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Posts:
204
Registered:
12/6/04


RE: Trisection
Posted:
Jul 15, 2002 2:05 PM


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. In response to John's suggestion of relettering (below) I am reposting my construction based on AOB.
>[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 ÃÂ¼ and ÃÂ½ 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 wonÃÂt) 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 EÃÂ.
Line OEÃÂ is a good trisection. However this is only the start. Repeating the process from step 4 using AEÃÂ as the arc radius results in a trisection to within 10E11 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 AC. This give a longer working line for the construction.



