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