Find the grey and black areas¶
Question¶
The figure below shows a butterfly badge design bounded by a square with side-length of 2. The design is comprised of four touching semi-circles and two inscribed circles. Find the grey and black areas.
![\begin{scope}[x=1.5in, y=1.5in, thin]
%% grey area
\begin{scope}[fill=gray]
\clip
(2,0) arc[start angle=270, end angle=90, radius=1] -- (2,0) %% right
(0,0) arc[start angle=-90, end angle=90, radius=1] -- (0,0); %% left
\fill (0,0) arc[start angle=180, end angle=0, radius=1]; %% bottom
\end{scope}
%% black area
\pgfmathsetmacro{\holeradius}{1-1/sqrt(2)}
\begin{scope}[fill=black, even odd rule]
\clip
(2,0) arc[start angle=270, end angle=90, radius=1] -- cycle %% right
(0,0) arc[start angle=-90, end angle=90, radius=1] -- cycle %% left
(0.5,1.5) circle (\holeradius) (1.5,1.5) circle (\holeradius);
\fill (0,2) arc[start angle=180, end angle=360, radius=1]; %% top
\end{scope}
%% boundaries
\draw (0,0) rectangle (2,2);
\draw (0,2) arc[start angle=180, end angle=360, radius=1]; %% top
\draw (2,0) arc[start angle=270, end angle=90, radius=1]; %% right
\draw (0,0) arc[start angle=180, end angle=0, radius=1]; %% bottom
\draw (0,0) arc[start angle=-90, end angle=90, radius=1]; %% left
%% quadrant
\begin{scope}[xshift=3.5in, yshift=0.75in]
\node[below left] at (0,0) {$A$};
\node[above left] at (0,1) {$B$};
\node[above right] at (1,1) {$C$};
\node[below right] at (1,0) {$D$};
\pgfmathsetmacro{\eradius}{sqrt(2)-1}
\node[right] at ($ (1,0) + (xy polar cs:angle=135, radius=\eradius)$) {$E$};
\draw (0,0) rectangle (1,1); %% boundary
\draw (0,0) arc[start angle=180, end angle=90, radius=1]; %% upper circle
\draw (0,0) arc[start angle=-90, end angle=0, radius=1]; %% lower circle
\draw[-latex] (1,0) -- ++ (xy polar cs:angle=135, radius=1) node [above left] {$R$};
\end{scope}
\end{scope}](../../_images/tikz-969c5b93fbded2e44c0877dc34a008f38834b67f.png)
Solution¶
Consider a single quadrant \(ABCD\) of the larger figure above. This quadrant is, by symmetry, congruent to any other quadrant. The grey area is twice that of the area bounded by the two arcs \(ARC\) and \(AEC\). This area, being the overlap of the circle segments \(ABC\) and \(ADC\), and since \(|DR| = |AD| = |AB| = 1\), is given by:
The grey area is therefore \(2A_1\) or
Computing the black area above is easy once the diameter \(|RE|\) has been computed. By symmetry, \(|DR| = |BE| = 1\). Since \(|BD| = \sqrt{2}\) and \(B\), \(R\), \(E\) and \(D\) are co-linear, it follows that
The radius of the circles within the black area are therefore both \(1 - 1/\sqrt{2}\) and hence the black area is given by
