# 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