```Date: May 28, 2002 1:38 PM
Author: John Conway
Subject: Re: Angle Trisection

On Sun, 26 May 2002, Earle Jones wrote:> 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 be> trisected using the ruler and compass. However, some angles still can be> triseced. But which angles can be, and which cannot be trisected? I have> found no answer in literature.   Algebraically, this question is equivalent to that of decidingwhether the cubic equation    3s - 4s^3 = sin(theta)(one of whose roots is sin(theta/3) ) has a root that's a rational function of sin(theta)  (if so, the other roots can be found by solvinga quadratic, so are also constructible).   If the structure of the field generated by sin(theta) is known,this is not hard to decide.> My ideas are:> > 1. If cos(Alpha) is transcendental, then construction is impossible.   This is indeed the case, since the field generated by any onetranscendental number is isomorphic to that generated by any other.> 2. If Alpha/Pi is rational, Alpha=p/q*Pi, then Alpha cannot be trisected> if q=3k and can be trisected if q=3k+1 or q=3k+2 (see example with Pi/7> below).   These statements too, are provable.> 3. If cos(Alpha) is algebraic, but Alpha/Pi is irrational - I have > samples where 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 no> ideas about how to determine, whether the angle can be trisected in the> case (3).   Here's the way - I'll use sines rather than cosines since I started that way before reading the rest of the letter, but you could use cosinesif you like - it doesn't matter.    Write  sin(theta) = P/Q,  where  P  and  Q  are algebraic integers.Then any root that's in the field necessarily has the form  p/q, wherep  and  q  are algebraic integers in the field, for which  p  divides  P, and  q  divides  4Q.  This leaves only finitely many possibilities,up to units, so we can suppose the typical one is  pu/q.  Plug thisinto the equation, and you'll get a cubic in u, which will have a unitroot if and only if it takes the form        Au^3 + Bu^2 + Cu + D = 0in which  A,B,C,D  are algebraic integers of which  A  and  D  are units.   This requires, of course, the ability to find all the (ideal) divisorsof a given algebraic integer in a given field, and to detect which ofthem are principal.  Ways to do these things are given in good books onAlgebraic Number Theory and Galois Theory (but they're not all that easy!).   John Conway
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