Math Forum Discussions

Earle Jones

Posts: 92
Registered: 12/6/04

Angle Trisection
Posted: May 25, 2002 11:15 PM

Plain Text

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