Prove that if n is not a multiple of 3, then the angle p/n can be trisected with ruler and compasses.
Solution
The key is to use π/3. Since n and 3 are relatively prime we can find integers a and b such that 3a + nb = 1. Hence aπ/n + bπ/3 = π/(3n). So take a circle with arcs subtending π/n and π/3 at the center. Start at a point X on the circumference, mark off |a| arcs subtending π/n in one direction, then mark off |b| arcs subtending π/3 in the other direction. We end up at a point Y with XY subtending π/3n at the center.
© John Scholes
jscholes@kalva.demon.co.uk
21 Aug 2002