Transparent clipping - make transparent center in tikz image

Question

I have tried to replicate a logo using tikz. As you can see the logo consists of seven triangles in a circle, with concave edges toward the center of the circle. To do this I have drawn the triangles and superimposed a white circle in the center. This works, but only if and when the background is white. I would love to make it transparent

I was hoping to use clip coupled with even odd rule to make this work, but that does not appear possible, so I am not sure how to do this in a simple, efficient way.

\definecolor{RFFgray}{HTML}{62615a}
\definecolor{RFFgreen}{HTML}{91b947}
\begin{tikzpicture}
    \foreach \i in {90,141.4,192.86,244.29,295.71,347.14}
    {
    \draw[fill,RFFgray] (\i:1)--++(122+\i:.5)--($(\i:1)+(\i-122:.5)$)--cycle;
    }    
    \draw[fill,RFFgreen] (38.57:1)--++(122+38.57:.5)--($(38.57:1)+(38.57-122:.5)$)--cycle;
    \draw[white,fill](0,0)circle(.87);
\end{tikzpicture}

Answer 1

You always can draw only the desired part: the 'triangles'. For example, with a \pic:

\documentclass[border=2mm,tikz]{standalone}
\usetikzlibrary{calc}
\tikzset
{%
  pics/my triangle/.style={%
    code={\path[pic actions] (0:1) -- (180/7:0.87) arc (180/7:-180/7:0.87) -- cycle;}
    }
}

\begin{document}
\definecolor{RFFgray} {HTML}{62615a}
\definecolor{RFFgreen}{HTML}{91b947}
\begin{tikzpicture}
  \foreach \i in {1,3,4,...,7}
    \pic[rotate=-360/7+360*\i/7,fill=RFFgray] {my triangle};
  \pic[rotate=360/7,fill=RFFgreen] {my triangle};
\end{tikzpicture}
\end{document}

Edit: As Anis suggests, here you can se the 'transparency'. Of course, there is nothing transparent here. Only what you can see is drawn.

\documentclass[border=2mm,tikz]{standalone}
\usetikzlibrary{calc}
\tikzset
{%
  pics/my triangle/.style={%
    code={\path[pic actions] (0:1) -- (180/7:0.87) arc (180/7:-180/7:0.87) -- cycle;}
    }
}

\begin{document}
\begin{tikzpicture}
  \node {\includegraphics[width=3.5cm]{example-image-duck}};
  \foreach \i in {1,3,4,...,7}
    \pic[rotate=-360/7+360*\i/7,fill=blue!50] {my triangle};
  \pic[rotate=360/7,fill=red!50] {my triangle};
\end{tikzpicture}
\end{document}

Answer 2

It uses the TikZ command arc. There are two constants \cR and \cA that controls the dimensions of the triangles in case you need to modify them.

You can modify also the number of triangles.

The code

\documentclass[11pt, margin=.5cm]{standalone}
\usepackage{tikz}
\usetikzlibrary{math}
\begin{document}

\definecolor{RFFgray}{HTML}{62615a}
\definecolor{RFFgreen}{HTML}{91b947}


\begin{tikzpicture}[evaluate={%
    integer \N; \N = 7;
    real \r, \cR, \d, \cD;
    \r = 1;
    \cR = .87;
    \cD = .33;
    \d = \cD*360/\N;
  }]
  \fill[violet!50] (-135: \r+.6) rectangle (45: \r+.6);
  \foreach \i [parse=true, evaluate=\i as \a using {(\i -1)*(360/\N) +90}]
  in {1, 2, ..., \N-1}{%
    \filldraw[RFFgray]
    (\a: \r) -- (\a -\d: \r*\cR) arc (\a -\d: \a +\d: \r*\cR) -- (\a: \r);
  }
  \filldraw[RFFgreen]
  (90 -360/\N: \r) -- (90 -360/\N -\d: \r*\cR)
  arc (90 -360/\N -\d: 90 -360/\N +\d: \r*\cR) -- (90 -360/\N: \r);

  \begin{scope}[xshift=2.5cm]
    \foreach \i [parse=true, evaluate=\i as \a using {(\i -1)*(360/\N) +90}]
    in {1, 2, ..., \N-1}{%
      \filldraw[RFFgray]
      (\a: \r) -- (\a -\d: \r*\cR) arc (\a -\d: \a +\d: \r*\cR) -- (\a: \r);
    }
    \filldraw[RFFgreen]
    (90 -360/\N: \r) -- (90 -360/\N -\d: \r*\cR)
    arc (90 -360/\N -\d: 90 -360/\N +\d: \r*\cR) -- (90 -360/\N: \r);
  \end{scope}
\end{tikzpicture}

\end{document}

Answer 3

You can use an inverted clip, e.g. from https://tex.stackexchange.com/a/138918/36296

\documentclass{standalone}

\usepackage{tikz}
\usetikzlibrary{calc}

%\pagecolor{red}

\tikzset{
    invclip/.style={
        clip, insert path={{(current page.north east) rectangle (current page.south west)}}},
}

\begin{document}

\definecolor{RFFgray}{HTML}{62615a}
\definecolor{RFFgreen}{HTML}{91b947}
\begin{tikzpicture}
    \begin{pgfinterruptboundingbox}
      \path[invclip] (0,0)circle(.87);
    \end{pgfinterruptboundingbox}
    \foreach \i in {90,141.4,192.86,244.29,295.71,347.14}
    {
    \draw[fill,RFFgray] (\i:1)--++(122+\i:.5)--($(\i:1)+(\i-122:.5)$)--cycle;
    }    
    \draw[fill,RFFgreen] (38.57:1)--++(122+38.57:.5)--($(38.57:1)+(38.57-122:.5)$)--cycle;
\end{tikzpicture}

\end{document}