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 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.

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 trisected

if q=3k and can be trisected if q=3k+1 or q=3k+2 (see example with Pi/7

below).

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 are few examples of angles that can be triseced, but this is not

obvious:

1. Pi/7 can be trisected since Pi/21 = Pi/3 - 2*Pi/7

2. 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 since

arctan((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 for

tan(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

some

ideas, how to prove this), but first cannot be trisected, and second can

be...

Any suggestions?

Thank you!

Sergei Markelov markelov@mccme.ru