polygons

[wildrabbitt]

Hi,

is it true that if a polygon with n sides can be constructed with ruler and compass then $\frac{\pi}{n}$ can be worked out in surd form?

Will

[CRGreathouse]

I think the result you're looking for is that a regular n-gon can be constructed with a compass and straightedge if and only if cos(2*pi/n) is constructible with field operations +, -, *, / and extraction of square roots.

[wildrabbitt]

thanks sounds like it.

Is what you're saying the same as

....if cos(2pi/n) is the solution of an algebraic equation?

[a1call]

n-gon is constructible using straightedge and compass if and only if the odd prime factors of
n
n (if any) are distinct Fermat primes.[167] Likewise, a regular
n
n-gon may be constructed using straightedge, compass, and an angle trisector if and only if the prime factors of
n
n are any number of copies of 2 or 3 together with a (possibly empty) set of distinct Pierpont primes, primes of the form
2
a
3
b
1
{\displaystyle 2^{a}3^{b}+1}.[169]
It is possible to partition any convex polygon into
n
n smaller convex polygons of equal area and equal perimeter, when
n
n is a power of a prime number, but this is not known for other values of
n
n.[170]

Please see article for unbroken version. But basically this is related to the fact that a circle can only be divided into n equal sections if and only if n can be factored into single-powers of Fermat primes and 2 (using a straight edge and compass). So a triangle, square, pentagon, hexagon and octagon are possible while a 9 sided regular polygon is not.

https://en.wikipedia.org/wiki/Prime_number

[a1call]

Here is a better source:

In 1796 (when he was 19 years old), Gauss gave a sufficient condition for a regular n-gon to be constructible, which he also conjectured (but did not prove) to be necessary, thus showing that regular n-gons were constructible for n=3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48, 51, 60, 64, ... (OEIS A003401).

http://mathworld.wolfram.com/ConstructiblePolygon.html

[CRGreathouse]

Yes, and this result in turn is due to Galois theory.

[Nick]

Originally Posted by wildrabbitt

thanks sounds like it.

Is what you're saying the same as

....if cos(2pi/n) is the solution of an algebraic equation?

No, you're only allowed to use square roots (or square roots of square roots, etc.)

Perhaps a better way to think of it is that you can construct the regular n-sided polygon with ruler and compasses if and only if \(\phi(n)\) is a power of two.

[Dr Sardonicus]

The important thing is alluded to in post #2 to this thread: numbers constructible with compass and straightedge are those that can be formed (starting with 1 and 0, or with the integers), using finitely many additions, subtractions, multiplications, divisions (not by zero!) and extraction of square roots.

Note that, for any positive integer n, cos(2*pi/n) satisfies a polynomial of degree at least 1 with rational coefficients. In fact, 2*cos(2*pi/n) satisfies a monic polynomial with integer coefficients, i.e. is a "algebraic integer." If n > 2, the degree of this polynomial is eulerphi(n)/2, half the degree of the cyclotomic polynomial for the primitive n^th roots of unity.

If z = exp(i*t), then x = z + 1/z = 2*cos(t). The polynomial for 2*cos(n*t) in terms of x may be expressed

P_n(x) = z^n + 1/z^n. By multiplying P_n(x) = z^n + 1/z^n by x = z + 1/z, we easily obtain the recurrence

P_n+1(x) = x*P_n(x) - P_n-1(x), so that

P₀(x) = 2, P₁(x) = x, P₂(x) = x^2 - 2, P₃(x) = x^3 - 3*x, etc.

Then 2*cos(2*pi/n) satisfies P_n(x) = 2*cos(2*pi), i.e. P_n(x) = 2. The minimum polynomial of 2*cos(2*pi/n) will be a proper factor of P_n(x) - 2.

Galois theory tells us that the roots of a polynomial equation are constructible by compass and straightedge (i.e. using finitely many additions, subtractions, multiplications, divisions (not by zero!) and extraction of square roots) precisely when the Galois group of the polynomial is a 2-group.

Fortuitously, the (irreducible factors of) the polynomials P_n(x) have Abelian Galois groups, which means in particular that the order of the Galois groups of the irreducible factors are equal to the degrees. So the criterion for constructibility becomes, eulerphi(n) is a power of 2. And this means that n is either a power of 2, or a power of 2 times the product of distinct Fermat primes.

Legend has it that Gauss's discovery of the constructibility of the regular 17-gon led to his decision to study mathematics rather than languages (philology). He was gifted in languages; legend also has it that he learned and mastered Russian when well into his fifties.

[a1call]

Originally Posted by Dr Sardonicus

The important thing is alluded to in post #2

Like I have said before, If he would only learn how to use Pari-GP.

Thank you for the nice clarification Dr Sardonicus.

[wildrabbitt]

Hey Guys, thanks very much for all the feedback.

I need to understand all of that. What I did understand will help me write the latex document I started on last night in order to try to inspire a yougn student I'm trying to help with maths.

Thanks. A great response.