1

Is it possible to use tikz to make the following figure:

enter image description here

\documentclass{article}
\usepackage{tikz}
\begin{document}

% Radius of regular polygons
\newdimen\R
\R=5cm

\begin{tikzpicture}
    \draw [black!20] circle (\R) ;
    \draw[black!20] circle (200pt)

    \draw[xshift=0\R] (0:\R) \foreach \x in {72,144,...,359} {
            -- (\x:\R)
        } -- cycle (90:\R) node[above] {} ;
    
\end{tikzpicture}

\end{document}
4
  • Yes, it is. But please show us your approach so far as a compilable minimal example. – Oleg Lobachev Sep 29 '20 at 22:32
  • Or ask your question to the Doctor Who. ;-) :-) – projetmbc Sep 29 '20 at 22:42
  • @projetmbc That is a solution but we are not in a good term right now! :P – Dalek Sep 29 '20 at 22:44
  • @OlegLobachev is this enough? – Dalek Sep 29 '20 at 22:56
1

Only for fun and as a starting point. Only some text on the picture was shown.

\documentclass[margin=3mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric, intersections, decorations.text}
\begin{document}

% Radius of regular polygons
\def\R{5cm}
%\R=5cm

\begin{tikzpicture}
    \path [draw,name path=c1,blue!20,fill=gray!20] circle (\R) ;
    \path [draw,name path=c2,blue!20,fill=gray!30] circle (\R-1cm);
    \path [draw,name path=c3,blue!20,fill=gray!40] circle (\R-2cm);
    
    \path [name path=l1](0,0)--++(-10:5);
    \path [name path=l2](0,0)--++(-30:5);
    \path [name intersections={of= c3 and l1, by=A}];
    \path [name intersections={of= c2 and l1, by=B}];
    \draw [blue] (A)--(B);
    \path [name intersections={of= c3 and l2, by=C}];
    \path [name intersections={of= c2 and l2, by=D}];
    \draw [blue](C)--(D);
    
    \path [name path=l3](0,0)--++(190:5);
    \path [name path=l4](0,0)--++(210:5);
    \path [name intersections={of= c3 and l3, by=E}];
    \path [name intersections={of= c2 and l3, by=F}];
    \draw [blue] (E)--(F);
    \path [name intersections={of= c3 and l4, by=G}];
    \path [name intersections={of= c2 and l4, by=H}];
    \draw [blue](G)--(H);
    
    \path [name path=l5](0.5,-5)--(0.5,5);
    \path [name path=l6](-0.5,-5)--(-0.5,5);
     \path [name intersections={of= c3 and l5, by={I,J}}];
    \path [name intersections={of= c2 and l5, by={K,L}}];
    \draw [blue] (J)--(L);
    \path [name intersections={of= c3 and l6, by={II,JJ}}];
    \path [name intersections={of= c2 and l6, by={KK,LL}}];
    \draw [blue](JJ)--(LL);
    
    \path [name path=l7](0,0)--++(40:5);
    \path [name path=l8](0,0)--++(60:5);
    \path [name intersections={of= c3 and l7, by=M}];
    \path [name intersections={of= c1 and l7, by=N}];
    \draw [blue] (M)--(N);
    \path [name intersections={of= c3 and l8, by=O}];
    \path [name intersections={of= c1 and l8, by=P}];
    \draw [blue] (O)--(P);  
    
    \path [name path=l9](0,0)--++(130:5);
    \path [name path=l10](0,0)--++(150:5);
    \path [name intersections={of= c3 and l9, by=R}];
    \path [name intersections={of= c1 and l9, by=S}];
    \draw [blue] (R)--(S);
    \path [name intersections={of= c3 and l10, by=T}];
    \path [name intersections={of= c1 and l10, by=U}];
    \draw [blue] (T)--(U);  
    \node[draw,minimum size=4cm,inner sep=0pt,regular polygon,regular polygon sides=5,rotate=180] (a) {};
    \node at ([yshift=2.5mm]a.center){Start};
     \node at ([yshift=-2.5mm]a.center){1};
    \path [postaction={decorate,decoration={raise=0ex,text along path, reverse path,text align=center, text={TR}}}] (40:4.5cm) arc (40:60:4.5cm);
     \path [postaction={decorate,decoration={raise=-1ex,text along path, reverse path,text align=center, text={6}}}] (40:4.3cm) arc (40:60:4.3cm);
      \path [postaction={decorate,decoration={raise=-1ex,text along path, reverse path,text align=center, text={5}}}] (40:3.6cm) arc (40:60:3.6cm);
      \node at (-15:3.5){R};
      \node at (-25:3.55){3};
      \node at (-90:3.5){Cue};
      \node at (-90:3.85){2};
 
\end{tikzpicture}

\end{document}

enter image description here

1
  • Thanks a lot for the answer. I was wondering whether it is possible to remove small part of the lines in each corner of pentagon which are in front of each entrance for cue, left, etc or not? – Dalek Oct 1 '20 at 0:10

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