Circle Sieve

I used a recursive definition instead of a classic series/infinite sum.

$A_{remaining} = 7\pi{}\cdot{}\left(\frac{r}{3}\right)^2 = \frac{7}{9}\pi{}r^2$

$A_{fill} = \pi{}r^2-A_{remaining}=\frac{2}{9}\pi{}r^2$

We square the remaining area of the original because purple only every other iteration and yes i know i stopped using pi r squared just dont talk about it

$A_{purple} = A_{fill}+A_{remaining}^2\cdot{}A_{purple} = \frac{2}{9}+\left(\frac{7}{9}\right)^2{}A_{purple}$

$A_{purple} = \frac{2}{9}+\frac{49}{81}A_{purple}$

$A_{purple}-\frac{49}{81}A_{purple} = \frac{2}{9}$

$\frac{32}{81}A_{purple} = \frac{18}{81}$

and finally

$A_{purple} = \frac{18}{32} = \frac{9}{16}$

also we knwo the blue area is the area of the entire thing minus the purple area so

$A_{blue} = 1-A_{purple} = \frac{7}{16}$

You have 9/16 against 7/16 the simplified ratios term things are 7 and 9 and there sums 16. Done.