TikZ Fill for Fraction Diagram

Question

I'm trying to shade in one third of this circle. Obviously, it's not working. What did I do wrong?

I got no error messages for either of my attempts in the code below, but neither filled anything in. Both just gave this circle.

\documentclass[12pt,letterpaper]{article}

\usepackage{tikz}
\usepackage{xcolor}

\begin{document}

How to shade in $\frac{1}{3}$ of this?

\begin{tikzpicture}
\draw (0,0) circle (3cm);
\draw (90:3)--(0,0);
\draw (210:3)--(0,0);
\draw (330:3)--(0,0);
\fill[gray] arc[start angle=-30, end angle =90, radius=3cm];
\end{tikzpicture}

% In this attempt, I put the color=gray part in the same square brackets as all the other stuff.
\begin{tikzpicture}
\draw (0,0) circle (3cm);
\draw (90:3)--(0,0);
\draw (210:3)--(0,0);
\draw (330:3)--(0,0);
\fill arc[color=gray, start angle=-30, end angle =90, radius=3cm];
\end{tikzpicture}

\end{document}

Answer 1

Considering the intended application, the following may be a useful starting point...

\documentclass[tikz, border=5]{standalone}
\newcount\segmentsleft
\tikzset{pics/.cd,
  circle fraction/.style args={#1/#2}{code={%
\segmentsleft=#1\relax
\pgfmathloop
\ifnum\segmentsleft<1\else
\ifnum\segmentsleft<#2 \edef\n{\the\segmentsleft}\else\def\n{#2}\fi
\begin{scope}[shift={(\pgfmathcounter,0)}]
\foreach \i [evaluate={\a=360/#2*(\i-1)+90;}] in {1,...,\n}
  \fill[fill=gray] (0,0) -- (\a:3/8) arc (\a:\a+360/#2:3/8) -- cycle;
\draw circle [radius=3/8];
\ifnum#2>1
  \foreach \i [evaluate={\a=360/#2*(\i-1);}] in {1,...,#2}
    \draw (0,0) -- (90+\a:3/8);
\fi
\end{scope}
\advance\segmentsleft by-#2
\repeatpgfmathloop
  }}
}

\begin{document}
\begin{tikzpicture}
\foreach \numerator/\denominator [count=\y] 
  in {1/1, 1/3, 2/4, 3/5, 8/8, 4/1, 10/3, 20/6, 30/7, 40/15}{
  \node at (-1/2,-\y) {$\frac{\numerator}{\denominator}$};
  \pic  at (0, -\y) {circle fraction={\numerator/\denominator}};
}
\end{tikzpicture}

For more general shapes, one can assume that each division is the same shape (if it isn't it's going to get tricky). So, the minimum that is required is

• code to shift to the position for the "containing" shape (e.g., circle)
• code to shift to the appropriate position for the ith shape-division
• code to determine how the ith shape-division is drawn
• code to draw each shape-division (e.g., circular sector).

Here is a reasonably general solution illustrated with a triangle style:

\documentclass[tikz, border=5]{standalone}
\newcount\tikzfractiondenominator
\newcount\tikzfractionnumerator
\def\tikzfractionempty{}
\let\tikzfractionstyle=\tikzfractionempty
\newif\iftikzfractionfill
\tikzset{pics/.cd,
  fraction/.style={%
    code={%
      \tikzset{pics/fraction/.cd, #1}%
      \pgfmathparse{int(ceil(\tikzfractionnumerator/\tikzfractiondenominator))}%
      \let\tikzfractionshapetotal=\pgfmathresult
      \ifx\tikzfractionstyle\tikzfractionempty
      \else%
        \pgfmathloop
          \ifnum\tikzfractionnumerator<1
        \else
          \pgfmathsetmacro\tikzfractionproper{int(\tikzfractionnumerator?\tikzfractionnumerator:\tikzfractiondenominator)}%
          \foreach \tikzfractionsegmentnumber in {1,...,\tikzfractiondenominator}{%
            \ifnum\tikzfractionsegmentnumber>\tikzfractionproper\relax%
              \tikzfractionfillfalse%
            \else%
              \tikzfractionfilltrue%
            \fi%
            \let\tikzfractionshapenumber=\pgfmathcounter%
            \begin{scope}
              \tikzset{pics/fraction/\tikzfractionstyle/shape position/.try}%
              \tikzset{pics/fraction/\tikzfractionstyle/segment position/.try}%
              \tikzset{pics/fraction/\tikzfractionstyle/segment draw/.try}%
            \end{scope}
          }% 
          \advance\tikzfractionnumerator by-\tikzfractiondenominator%
        \repeatpgfmathloop%
      \fi%
    }
  },
  fraction/.cd,
    style/.store in=\tikzfractionstyle,
    numerator/.code=\pgfmathsetcount\tikzfractionnumerator{#1},
    denominator/.code=\pgfmathsetcount\tikzfractiondenominator{#1},
    fraction/.style args={#1/#2}{%
      /tikz/pics/fraction/.cd,
        numerator={#1}, denominator={#2}
    }
}
\tikzset{%
  /tikz/pics/fraction/triangles/.cd,
    shape position/.code={
      \pgfmathsetmacro\y{sqrt(\tikzfractiondenominator)}
      \tikzset{
        shift=(0:{(\tikzfractionshapenumber-1)*\y}),
        shift={(0,\y/4)},
      }
    },
    segment position/.code={
      \let\i=\tikzfractionsegmentnumber
      \pgfmathsetmacro\z{int(sqrt(\i-1))}
      \pgfmathsetmacro\q{\i-(\z)^2}
      \tikzset{
        shift={({sin(60) * (\q-\z) / 2}, {-\z*0.75 -mod(\q,2)*cos(60)/2})},
        rotate={mod(\q-1,2)*180}
      }
    },
    segment draw/.code={
      \iftikzfractionfill
        \tikzset{triangle fill/.style={blue!50!cyan!50}}
      \else
        \tikzset{triangle fill/.style={gray!20}}
      \fi
      \fill [triangle fill] (90:0.45) -- (210:0.45) -- (330:0.45) -- cycle;
    }
}
\begin{document}
\begin{tikzpicture}
\foreach \numerator/\denominator [count=\y]  in {1/1, 2/4, 13/9}{
\tikzset{shift=(270:\y*2)}
\pic {fraction={style=triangles, fraction={\numerator/\denominator}}};
\node at (-1,0)  {$\frac{\numerator}{\denominator}$};
}
\end{tikzpicture}
\end{document}

