Generalize TikZ Fraction Diagrams to any n-sided Polygon

Question

I would like to extend Mark Wibrow's answer in this question to general polygons. The user would be able to type something like

\begin{tikzpicture}
\pic {fraction={style=8-gon, segment=radius, color=gray, fraction={5/8}}};
\end{tikzpicture}

to get an octagon with segments drawn as radii, or

\begin{tikzpicture}
\pic {fraction={style=5-gon, segment=apothem, color=gray, fraction={12/5}}};
\end{tikzpicture}

to get a pentagon with segments drawn as apothems.

In Mark Wibrow's answer he has described how to create styles for circles, triangles, and flower petals. I would like to be able to generate the following types of fraction diagrams for any n-gon:

(Left: segments drawn as radii. Right: segments drawn as apothems)

I have very little experience with TikZ - I created the above graphics using Microsoft Paint. Thank you in advance for your help!

Answer 1

Just use a regular polygon node and add a path picture. There is a option for nontrivial greatest common divisors and the case in which the fraction equals 1/2.

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{shapes.geometric}
\newif\ifgcd
\begin{document}
\begin{tikzpicture}[ngon fraction/.style args={#1/#2}{regular polygon,
    minimum size=\pgfkeysvalueof{/tikz/ngon size},
    regular polygon sides=#2,draw,path picture={\ifodd#2
     \pgfmathsetmacro{\mystartangle}{90-360/#2}
    \else
     \pgfmathsetmacro{\mystartangle}{0}
    \fi
    \pgfmathtruncatemacro{\itest}{ifthenelse(#2/#1==2,1,0)}
    \ifnum\itest=1
     \foreach \X in {0,2,...,#2}
        {\draw[fill=gray!20]  (\mystartangle+\X*360/#2:\pgfkeysvalueof{/tikz/ngon size})
            -- (0,0) --  (\mystartangle+\X*360/#2+360/#2:\pgfkeysvalueof{/tikz/ngon size});}
    \else
     \fill[gray!20] (0,0) -- (\mystartangle:\pgfkeysvalueof{/tikz/ngon size}) arc[start angle=\mystartangle,end
     angle={\mystartangle+#1*360/#2},radius=\pgfkeysvalueof{/tikz/ngon size}];
     \ifgcd
      \pgfmathtruncatemacro{\mygcd}{gcd(#1,#2)}
      \pgfmathtruncatemacro{\myupper}{#2/\mygcd}
      \foreach \X in {1,...,\myupper}
      {\draw (0,0) -- (\mystartangle+\mygcd*\X*360/#2:\pgfkeysvalueof{/tikz/ngon size});}
     \else
      \foreach \X in {1,...,#2}
        {\draw (0,0) -- (\mystartangle+\X*360/#2:\pgfkeysvalueof{/tikz/ngon size});}
     \fi 
    \fi 
    }},gcd/.is if=gcd,apothem/.style={shape border rotate=180/#1},
    ngon size/.initial=2cm
    ]
  \path (0,0) node[ngon fraction=1/4,]{}
     (3,0) node[ngon fraction=1/4,apothem=4]{}
     (0,-3) node[ngon fraction=4/5,rotate=108]{}
     (3,-3) node[ngon fraction=4/5,rotate=108,apothem=5]{}
     (0,-6) node[ngon fraction=4/6,rotate=150]{}
     (3,-6) node[gcd,ngon fraction=4/6,rotate=150,apothem=6]{}
     (0,-9) node[ngon fraction=4/8,shape border rotate=360/16,rotate=360/16]{}
     (3,-9) node[ngon fraction=4/8]{}
     ;
\end{tikzpicture}
\end{document}

Answer 2

Schrödinger's cat answer is perfect. I give only a way with tkz-euclide to avoid complications. It's a test to see if it's possible... There is some work to complete and get the link between fraction and polygon. The solution with apothem can be made in the same way.

\documentclass[]{article} 
\usepackage{tkz-euclide}
\parindent=0pt
\begin{document} 

 \foreach \i in {3,...,7}
{   \begin{tikzpicture}
        \tkzDefPoints{0/0/P0,2/0/P1}
      \tkzDefRegPolygon[center,sides=\i](P0,P1)
        \tkzDrawPolygon(P1,P...,P\i)
        \tkzFillPolygon[gray!20](P0,P...,P\i)
        \foreach \j in {1,...,\i} {\tkzDrawSegment[black](P0,P\j)}
    \end{tikzpicture}\\}
\end{document}

Now with this :

\documentclass[]{article} 
\usepackage{tkz-euclide}
\parindent=0pt
\begin{document} 

     \foreach \i  in {3,...,7}
    {   \begin{tikzpicture}
            \tkzDefPoints{0/0/P0,0/0/Q0,2/0/P1}
            \tkzDefMidPoint(P0,P1) \tkzGetPoint{Q1}
            \tkzDefRegPolygon[center,sides=\i](P0,P1)
            \tkzDefMidPoint(P1,P2) \tkzGetPoint{Q1}
            \tkzDefRegPolygon[center,sides=\i,name=Q](P0,Q1)
            \tkzDrawPolygon(P1,P...,P\i)
            \tkzFillPolygon[gray!20](Q0,Q1,P2,Q2)
            \foreach \j in {1,...,\i} {\tkzDrawSegment[black](P0,Q\j)}
        \end{tikzpicture}\\}
\end{document}