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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
.
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