Thanks for the hard work Eric.

> 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)

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 ¼ and ½ 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 won’t) stop, your first guess was an exact

trisection.

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

2.

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

mailto:Eric.Bainville@cabri.com - http://www.cabri.com

mailto:Eric.Bainville@cabri.com - http://www.cabri.com