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 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!

Regards, John Conway