Circles Ad Infinitum

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:

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.