Reusing the fraction pic defined above (not shown below), it is then possible to be a bit more extravagant:

\tikzset{%
  /tikz/pics/fraction/petals/.cd,
    shape position/.code={
      \tikzset{
        shift=(360/\tikzfractionshapetotal*\tikzfractionshapenumber:2)
      }
    },
    segment position/.code={
      \tikzset{
        rotate=(360/\the\tikzfractiondenominator*\tikzfractionsegmentnumber)
      }
    },
    segment draw/.code={
      \iftikzfractionfill
        \tikzset{petal/.style={bottom color=purple, top color=pink}}
      \else
        \tikzset{petal/.style={bottom color=yellow!50, top color=orange!50}}
      \fi
      \pgfmathparse{180/\tikzfractiondenominator}%
      \let\r=\pgfmathresult
      \path [petal] (0:0) [rounded corners=1ex] -- 
        (-\r:0.5) -- (0:.75) -- (\r:0.5) -- cycle;
    }
}
\begin{tikzpicture}
\pic {fraction={style=petals, fraction={53/8}}};
\node {$\frac{53}{8}$};
\end{tikzpicture}

Answer 2

There are really two problems. The first is that the arc construction must follow a coordinate specification. You cannot start an arc from nowhere.

\documentclass[tikz,border=10pt,multi,12pt]{standalone}
\begin{document}
\begin{tikzpicture}
  \draw (0,0) circle (3cm)
   (90:3)--(0,0)
   (210:3)--(0,0)
   (330:3)--(0,0);
  \fill [gray] (330:3) arc[start angle=-30, end angle =90, radius=3cm];
\end{tikzpicture}
\end{document}

This is better:

TikZ is now filling a path - albeit not the path you'd like. Basically what is happening is that it is constructing the path which is the arc.

This path (shown in red) is open - it is not closed. So there is, technically, not much to fill. What TikZ does in this case, since you've requested it be filled is to close the path in the most expedient way:

And this is what TikZ fills. So you need to tell it explicitly how to close the path:

\documentclass[tikz,border=10pt,multi,12pt]{standalone}
\begin{document}
\begin{tikzpicture}
  \draw (0,0) circle (3cm)
   (90:3)--(0,0)
   (210:3)--(0,0)
   (330:3)--(0,0);
  \fill [gray] (330:3) arc[start angle=-30, end angle =90, radius=3cm] -- (0,0) -- cycle;
\end{tikzpicture}
\end{document}

This is much better but the fill will be filled over the paths drawn earlier. Best is to fill the area before drawing the other paths:

\documentclass[tikz,border=10pt,multi,12pt]{standalone}
\begin{document}
\begin{tikzpicture}
  \fill [gray] (330:3) arc[start angle=-30, end angle =90, radius=3cm] -- (0,0) -- cycle;
  \draw (0,0) circle (3cm)
   (90:3)--(0,0)
   (210:3)--(0,0)
   (330:3)--(0,0);
\end{tikzpicture}
\end{document}

Answer 3

Since the problem is solved in the other answers with an detailed explanation in cfr's answer, the optimization of the drawing remains. The whole image can be filled and drawn in one \path command. The trick is based on using the nonzero filling rule.

\documentclass[12pt,letterpaper]{article}

\usepackage{tikz}
\usepackage{xcolor}

\begin{document}
\begin{tikzpicture}
  \draw[fill=gray, radius=3cm]
    (210:3cm) -- (0, 0) circle[]
    -- (-30:3cm) arc[start angle=-30, end angle=90] -- cycle
  ;
\end{tikzpicture}
\end{document}   

Answer 4

You need a path enclosing an area to fill, not just the arc (fill first, draw later); for example:

\documentclass[12pt,letterpaper]{article}
\usepackage{tikz}

\begin{document}

\begin{tikzpicture}
\fill[gray]
  (0,0) --
  (-30:3) 
  arc[start angle=-30, end angle =90, radius=3cm] -- 
  cycle;

\draw 
  (0,0) circle (3cm)
  (90:3)--(0,0)
  (210:3)--(0,0)
  (330:3)--(0,0);
\end{tikzpicture}

\end{document}