Difference between revisions of "Three Greek problems of antiquity"

(Doubling of a Cube: This is wrong - x^3=2 for counter.)
(Um cos 20 is algebraic)
Line 3: Line 3:
 
All three constructions have been shown to be impossible.  
 
All three constructions have been shown to be impossible.  
 
==Trisection of the General Angle==
 
==Trisection of the General Angle==
Statement: Given an angle, construct by means of [[straight edge]] and [[compass]] only, an angle one-third its measure. This is impossible because the non-[[algebraic number]] <math>\cos 20^{\circ}</math> would have to be constructable in this case.
+
Statement: Given an angle, construct by means of [[straight edge]] and [[compass]] only, an angle one-third its measure. This is impossible because the number <math>\cos 20^{\circ}</math> would have to be constructable in this case, and constructions can only construct [[square roots]] and arithmetic operations.
  
 
==Doubling of a Cube==
 
==Doubling of a Cube==

Revision as of 14:20, 7 June 2009

The three Greek problems of antiquity were some of the most famous unsolved problems in history. They were first posed by the Greeks but were not settled till the advent of Abstract algebra and Analysis in modern times.

All three constructions have been shown to be impossible.

Trisection of the General Angle

Statement: Given an angle, construct by means of straight edge and compass only, an angle one-third its measure. This is impossible because the number $\cos 20^{\circ}$ would have to be constructable in this case, and constructions can only construct square roots and arithmetic operations.

Doubling of a Cube

Given a cube, construct by means of straight edge and compass only, a cube with double the volume. This is impossible, because the number $\sqrt[3]{2}$ would have to be constructed, and constructions can only construct square roots and arithmetic operations.

Squaring of a Circle

Given a circle, construct by means of straight edge and compass only, a square with area same as that of the circle.

This was shown to be impossible, as the transcendental number $\sqrt{\pi}$ would have to be constructable in this case.

See Also

Invalid username
Login to AoPS