> I have algebraically checked this construction. It will effectively converge > to the trisection, with a cubic convergence rate. More precisely, if we take A > as the origin and C as the unit point on the X axis, 3*t the angle to trisect, > if h is the difference in abscissa between the first guess and the result > (i.e. the > difference in cosine), and h' the same difference after one iteration, we have: > > h' = h^3 / (48*sin(t)^2) + O(h^4)
That's a really quite remarkable convergence rate, on which Mark is to be congratulated!
> About your question on the imprecision of the practical construction, one may > take the other intersection D' of the circle centered at C and the line BC. > Line ED' > intersects the circle of radius 3 and center A in two points G1 and G2, > providing > two other points E1' and E2' on the unit circle. With the original point > E', these two > points form a nearly equilateral triangle corresponding to the three > solutions of the trisection problem.
I'm sorry that I can't follow this, because I didn't keep Mark's original posting, and can't remember the lettering he used. Would you mind editing your modification into his construction? [It might also be a good idea to change the lettering, since, to the extent that there's a standard, it's usually an angle AOB that gets trisected.]
Thanks very much for doing this calculation, which I started myself but found too laborious!