Lill's Angle Trisection

ϕ

π

3

z

1.879

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ϕ = π/3

P(z) =

3

z

-3z-1

P(z)	 L 4 L' 3 	z
-0.00292556	0.00292556	1.879

This Demonstration shows Lill's angle trisection. The angle

ϕ

to be trisected is marked with a green arc.

Consider the trigonometric identity for

cos(3α)

in the form

3

(2cosα)

-3(2cosα)-2cos(3α)=0

.

Substitute

z=2cosα

and

3α=ϕ

to get

P(z)=

3

z

-3z-2cosϕ=0

.

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

ϕ

(the green arc), we must find

z

such that



L

4

L'

3

=0

(i.e. the length of the thick red segment is 0, thus the points

L

4

and

L'

3

coincide). Then we find

cos(ϕ/3)=z

. From

cos(ϕ/3)

, we can construct

ϕ/3

.

In the special case

ϕ=π/9

, we get

3

z

-3z-1=0

.