At 08:59 16/07/2002 -0400, Mark Stark wrote:
Thanks for the hard work Eric.

> I have algebraically checked this construction. It will effectively
> 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
> (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)

I was unable to follow your suggestion for finding a longer
construction line so I found my own (see below).
In response to John's suggestion of relettering, I am reposting my
construction based on AOB.

Actually, I was wrong. Algebraically, if you do the same constructions
choosing all possible intersection points (a circle / line intersection will
give generally 2 points ; in your construction, you specifiy which one
should be chosen), you will obtain the other solutions to the trisection
problem, with angles T/3, T/3 + 120deg, and T/3 - 120deg. But your
method gives the exceptional cubic order only for this choice of intersection
points.  Actually, my Maple script was not able to give formally the order of
convergence for the other points, but numerically it is so.

>[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.]

Near Exact Trisection:
1. Start with an unknown angle <90 deg., label the vertex O.
2. Draw an arc with origin at O crossing both lines of the angle at
points A and B.
3. Draw line AB making an isosceles triangle.
4. Using point A as the origin, draw an arc crossing line AB and the
earlier arc somewhere between and way between points A and B.
Label where this new arc crosses line AB point D.
Label where this new arc crosses the first arc point E.
5. Draw line DE and extend it well past O .  If  line DE passes
exactly through point O (it wont) stop, your first guess was an exact
6. Extend line OA well past point O, step off  3 times length OA from
point O and label the new point F.
7. Swing an arc of length OF with O as the origin that crosses the
extended line DE near point F.
Label the intersection G.
8. Draw line GO and extend it to intersect the original arc from step
Label the intersection E.

Line OE is a good trisection. However this is only the start.
Repeating the process from step 4 using AE as the arc radius results
in a trisection to within 10E-11 degrees. Each subsequent iteration
improves the trisection by several orders of magnitude.

It takes an extra step, but an alternate construction using a line
 from point B perpendicular to line OA (line AB is no longer necessary)
yields exactly the same results if your step 4 circle has its origin
at point B, is between 1/2 and 3/4 angle AOB, and point D is on the
new perpendicular line instead of line AB. This gives a longer working
line for the construction.

This is great ! I checked algebraically that you obtain the same line
with the two versions of point D, so this modified construction will also
converge to the trisection with cubic rate.

-- Eric

Eric Bainville - Product Development Manager - CabriLog -