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)
For the iteration process, it would have been better to first choose point E on the circle of center A by B and C, since the construction gives another point on this circle. Then D would be the intersection of circle centered in C by E and the initial circle.
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.
At 16:05 21/06/2002 -0400, John Conway wrote: >On 21 Jun 2002, Mark Stark wrote: > > > Thanks for the kind words John. I'm afraid this is a case of "have CAD > > system -will play". It's the result of taking the word impossible as a > > challenge. The accuracy does vary with the size of the given angle and > > the quality of the first guess. The smaller the given angle the > > better. My error calculations were based on a medium sized angle (60 > > degrees) and fair first guess (25 degrees). However, even a 90 deg. > > angle and a poor guess (45 degrees) yields a first iteration of 30.021 > > deg. and a second iteration within 10E-7. > > It's nice to hear from you. I didn't want to say so in my first >message, but it's a bit unfortunate that, although the construction >theoretically has this high accuracy, in practice it's going to be >weak because you must produce the short line ED to the decidedly >lone one EF. I expect this defect can be cured in some simple way, >and hope you'll work on this. > > Here's a question: For a given angle CAB, let D (on the >straight line CB) and E (on the arc), vary in your manner (so >that CD = CE). Then what's the envelope of the line DE ? > > The way my thoughts are running is this: if all such lines >that are tolerably near to the correct one were "understandable", >then one could use probably this to give an alternative finish to >the construction. > > The neusis construction I was thinking of is this. Make your >original circle be a unit one, and draw also the unit circle >centered at C. Then adjust your ruler, which has two marks X >and Y one unit apart, so that X lies on this latter circle, >Y on the straight line CB, and so that it also passes through A. >Then the ruler will trisect the angle CAB. > > This has the advantage over Archimedes' one (if you don't know >that, you can find it in "The Book of Numbers", which I wrote >jointly with Richard Guy), that it trisects the given angle "in >situ", as it were. It therefore combines very nicely with my >angle-trisector construction for the regular heptagon (which you >can also find in The BoN). However, I want to look at that >again, because for the version in The BoN, the angle to be >trisected is inconveniently small, and because there may also >be a nice way to economize by using something twice (once in >the angle-trisection, and once in the ensuing heptagon construction). > > John Conway >