Lill's Angle Trisection
Lill's Angle Trisection
This Demonstration shows Lill's angle trisection. The angle to be trisected is marked with a green arc.
ϕ
Consider the trigonometric identity for in the form
cos(3α)
3
(2cosα)
Substitute and to get
z=2cosα
3α=ϕ
P(z)=-3z-2cosϕ=0
3
z
This equation has the solutions
2cos
ϕ
3
2cos+
ϕ
3
2π
3
2cos+
ϕ
3
4π
3
By Lill's construction,
P(z)=
L
4
L'
3
so given (the green arc), we must find such that (i.e. the length of the thick red segment is 0, thus the points and coincide). Then we find . From , we can construct .
ϕ
z
=0
L
4
L'
3
L
4
L'
3
cos(ϕ/3)=z
cos(ϕ/3)
ϕ/3
In the special case , we get -3z-1=0.
ϕ=π/9
3
z