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sci.math FAQ: The Trisection of an Angle


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Archive-name: sci-math-faq/trisection
Last-modified: February 20, 1998
Version: 7.5

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                         The Trisection of an Angle
                                       
   Theorem 4. The trisection of the angle by an unmarked ruler and
   compass alone is in general not possible.
   
   This problem, together with Doubling the Cube, Constructing the
   regular Heptagon and Squaring the Circle were posed by the Greeks in
   antiquity, and remained open until modern times.
   
   The solution to all of them is rather inelegant from a geometric
   perspective. No geometric proof has been offered [check?], however, a
   very clever solution was found using fairly basic results from
   extension fields and modern algebra.
   
   It turns out that trisecting the angle is equivalent to solving a
   cubic equation. Constructions with ruler and compass may only compute
   the solution of a limited set of such equations, even when restricted
   to integer coefficients. In particular, the equation for theta = 60
   degrees cannot be solved by ruler and compass and thus the trisection
   of the angle is not possible.
   
   It is possible to trisect an angle using a compass and a ruler marked
   in 2 places.
   
   Suppose X is a point on the unit circle such that angle XOE is the
   angle we would like to ``trisect''. Draw a line AX through a point A
   on the x-axis such that |AB| = 1 (which is the same as the radius of
   the circle), where B is the intersection-point of the line AX with the
   circle.
   
   
   Figure 7.1: Trisection of the Angle with a marked ruler
   
   Let theta be angle BAO. Then angle BOA = theta , and angle XBO = angle
   BXO = 2 theta 
   
   Since the sum of the internal angles of a triangle equals pi radians
   (180 degrees) we have angle XBO + angle BXO + angle BOX = pi ,
   implying 4 theta + angle BOX = pi . Also, we have that angle AOB +
   angle BOX + angle XOE = pi , implying theta + angle BOX + angle XOE =
   pi . Since both quantities are equal to pi we obtain
   
   4 theta + angle BOX = theta + angle BOX + angle XOE 
   
   From which
   
   3 theta = angle XOE 
   
   follows. QED.
-- 
Alex Lopez-Ortiz                                           alopez-o@unb.ca
http://daisy.uwaterloo.ca/~alopez-o                    Assistant Professor	
Faculty of Computer Science                    University of New Brunswick

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