Lill's Angle Trisection

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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 .


Contributed by: Izidor Hafner (September 2017)
Open content licensed under CC BY-NC-SA



Only in the third snapshot is the condition fulfilled.


[1] D. Kurepa, Higher Algebra, Book 2 (in Croatian), Zagreb: Skolska knjiga, 1965 pp. 1072–1074.

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