```Date: Jul 3, 2002 10:52 AM
Author: Jacques
Subject: Trisection (and pentasection and septasection etc)

Very interesting Mark.  I would like to share my schooltime methodwith you.  Unlike yours, it is very easy to prove why my method iswrong but, nevertheless, it is quite accurate.Start with any angle (<45 degrees) and label the vertex A.  Step offthree 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) tocut 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 Krespectively. 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 actuallythe length that constitutes a third of arc CJ but the difference isvirtually insignificant if angle A < 45 degrees.  Therefore, if anangle is more than, say 60 degrees, bisect it and do the operation onhalf 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 andwill fit slightly less than three times into curve DG.  Arc CF is 2/3arc DG (^A*AC = 2/3 of ^A*AD because AC is 2 units and AD is threeunits, ^A in radians of course).In the same way any arc can be subdivided into 5, 7 or whatever byjust drawing the appropriate number of concentric arcs.
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