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Solution to the problem of the (previous) week

Problem:

#./interaufg60_en.tex#As you can see from the figure below, $ n$ circles with radius $ r$ are inscribed in a circle with radius $ 1$.

\includegraphics{A755_vier_kreise}     \includegraphics{A755_drei_kreise}

Determine the radius $ r$ and the area $ A$ of the shaded region between the circles for $ n = 4$ and $ n = 3$.

Hint: In an equilateral triangle the medians divide each other in the ratio of $ 2 : 1$.


Answer:

$ n = 4$:     $ r=$ $ ,$    $ A=$
$ n = 3$:     $ r=$ $ ,$    $ A=$

(The results should be correct to four decimal places.)

Solution:

Case $ n = 4$:

The centers of the circles form a square with length $ l$ of the sides and length $ d$ of the diagonals. We have

$\displaystyle \frac{l}{d} = \frac{1}{\sqrt{2}} = \frac{2r}{2(1-r)}\quad \Rightarrow\quad r = \frac{1}{1+\sqrt{2}} \,. $

The area is

$\displaystyle A = (2r)^2 - 4\frac{\pi r^2}{4} = \frac{4-\pi}{(1+\sqrt{2})^2} \,. $


Case $ n = 3$:

The centers of the circles form an equilateral triangle with side length $ l$ and height $ h$.

$\displaystyle \dfrac{2}{3} h = 1 - r \quad \Rightarrow\quad h = \dfrac{3}{2}\,(1-r) $

$\displaystyle \frac{\strut l}{h} = \frac{1}{\sqrt{3}/2} = \frac{2r}{(3/2) (1-r)}\quad \Rightarrow\quad r = \frac{3}{3+2\sqrt{3}} \,. $

The area is

$\displaystyle A = \frac{\sqrt{3}}{4} (2r)^2 - 3\frac{\pi r^2}{6} = \left(\sqrt{3} - \frac{\pi}{2}\right) \frac{9}{(3+2\sqrt{3})^2} \,. $

[problem of the week]