Solution of tangential circles mentioned in the Extensions section.

We could not resist!

We have three concentric circles

r1(t) = 2 R1 sin(t), r2(t) = 2 R2 sin(t), and r3(t) = 2 R sin(t)

We want the regions 1, 2, and 3 to each be 1/3 Pi (2 R)^2. and since there are two variables, R1 and R2, we shall get two conditions in the two unknowns.

And now for the solution of the equations.

Hence we have the inner radii R1sol and R2sol.

We examine the ratio and see it looks familiar!!!

So we get the exact radii and ratios to confirm our suspicion of the ratio being the square root of 2. This suggests that if we take the extensions to more circles the ratios would have interesting ratios. Try it and see!!!

And we check to see the three areas are indeed 1/3 Pi (2R) R^2 = 4.18879 R^2. They are!!!

We plot a sample solution for visual effect.