. You just need to figure out how to specify mathematically the location of the smaller circles on the big one (or probably not on it).

\documentclass[pstricks,border=12pt]{standalone}
\begin{document}
\begin{pspicture}(-5,-5)(5,5)
    \psclip{\pscircle{5}}
        \foreach \a in {0,30,...,330}{\pscircle(5;\a){2}}
    \endpsclip
\end{pspicture}
\end{document}

answered Jan 26 '21 at 06:03

Display Name

46,933

Just FYI, in a hyperbolic diagram then the inner circles are meant to meet the outer one perpendicularly. So the centres aren't located on the outer circle but at the intersection of certain tangents to the outer circle. – Andrew Stacey Jan 26 '21 at 07:10
@AndrewStacey: Thank you for informing it. :-) – Display Name Jan 26 '21 at 10:43

Creating a Tessellated Hyperbolic Disk with Tikz

2 Answers2