# Difference between revisions of "Trisection of an angle"

m (better) |
m (ce) |
||

Line 1: | Line 1: | ||

{{TEX|done}}{{MSC|51M04|01A}} | {{TEX|done}}{{MSC|51M04|01A}} | ||

− | The problem of dividing an angle | + | The problem of dividing an angle into three equal parts. |

− | The special case of trisection using only ruler-and-compass construction was one of the classical problems of Antiquity. The solution of the problem of trisecting an angle reduces to finding rational roots of a cubic equation $4x^3-3x-\cos\phi=0$, where $x=\cos(\phi/3)$, which, in general, is not solvable by quadratic radicals: that is, the roots of the general cubic do not lie in the field of [[constructible number]]s. Thus, the problem of trisecting a general angle cannot be solved by means of ruler and compass, as was proved in 1837 by P. Wantzel. However, such a construction is possible for angles $m\cdot90^\circ/2^n$, where $n,m$ are integers. | + | The special case of trisection using only ruler-and-compass construction was one of the classical problems of Antiquity. The solution of the problem of trisecting an angle $\phi$ reduces to finding rational roots of a cubic equation $4x^3-3x-\cos\phi=0$, where $x=\cos(\phi/3)$, which, in general, is not solvable by quadratic radicals: that is, the roots of the general cubic do not lie in the field of [[constructible number]]s. Thus, the problem of trisecting a general angle cannot be solved by means of ruler and compass, as was proved in 1837 by P. Wantzel. However, such a construction is possible for angles $m\cdot90^\circ/2^n$, where $n,m$ are integers. |

The problem may be solved by using other means and instruments of construction (for example, the [[Dinostratus quadratrix]] or the [[conchoid]]). | The problem may be solved by using other means and instruments of construction (for example, the [[Dinostratus quadratrix]] or the [[conchoid]]). |

## Latest revision as of 15:05, 7 December 2014

2010 Mathematics Subject Classification: *Primary:* 51M04 *Secondary:* 01A [MSN][ZBL]

The problem of dividing an angle into three equal parts.

The special case of trisection using only ruler-and-compass construction was one of the classical problems of Antiquity. The solution of the problem of trisecting an angle $\phi$ reduces to finding rational roots of a cubic equation $4x^3-3x-\cos\phi=0$, where $x=\cos(\phi/3)$, which, in general, is not solvable by quadratic radicals: that is, the roots of the general cubic do not lie in the field of constructible numbers. Thus, the problem of trisecting a general angle cannot be solved by means of ruler and compass, as was proved in 1837 by P. Wantzel. However, such a construction is possible for angles $m\cdot90^\circ/2^n$, where $n,m$ are integers.

The problem may be solved by using other means and instruments of construction (for example, the Dinostratus quadratrix or the conchoid).

#### References

[1] | Yu.I. Manin, "Ueber die Lösbarkeit von Konstruktionsaufgaben mit Zirkel und Lineal" , Enzyklopaedie der Elementarmathematik , 4. Geometrie , Deutsch. Verlag Wissenschaft. (1969) pp. 205–230 (Translated from Russian) |

#### Comments

The problem of trisection of an angle, like duplication of the cube, is one of the problems dealt with in Galois theory, cf. also [a3].

A remarkable result on trisection of the angles of a triangle is F. Morley's theorem (1899), stating that the three points of intersection of the adjacent trisectors of the angles of an arbitrary triangle form an equilateral triangle (cf. [a1]).

#### References

[a1] | H.S.M. Coxeter, "Introduction to geometry" , Wiley (1961) |

[a2] | W.W.R. Ball, H.S.M. Coxeter, "Mathematical recreations and essays" , Dover, reprint (1987) |

[a3] | I. Stewart, "Galois theory" , Chapman & Hall (1973) pp. Chapt. 5 |

**How to Cite This Entry:**

Trisection of an angle.

*Encyclopedia of Mathematics.*URL: http://www.encyclopediaofmath.org/index.php?title=Trisection_of_an_angle&oldid=35448