(2) Cuts as secant lines meeting at one point.

We consider the polar function r = 2 R Cos[q] and require that the polar region BAC be 1/3 Pi R^2. We integrate from angle -a to angle a (where a = q/2).

We determine the value of a to be 0.268 radians which will give 1/3 the area of the circle.

We check that the "wedge" (BAC) has area 1/3 area of the circle (1/3 Pi R^2 or

1.0472 R^2) - and it does!

Finally, we check that the top piece above the wedge has area 1/3 the area of the circle and by symmetry the bottom piece also does.