The region \(A\) between concentric circles of radii \(R+r\), \(R-r\) contains \(n\) circles of radius \(r\). Each circle of radius \(r\) touches both of the larger circles as well as its two neighbours of radius \(r\), as shown in the figure. Find the relationship which must hold between \(n,R\) and \(r\).
Show that \(Y\), the total area of \(A\) outside the circle of radius \(r\) and adjacent to the circle of radius \(R-r\), is given by
\[
Y=nr\sqrt{R^{2}-r^{2}}-\pi(R-r)^{2}-n\pi r^{2}\left(\frac{1}{2}-\frac{1}{n}\right).
\]
Find similar expressions for \(X\), the total area of \(A\) outside the circles of radius \(r\) and adjacent to the circle of radius \(R+r\), and for \(Z\), the total area inside the circle of radius \(r\).
What value does \((X+Y)/Z\) approach when \(n\) becomes large?
Solution:
\par
The shown isoceles triangle has base \(2r\), and the two side lengths are \(R\). The angle at the center of the circle is \(\frac{2\pi}{n}\). The height of the triangle (by Pythagoras) is \(\sqrt{R^2-r^2}\) and so the area enclosed in the triangle is \(\frac12 2r \sqrt{R^2-r^2}\). The area in the three sectors are: \(\frac{\pi}{n}(R-r)^2\), two sets of \(\frac12(\frac12 \l \pi - \frac{2\pi}{n}\r)r^2 = \l\frac{1}{2} - \frac{1}{n} \r \frac12 \pi r^2\).
Therefore the remaining area \(Y/n\) is \(r \sqrt{R^2-r^2} -\frac{\pi}{n}(R-r)^2 - \l\frac{1}{2} - \frac{1}{n} \r \pi r^2\). Multiplying this by \(n\) we get the desired result.
For \(X\) we can look at the larger sector, which we obtain using the extension of this triangle. This has area \(\frac12 \frac{2 \pi}{n} (R+r)^2\). The areas to be removed are the area of the triangle: \(r \sqrt{R^2-r^2}\) and the areas of the two sectors, which have radii \(r\) and angles \(\pi - \l \frac12 -\frac1{n} \r\pi = \l \frac{1}{2} + \frac{1}{n} \r \pi\). Therefore the area for \(X/n\) is \(\frac{\pi}{n} (R+r)^2 - r \sqrt{R^2-r^2} - \l \frac{1}{2} + \frac{1}{n} \r \pi r^2\) and so \(X\) has area:
\[ X = \pi (R+r)^2 - nr \sqrt{R^2-r^2} - n\pi r^2\l \frac{1}{2} +\frac{1}{n} \r \]
\(Z = n\pi r^2\)
\begin{align*}
\frac{(X+Y)}{Z} &= \frac{\pi (R+r)^2 - nr \sqrt{R^2-r^2} - n\pi r^2\l \frac{1}{2} +\frac{1}{n} \r}{n \pi r^2 }\\
&= \qquad \frac{nr\sqrt{R^{2}-r^{2}}-\pi(R-r)^{2}-n\pi r^{2}\left(\frac{1}{2}-\frac{1}{n}\right)}{n \pi r^2} \\
&= \frac{4\pi R r - n \pi r^2}{n\pi r^2} \\
&= \frac{4R-nr}{n r} \\
&= \frac{4R}{nr} - 1
\end{align*}
Since \(\frac{r}{R} = \sin \frac{\pi}{n}\) we have:
\begin{align*}
\frac{(X+Y)}{Z} &= \frac{4R}{nr} - 1 \\
&= \frac{4}{n \sin \frac{\pi}n} - 1 \\
& \to \frac{4}{\pi} - 1
\end{align*}
