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The coordinates of an arc of a circle of length $\frac{2pi}{p}$ are an algebraic number, and when $p$ is a Fermat prime you can find it in terms of square roots.

Gauss said that the method applied to a lot more curves than the circle. Will you please tell if you know any worked examples of this (finding the algebraic points on other curves)?

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Am I correct, or are you saying that there exists a a rational number $p$ such that $\frac{2\pi}{p}$ is algebraic over the rationals?? –  Asaf Karagila Feb 15 '11 at 18:36
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I think he might mean the coordinates of the end points of the arc. It's easy to see from De Moivre's formula. Maybe a more appropriate word would be to use the word constructible? –  Raskolnikov Feb 15 '11 at 19:09
    
A circle is defined by the algebraic curve $x^2+y^2=1$ and the line of slope 1/p is given by the curve $x-py=0$. So the intersection is a (1 dimensional) affine variety over the rationals, hence its points lie in the algebraic numbers. Not sure if that's the kind of thing you're asking about. I'm not sure about the best way of generalizing the statement about finding it in terms of square roots. –  George Lowther Feb 15 '11 at 21:30
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Chapter 3 of Prasolov/Solovyev may be of interest. –  J. M. Apr 19 '11 at 18:06

3 Answers 3

up vote 6 down vote accepted
+100

Apparently the same exercise can be done for the lemniscate with the same result. For instance, see http://www.jstor.org/stable/2321821 where Theorem 2 states that

If the lemniscate can be divided in n parts with ruler and compass, then n is a power of two times a product of distinct Fermat primes.

The main difficulty, when compared to the better known theorem about the circle, appears to be the shift from circular functions (sin, cos) to elliptic functions. For instance one requires some sort of addition theorem for these functions.

This is only one more curve, but one that can be associated to the important elliptic integral $\int \frac{dt}{\sqrt{1-t^4}}$, making an appearance as the arc-length of the lemniscate. I'm guessing there is a wide class of curves that are associated to elliptics integrals this way, but I doubt that any of them would naturally be as interesting as the circle or the lemniscate.

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See this. In particular, the Gauss-Wantzel Theorem says the following:

Theorem. A regular $n$-gon can be constructed with compass and straightedge iff

  • $n$ is a Fermat prime
  • $n$ is a power of $2$
  • $n$ is a product of of a power of $2$ and distinct Fermat primes
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sorry my question is really about curves other than the circle. Good theorem though. –  quanta Jan 13 '11 at 18:15

To flesh out the comment I gave: Prasolov and Solovyev mention an example due to Euler and Serret: consider the plane curve with complex parametrization

$$z=\frac{(t-a)^{n+2}}{(t-\bar{a})^n (t+i)^2}$$

where $a=\frac{\sqrt{n(n+2)}}{n+1}-\frac{i}{n+1}$ and $n$ is a positive rational number.

The arclength function for this curve is $s=\frac{2\sqrt{n(n+2)}}{n+1}\arctan\,t$; since

$$\arctan\,u+\arctan\,v=\arctan\frac{u+v}{1-u v}$$

the division of an arc of this curve can be done algebraically (with straightedge and compass for special values).

Here are plots of these curves for various values of $n$:

Euler-Serret curves

Serret also considered curves whose arclengths can be expressed in terms of the incomplete elliptic integral of the first kind $F(\phi|m)$; I'll write about those later once I figure out how to plot these... (but see the Prasolov/Solovyev book for details)

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