Elizabeth sent us her solution:

The area of the square is $(2 r)^{2} = 4 r^{2}$ . The area of the circle is $π r^{2}$ , so the area of the red bit is $4 r^{2} - π r^{2} = r^{2} (4 - π)$ .
This time the radius is $r \sqrt{2}$ (using Pythagoras' Theorem), so replacing $r$ with $r \sqrt{2}$ in the above, we get the area of the blue bit to be $2 r^{2} (4 - π)$ .
Now the radius is $2 r$ , so the area of the orange part is $4 r^{2} (4 - π)$ .

The area is doubling each time, because each time the radius increases by a factor of

\sqrt{2}

(using Pythagoras), so the area increases by a factor of 2 each time.

Starting from the outside, the areas get smaller by a factor of 1 each time. So we need to work out $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ . But using the formula for the sum to infinity of a geometric progression with $a = 1$ and $r = \frac{1}{2}$ , this is $\frac{1}{1 - \frac{1}{2}} = 2$ . So the area is 2.

If you haven't come across the idea of a geometric progression, don't panic: here's a way to work out the sum.

Let's call the sum

S

, so

S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots

. Now what is half of

S

\frac{S}{2} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots

. But this is just

S - 1

. So

\frac{S}{2} = S - 1

, so

S = 2 S - 2

, so

S = 2