There are several ways to trisect a circular region, but this is one of the prettiest. Trisect the diameter, and then draw semicircles on it as shown. Your job is to prove that the three wavy regions have the same area.
(You may assume that the area of I equals the are of III.)
So, A1 + C2 ≟ B1 + B2 ≟ C1 + A2?
Suppose the diameter PS = 12.
Then A1 = 
(62)/2 – B1 = 18 – B1.
B1 = (42)/2 – C1 = 8 – C1
C1 = (22)/2 = 2
B1 = 8 – 2 = 6 and A1 = 18 – 6 =12.
Then A1 + C2 = 12 + 2 = 14
and B1 + B2 = 6 + 6 = 12 ≠ 14 UH OH!
Looks as if our pretty way doesn’t work. Or is there a mistake somewhere? Help us out!