A white square has a red circle inscribed in it, which has a white square inscribed in it, and so on. Find the red area A white  regular polygon with area $1$ has a red circle inscribed(the circle touches the edges of the square), with a white regular polygon with the same amount of sides as the first one(the polygon's corners touch the circle) and this pattern continues forever, creating a pattern like this:
(https://i.stack.imgur.com/IARUV.jpg)
1. What is the area that is coloured red if the polygon is a square?


*If you knew the amount of sides, what would be a formula to find the area coloured red, in terms of $s$ (for sides)

 A: As the part inside the second polygon is a scaled copy of the entire construction, you only need to consider the area between the first and second copy of the polygon.
If it's a square that's easy.
If it's not a square you need to consider whether the question is even well-defined, i.e. whether the ratio of the white/red areas is the same. (And if it is, you should be able to realise quite quickly what the answer must be)
A: Consider the larger red area. It is the area of the circle having radius $r=1$ , $C=\pi$ less the square having area $A=\sqrt{2}^2=2$
So its area is $R_1=\pi-2$
Now consider the similitude ratio between the inscribed and the circumscribed squares. The inner square has side $l_1=\sqrt{2}$ and the outer is $L_1=2$
The ratio is $r_1=\frac{\sqrt 2}{2}$ then the ratio for areas is $r_1^2=\frac{1}{2}$
As the red part become smaller, the ratio between them is constant and is $\frac{1}{2}$
Therefore the sum of the red areas is
$$R=R_1+\frac{1}{2}R_1+\frac{1}{2}\frac{1}{2}R_1+\ldots=R_1\sum_{n=0}^{\infty}\frac{1}{2^n}=2R_1$$
so the red area is $R=2(\pi-2)$
Hope this helps
