How to color a region?

Question

The following code

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections,positioning}

\begin{document}

\begin{tikzpicture}[radius=5]
\draw (0,0) rectangle (5,5);
\draw[name path=c1] (0,0) arc[start angle=-90,end angle=0] -- (5,5);
\draw[name path=c2] (5,0) arc[start angle=0,end angle=90] -- (0,5);
\draw[name path=c3] (5,5) arc[start angle=90,end angle=180] -- (0,5);
\draw[name path=c4] (0,5) arc[start angle=180,end angle=270] -- (5,0);
\begin{scope}
\fill [name intersections={of=c1 and c2}]
  (intersection-1) circle (2pt) node[label=right:$A$] {};
\fill [name intersections={of=c2 and c3}]
  (intersection-1) circle (2pt) node[label=above:$B$] {};
\fill [name intersections={of=c3 and c4}]
  (intersection-1) circle (2pt) node[label=left:$C$] {};
\fill [name intersections={of=c4 and c1}]
  (intersection-1) circle (2pt) node[label=below:$D$] {};
\end{scope}
\end{tikzpicture}

\end{document}

produces

How can I apply color only to the central region limited by the arcs crossing at A, B, C and D?

Answer 1

If you don't mind to have the paths twice in your picture you can use clipping to only draw the inner area. Clipping is local to the scope and accumulates. So you can take the figure as two overlapping "leaves" and clip to both, then fill the whole rectangle.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections,positioning}

\begin{document}

\begin{tikzpicture}[radius=5]
\begin{scope}
\clip (0,0) arc[start angle=-90,end angle=0]   -- (5,5)
            arc[start angle=90,end angle=180]  -- (0,0);
\clip (0,5) arc[start angle=180,end angle=270] -- (5,0)
            arc[start angle=0,end angle=90]    -- (0,5);
\fill [red] (0,0) rectangle (5,5);
\end{scope}
\draw (0,0) rectangle (5,5);
\draw[name path=c1] (0,0) arc[start angle=-90,end angle=0] -- (5,5);
\draw[name path=c2] (5,0) arc[start angle=0,end angle=90] -- (0,5);
\draw[name path=c3] (5,5) arc[start angle=90,end angle=180] -- (0,5);
\draw[name path=c4] (0,5) arc[start angle=180,end angle=270] -- (5,0);
\begin{scope}
\fill [name intersections={of=c1 and c2}]
  (intersection-1) circle (2pt) node[label=right:$A$] {};
\fill [name intersections={of=c2 and c3}]
  (intersection-1) circle (2pt) node[label=above:$B$] {};
\fill [name intersections={of=c3 and c4}]
  (intersection-1) circle (2pt) node[label=left:$C$] {};
\fill [name intersections={of=c4 and c1}]
  (intersection-1) circle (2pt) node[label=below:$D$] {};
\end{scope}
\end{tikzpicture}

\end{document}

Answer 2

I know a solution has already been accepted, but here is another solution. This one uses the even odd rule to fill regions. Another available rule is the nonzero rule. Note that I have to use a preaction to color the whole thing blue, then color the tips white.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections,positioning}

\begin{document}

\begin{tikzpicture}

\draw[name path=c1] (0,0) to[out=0,in=-90] (5,5) to [out=180,in=90]  (0,0) ;
\draw[name path=c2] (5,0) to[out=180,in=-90] (0,5) to[out=0,in=90] (5,0);

\fill[even odd rule,white,preaction={fill=blue}] 
    (0,0) to[out=0,in=-90] (5,5) to [out=180,in=90]  (0,0)
    (5,0) to[out=90,in=0] (0,5) to[out=-90,in=180] (5,0);

\path[name intersections={of=c1 and c2},fill=red]
    \foreach \i/\name/\pos in {1/$D$/below,2/$A$/right,3/$C$/left,4/$B$/above}{
        (intersection-\i) circle (2pt) node[\pos] {\name}
    };
\end{tikzpicture}
\end{document}

The output is

Answer 3

Use clips.

\documentclass{standalone}
% http://tex.stackexchange.com/q/17300/86
\usepackage{tikz}
\usetikzlibrary{intersections,positioning}

\begin{document}

\begin{tikzpicture}[radius=5]
\draw (0,0) rectangle (5,5);
\begin{scope}
\clip (0,0) arc[start angle=-90,end angle=0] arc[start angle=90,end angle=180];
\clip (5,0) arc[start angle=0,end angle=90] arc[start angle=180,end angle=270];
\fill[green] (0,0) rectangle (5,5);
\end{scope}
\draw[name path=c1] (0,0) arc[start angle=-90,end angle=0];
\draw[name path=c2] (5,0) arc[start angle=0,end angle=90];
\draw[name path=c3] (5,5) arc[start angle=90,end angle=180];
\draw[name path=c4] (0,5) arc[start angle=180,end angle=270];
\begin{scope}
\fill [name intersections={of=c1 and c2}]
  (intersection-1) circle (2pt) node[label=right:$A$] {};
\fill [name intersections={of=c2 and c3}]
  (intersection-1) circle (2pt) node[label=above:$B$] {};
\fill [name intersections={of=c3 and c4}]
  (intersection-1) circle (2pt) node[label=left:$C$] {};
\fill [name intersections={of=c4 and c1}]
  (intersection-1) circle (2pt) node[label=below:$D$] {};
\end{scope}
\end{tikzpicture}

\end{document}

(Sadly, in upgrading my machine, I appear to have lost the ability to easily convert pdfs to images so you'll have to wait for the picture!)

Note that I filled the region before drawing it. Otherwise, the fill goes over (half) the lines and doesn't look so good.

(Your syntax for the arcs was a little odd ... did you mean to put in the --s afterwards? I wasn't sure why they were there.)

Answer 4

\documentclass{scrartcl}
\usepackage{tkz-euclide} 
\usetkzobj{all}
\begin{document}
\begin{tikzpicture}
\tkzDefPoint(0,0){A} \tkzDefPoint(8,0){B}
\tkzDefPoint(8,8){C} \tkzDefPoint(0,8){D}
\tkzDrawPolygon(A,B,C,D) 
\begin{scope}
 \tkzClipCircle(B,C) \tkzClipCircle(D,C)
 \tkzClipCircle(A,B) \tkzClipCircle(C,D)  
 \tkzDrawPolygon[fill=magenta!40](A,B,C,D) 
\end{scope}
\tkzClipPolygon(A,B,C,D)
\tkzDrawCircle(B,C) \tkzDrawCircle(D,C)
\tkzDrawCircle(A,B) \tkzDrawCircle(C,D)    
\tkzInterCC(A,B)(B,A) \tkzGetFirstPoint{I}
\tkzInterCC(B,A)(C,B) \tkzGetFirstPoint{J}
\tkzInterCC(C,B)(D,C) \tkzGetFirstPoint{K}
\tkzInterCC(D,C)(A,B) \tkzGetFirstPoint{H}
\tkzDrawPoints(I,J,K,H)
\tkzLabelPoint[above](I){$I$}
\tkzLabelPoint[below](K){$K$}
\tkzLabelPoint[right](H){$H$}
\tkzLabelPoint[left](J){$J$}   
\end{tikzpicture}
\end{document}