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Lill's Angle Trisection

This Demonstration shows Lill's angle trisection. The angle to be trisected is marked with a green arc.
Consider the trigonometric identity for in the form
.
Substitute and to get
.
This equation has the solutions
,
,
.
By Lill's construction,
,
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 .
In the special case , we get .

SNAPSHOTS

  • [Snapshot]
  • [Snapshot]
  • [Snapshot]

DETAILS

Only in the third snapshot is the condition fulfilled.
Reference
[1] D. Kurepa, Higher Algebra, Book 2 (in Croatian), Zagreb: Skolska knjiga, 1965 pp. 1072–1074.
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