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- From: alopez-o@neumann.uwaterloo.ca (Alex Lopez-Ortiz)
- Subject: sci.math FAQ: The Trisection of an Angle
- Summary: Part 18 of 31, New version
- Originator: alopez-o@daisy.uwaterloo.ca
- Message-ID: <Ep1yKz.B04@undergrad.math.uwaterloo.ca>
- Sender: news@undergrad.math.uwaterloo.ca (news spool owner)
- Approved: news-answers-request@MIT.Edu
- Date: Fri, 27 Feb 1998 19:38:59 GMT
- Expires: Sun, 1 Mar 1998 14:55:55 GMT
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- Xref: senator-bedfellow.mit.edu sci.math:242797 news.answers:124266 sci.answers:7875
-
- Archive-name: sci-math-faq/trisection
- Last-modified: February 20, 1998
- Version: 7.5
-
-
-
-
- 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
-