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Re: Angle Trisection
Posted:
May 28, 2002 1:38 PM


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 wellknown, 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 deciding whether 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 solving a 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 one transcendental 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 cosines if 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, where p 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 this into the equation, and you'll get a cubic in u, which will have a unit root if and only if it takes the form
Au^3 + Bu^2 + Cu + D = 0
in 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) divisors of a given algebraic integer in a given field, and to detect which of them are principal. Ways to do these things are given in good books on Algebraic Number Theory and Galois Theory (but they're not all that easy!).
John Conway



