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Re: Trisection
Posted:
Jul 3, 2002 10:52 AM
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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
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