```Date: Jul 3, 2002 10:52 AM
Author: John Conway
Subject: Re: Trisection

On Fri, 28 Jun 2002, Eric Bainville wrote:> 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 youmind editing your modification into his construction?  [It might also bea 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 myselfbut found too laborious!    Regards,  John Conway
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