Jacques
Posts:
35
Registered:
12/4/04


Trisection (and pentasection and septasection etc)
Posted:
Jul 3, 2002 10:52 AM


Very interesting Mark. I would like to share my schooltime method with you. Unlike yours, it is very easy to prove why my method is wrong but, nevertheless, it is quite accurate. Start with any angle (<45 degrees) and label the vertex A. Step off three equal distances on one leg and label them B, C and D. With A as centre draw three concentric arcs (radii AB, AC and AD) to cut the to cut the other leg in E, F and G resp. Bisect angle A to cut the three arcs BE, CF and DG in H, J and K respectively. With compass step off distance CJ on arc DG; it trisects the arc & thereby trisects angle A. It is easy to see that arc CJ is actually the length that constitutes a third of arc CJ but the difference is virtually insignificant if angle A < 45 degrees. Therefore, if an angle is more than, say 60 degrees, bisect it and do the operation on half the angle and double afterwards.
Arc BE is also a third of arc DG but because of the greater curvature ("fatter" curve) distance BE is significantly shorter than arc BE and will fit slightly less than three times into curve DG. Arc CF is 2/3 arc DG (^A*AC = 2/3 of ^A*AD because AC is 2 units and AD is three units, ^A in radians of course).
In the same way any arc can be subdivided into 5, 7 or whatever by just drawing the appropriate number of concentric arcs.

