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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.

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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 .

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

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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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