Colour-blind readers may find it harder than others to solve this problem posed by Alex Bellos in his entertaining fortnightly puzzle column in the Guardian newspaper.
Prove that area of the blue intersection of the two semi-circles is the same as the red area of the quadrant outside them.
If one sets the side-length of the larger section to two, that makes it's area:
ReplyDelete(2^2(Pi))/4=(Pi)
The smaller semicircles can be combined to make a circle of half the larger section's side length, which makes it's area:
1^1(Pi)=(Pi)
We can therefore see that the areas of both the section and none-overlapping semicircles are equivalent. We can therefore conclude that however much space is 'given back' to the section must be taken away from the circles where they overlap.