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Re: Trisection
Posted:
Jul 3, 2002 10:52 AM
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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.
-- Eric
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
>
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