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## 'Circles Ad Infinitum' printed from http://nrich.maths.org/

Thank you Tom
from Wolgarston High School, Hidayah who does not name a school,
Hamish and Jared from **Ulverston Victoria High School, Eduardo from The
British School, Manila, Justin** from Skyview High School, Billings, MT
USA and **Fabian** from Sabino High School in Tucson, Arizona,
U.S.A. for sending your very good solutions to this problem.You all
tackled the problem in a similar way.

Let $C$ be the centre of the central circle. Then $PS$ is a
tangent to this circle. Then triangle $CSP$ is a 30-60-90 triangle
and, given that $CS = 1 $ cm, it follows that $PS = \sqrt{3}$ cm
and $PC = 2$ cm. Hence the height of the largest equilateral
triangle is $3$ cm. ($PC$ + radius of $ 1$ cm.), the height of the
next size equilateral triangle is $1$ cm. ($PC$ - radius of $1$
cm.) and the radius of the next size circle is $1/3$ cm.

Each circle is scaled down by a linear scale factor of $1/3$
and by an area scale factor of $1/9$.

There is one central circle and at each stage after that 3 new
circles are added.

The 'total circumference' is the sum of the circumferences of
all the circles, given by:

Total circumference = $ 2\pi + 3 ({ 2\pi\over 3} + {2\pi\over
9} + {2\pi \over 27}+ \ldots ) $

where the dots denote that the sum goes on for ever. Summing
this infinite geometric series gives: $$ \mbox{Total circumference}
= 2\pi + 2\pi (1 + {1\over 3} + {1\over 9} + {1\over 27} + \ldots )
= 2\pi + \frac{2\pi}{1 - 1/3} = 5\pi $$ The 'total area' is the sum
of the areas of all the circles, given by:

Total area = $ \pi + 3 ({\pi\over 9} + {\pi\over 9^2} + {\pi\over
9^3} + \ldots )$

where the dots denote that the sum goes on for ever. Summing
this infinite geometric series gives: $$ \mbox{Total area} = \pi +
{\pi\over 3} (1 + {1\over 9} + {1\over 9^2} + \ldots ) = \pi +
\frac{\pi}{3(1 - 1/9)} = \frac{11\pi}{8} $$ Clearly, the total of
the circumferences grows faster than the total of the areas because
the scale factor 1/3 is bigger than the scale factor 1/9.