```Date: May 25, 2002 11:15 PM
Author: Earle Jones
Subject: Angle Trisection

I found this on sci.math and thought this group might be interested:(Original posted by Sergei Markelov markelov@mccme.ru)***It is well-known, that some angles (for instance, Pi/3) cannot betrisected using the ruler and compass. However, some angles still can betriseced. But which angles can be, and which cannot be trisected? I havefound no answer in literature.My ideas are:1. If cos(Alpha) is transcendental, then construction is impossible.2. If Alpha/Pi is rational, Alpha=p/q*Pi, then Alpha cannot be trisectedif q=3k and can be trisected if q=3k+1 or q=3k+2 (see example with Pi/7below).3. If cos(Alpha) is algebraic, but Alpha/Pi is irrational - I have sampleswhere Alpha can be triseced, and where Alpha cannot be trisected(artan(11/2) can be trisected, arctan(1/2) cannot, see below).I have the proof of (1), have some ideas about how to prove (2), and noideas about how to determine, whether the angle can be trisected in thecase (3).Here are few examples of angles that can be triseced, but this is notobvious:1. Pi/7 can be trisected since Pi/21 = Pi/3 - 2*Pi/72. arctan(11/2) can be trisected since arctan(11/2) / 3 = arctan(1/2)3. arctan((1+3*2^(1/3))/5) can be trisected sincearctan((1+3*2^(1/3))/5) = 3*arctan(2^(1/3)-1)Both these formulas can be proved using tan(3*arctan(x)) = (3*x-x^3)/(1-3*x^2) However, arctan(1/2) cannot be trisected, because minimal polynom fortan(arctan(1/2)/3) is 2*x^3-3*x^2-6*x+1 and it is irreducible over Q.Both arctan(1/2) and arctan(11/2) are incommensurably with Pi (I have someideas, how to prove this), but first cannot be trisected, and second canbe...Any suggestions?Thank you!Sergei Markelov markelov@mccme.ru
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