Elizabeth sent us her solution:

  1. The area of the square is (2r)2=4r2. The area of the circle is pr2, so the area of the red bit is 4r2-pr2=r2(4-p).
  2. This time the radius is rÖ2 (using Pythagoras' Theorem), so replacing r with rÖ2 in the above, we get the area of the blue bit to be 2r2(4-p).
  3. Now the radius is 2r, so the area of the orange part is 4r2(4-p).
The area is doubling each time, because each time the radius increases by a factor of Ö2 (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+ 1
2
 + 1
4
 + 1
8
 +¼

. But using the formula for the sum to infinity of a geometric progression with a=1 and
r= 1
2

, this is
1
1- 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+ 1
2
 + 1
4
 + 1
8
 +¼

. Now what is half of S?
S
2
 = 1
2
 + 1
4
 + 1
8
 +¼

. But this is just S-1. So
S
2
 =S-1

, so S=2S-2, so S=2.