Year: 1987
Paper: 1
Question Number: 2
Course: LFM Pure
Section: Coordinate Geometry
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
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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?
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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*}