How to divide a circle into three equal circles using PGF/TikZ?

Question

I want to draw the following figure:

The accepted answer of SebGlav from this post gives some code.

The code (due to SebGlav):

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{calc}
\begin{document}
\newcommand{\RadiusA}{8} %Define the radius of big-boundary circle
     \pgfmathsetmacro{\RadiusB}{\RadiusA/(1+2/sqrt(3))}
     \begin{tikzpicture}
         \draw[] (0,0) circle[radius=\RadiusA cm];
         \foreach \ang [count = \i from 1] in {90,210,330}
             \draw[] (\ang:\RadiusA-\RadiusB) coordinate (center-\i) circle[radius=\RadiusB cm];
         \pgfmathsetmacro{\RadiusC}{\RadiusB/(1+2/sqrt(3))}
         \foreach \i in {1,2,3}
             \foreach \ang in {90,210,330}
                 \draw[] ($(center-\i)+(\ang:\RadiusB-\RadiusC)$) circle[radius=\RadiusC cm];
     % Divide each small circle into 3 equal circles
     \foreach \i in {1,2,3}
         \foreach \ang in {90,210,330}
             \foreach \j in {0,1,2}
                 \draw[] ($(center-\i)+(\ang:\RadiusB-\RadiusC)+(120*\j:\RadiusB/3)$) circle[radius=\RadiusB/6];
                 %this is the main governng equation for small circles
 \end{tikzpicture}

\end{document}

plotted the following figure:

But the above figure can further be improved by improving the last governing equation

\draw[] ($(center-\i)+(\ang:\RadiusB-\RadiusC)+(120*\j:\RadiusB/5)$) circle[radius=\RadiusB/7];

as follows:

The later figure is much better, but still not what I require. We need to make sure the three small circles touch each other, as in the image in my question above.

I think we have to adjust the main governing equation at the end of the code. I appreciate suggestions.

Answer 1

An approach to this could be:

• calculate the ratio between the larger radius and the smaller radius of the circles, this happens to be 1/(1+(2/sqrt(3))).
• define a function in order to simplify things in the code.
• repeat and stick the function inside itself for smaller circles.

It should be possible to simplify this even further by self-referential (recursive) code, but I leave it as it is to make the approach more visible:

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[
        declare function={
            sr(\r) = \r/(1+2/sqrt(3));
        }
    ]
\draw (0:0) circle[radius=5];

\foreach \a in {90,210,330} {
    \draw (\a:{5-sr(5)}) circle[radius={sr(5)}];

    \begin{scope}[shift={(\a:{5-sr(5)})}]
    \foreach \a in {90,210,330} {
        \draw (\a:{sr(5)-sr(sr(5))}) circle[radius={sr(sr(5))}];

        \begin{scope}[shift={(\a:{sr(5)-sr(sr(5))})}]
        \foreach \a in {90,210,330} {
            \draw (\a:{sr(sr(5))-sr(sr(sr(5)))}) circle[radius={sr(sr(sr(5)))}];
        }
        \end{scope}

    }
    \end{scope}

}


\end{tikzpicture}
\end{document}

---

As stated above, it is possible to create a recursive function here (thanks to Qrrbrbirlbel for their input):

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[
        declare function={
            sr(\r,\i) = \r/((1+2/sqrt(3))^\i);
        }
    ]
\draw (0:0) circle[radius=5];

\NewDocumentCommand{\DrawThreeCircles}{ m O{0:0} O{0} m }{
    \begin{scope}[shift={(#2)}]
        \foreach \a in {90,210,330} {
            \draw (\a:{sr(#1,#3)-sr(#1,{#3+1})}) circle[radius={sr(#1,{#3+1})}];
            \ifnum#3<#4
                \pgfmathtruncatemacro{\x}{#3}
                \pgfmathtruncatemacro{\y}{#3+1}
                \DrawThreeCircles{#1}[\a:{sr(#1,\x)-sr(#1,\y)}][\y]{#4}
            \fi
        }
    \end{scope}
}

% first argument: radius of largest circle
% second argument: number of iterations
\DrawThreeCircles{5}{3}


\end{tikzpicture}
\end{document}