\section*{Problem 16 - V01016}

For a circle with radius \(r\), four larger concentric circles are to be drawn successively so that each resulting annulus has the same area as the original circle.

a) Express the radii of the four additional circles \(r_{1}, r_{2}, r_{3}, r_{4}\) in terms of the original radius \(r\) in general!

b) Perform the calculation and drawing for \(r=20 \mathrm{~mm}\).
}

The radius of the smallest circle is according to b) \(r=2 \mathrm{~cm}\). The radii of the remaining concentric circles, whose corresponding circular rings are to have the same area as the initial circle, are \(r_{1}, r_{2}, r_{3}\) and \(r_{4}\). It is required that

\[
\begin{array}{lll}
F_{1}=\pi r^{2} \quad ; \quad F_{2}=\pi\left(r_{1}^{2}-r^{2}\right) & ; \quad F_{3}=\pi\left(r_{2}^{2}-r_{1}^{2}\right) \\
& F_{4}=\pi\left(r_{3}^{2}-r_{2}^{2}\right) & ; \quad F_{5}=\pi\left(r_{4}^{2}-r_{3}^{2}\right)
\end{array}
\]

This means \(\pi r^{2}=\pi\left(r_{1}^{2}-r^{2}\right)\); dividing by \(\pi\) then yields

\[
\begin{aligned}
& r^{2}=r_{1}^{2}-r^{2} \Rightarrow 2 r^{2}=r_{1}^{2} \Rightarrow r_{1}=r \sqrt{2} \\
& r_{1}^{2}-r^{2}=r_{2}^{2}-r_{1}^{2} \Rightarrow 2 r_{1}^{2}-r^{2}=r_{2}^{2} \Rightarrow r_{2}=r \sqrt{3} \\
& r_{2}^{2}-r_{1}^{2}=r_{3}^{2}-r_{2}^{2} \Rightarrow 2 r_{2}^{2}-r_{1}^{2}=r_{3}^{2} \Rightarrow r_{3}=r \sqrt{4} \\
& r_{3}^{2}-r_{2}^{2}=r_{4}^{2}-r_{3}^{2} \Rightarrow 2 r_{3}^{2}-r_{2}^{2}=r_{4}^{2} \Rightarrow r_{4}=r \sqrt{5}
\end{aligned}
\]

The circles have the following radii: \(r=2 \mathrm{~cm}, r_{1}=2 \sqrt{2} \approx 2.8 \mathrm{~cm}, r_{2}=2 \sqrt{3} \approx 3.5 \mathrm{~cm}, r_{3}=2 \sqrt{4}=4 \mathrm{~cm}\) and \(r_{4}=2 \sqrt{5} \approx 4.5 \mathrm{~cm}\).

\section*{Adapted from \([6]\)}