Editing Angle trisection (section)

== Proof of impossibility ==
[[File:Lineale.jpg|thumb|[[Ruler]]s. The displayed ones are marked — an ideal [[straightedge]] is un-marked|left]]
[[File:Zirkel.jpg|thumb|Compasses|left]]
[[Pierre Wantzel]] published a proof of the impossibility of classically trisecting an arbitrary angle in 1837.<ref>{{cite journal|last=Wantzel|first=P M L|title=Recherches sur les moyens de reconnaître si un problème de Géométrie peut se résoudre avec la règle et le compas.|journal=Journal de Mathématiques Pures et Appliquées|date=1837|volume=2|series=1|pages=366–372|url=http://math-doc.ujf-grenoble.fr/JMPA/PDF/JMPA_1837_1_2_A31_0.pdf#2 |archive-url=https://ghostarchive.org/archive/20221009/http://math-doc.ujf-grenoble.fr/JMPA/PDF/JMPA_1837_1_2_A31_0.pdf#2 |archive-date=2022-10-09 |url-status=live|access-date=3 March 2014}}</ref> Wantzel's proof, restated in modern terminology, uses the concept of [[field extension]]s, a topic now typically combined with [[Galois theory]]. However, Wantzel published these results earlier than [[Évariste Galois]] (whose work, written in 1830, was published only in 1846) and did not use the concepts introduced by Galois.<ref>For the historical basis of Wantzel's proof in the earlier work of Ruffini and Abel, and its timing vis-a-vis Galois, see {{citation|title=History of Mathematics: A Supplement|first=Craig|last=Smorynski|publisher=Springer|year=2007|isbn= 9780387754802|page=130|url=https://books.google.com/books?id=_zliInaOM8UC&pg=PA130}}.</ref>

The problem of constructing an angle of a given measure {{math|''θ''}} is equivalent to constructing two segments such that the ratio of their length is {{math|cos&nbsp;''θ''}}. From a solution to one of these two problems, one may pass to a solution of the other by a compass and straightedge construction. The [[triple-angle formula]] gives an expression relating the cosines of the original angle and its trisection: {{math|cos&nbsp;''θ''}}&nbsp;=&nbsp;{{math|4 cos<sup>3</sup> {{sfrac|''θ''|3}} − 3 cos {{sfrac|''θ''|3}}}}.

It follows that, given a segment that is defined to have unit length, the problem of angle trisection is equivalent to constructing a segment whose length is the root of a [[cubic polynomial]]. This equivalence reduces the original geometric problem to a purely algebraic problem.

Every rational number is constructible. Every [[irrational number]] that is  [[constructible number|constructible]] in a single step from some given numbers is a root of a [[polynomial]] of degree 2 with coefficients in the [[field (mathematics)|field]] generated by these numbers. Therefore, any number that is constructible by a sequence of steps is a root of a [[minimal polynomial (field theory)|minimal polynomial]] whose degree is a [[power of two]]. The angle {{math|{{sfrac|π|3}}}} [[radian]]s (60 [[degree (angle)|degree]]s, written 60°) is [[equilateral triangle|constructible]].  The argument below shows that it is impossible to construct a 20° angle. This implies that a 60° angle cannot be trisected, and thus that an arbitrary angle cannot be trisected.

Denote the set of [[rational numbers]] by {{math|'''Q'''}}. If 60° could be trisected, the degree of a minimal polynomial of {{math|cos 20°}} over {{math|'''Q'''}} would be a power of two. Now let {{math|''x'' {{=}} cos 20°}}. Note that {{math|cos 60°}} = {{math|cos {{sfrac|π|3}}}} = {{math|{{sfrac|1|2}}}}. Then by the triple-angle formula, {{math|cos {{sfrac|π|3}} {{=}} 4''x''<sup>3</sup> − 3''x''}} and so {{math|4''x''<sup>3</sup> − 3''x'' {{=}} {{sfrac|1|2}}}}. Thus {{math|8''x''<sup>3</sup> − 6''x'' − 1 {{=}} 0}}. Define {{math|''p''(''t'')}} to be the polynomial {{math|''p''(''t'') {{=}} 8''t''<sup>3</sup> − 6''t'' − 1}}.

Since {{math|''x'' {{=}} cos 20°}} is a root of {{math|''p''(''t'')}}, the minimal polynomial for {{math|cos 20°}} is a factor of {{math|''p''(''t'')}}.  Because {{math|''p''(''t'')}} has degree 3, if it is reducible over by {{math|'''Q'''}} then it has a [[rational root]].  By the [[rational root theorem]], this root must be {{math|±1, ±{{sfrac|1|2}}, ±{{sfrac|1|4}}}} or {{math|±{{sfrac|1|8}}}}, but none of these is a root.  Therefore, {{math|''p''(''t'')}} is [[irreducible polynomial|irreducible]] over by {{math|'''Q'''}}, and the minimal polynomial for {{math|cos 20°}} is of degree&nbsp;{{math|3}}.

So an angle of measure {{math|60°}} cannot be trisected.