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Yesterday, I talked with a friend about a problem where the solution would be an angle of $2$ radians (about $114.6°$). Then somehow the question arose whether such an angle would be constructible (with straight edge and compass), and it looks like no - but I had not suddenly an idea on how to prove it either way.

The "easily" constructible angles (e.g. the ones for which I would know how to construct them without long thinking) are some rational multiples of $\pi$: We know that we can half each angle, and also construct sums and differences of angles, and we have angles of $\frac\pi3$ and $\frac\pi2$ to start with. (Then there are some more from some constructible regular polygons).

This is enough to approximate any angle, but does not really help to construct any rational (or even algebraic) angles (with the exception of the trivial angle $0$), since approximation is not a valid operation in constructive geometry.

Wikipedia says:

[...] The angles that are constructible form an abelian group under addition modulo $2\pi$ ([...]). The angles that are constructible are exactly those whose tangent (or equivalently, sine or cosine) is constructible as a number. [...]

It looks like I don't know enough about these functions to utilize this information.

The only angles of finite order that may be constructed starting with two points are those whose order is either a power of two, or a product of a power of two and a set of distinct Fermat primes.

Okay, these are the mentioned rational multiples of $\pi$.

In addition there is a dense set of constructible angles of infinite order.

All rational angles are angles of infinite order, thus if any would be constructible, it would be in this category. And of course, if we had any rational angle, we would get quite a lot other ones from halving and addition.

My question: Is there any rational angle $\alpha \in \mathbb Q \setminus\{0\}$ which is constructible, or is there a proof that no such angle exists?

If there are no rational ones, maybe any algebraic one?

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One interesting fact about the circle sector of angle 2 rad in the unit circle is that its area is one. I thought about using this to reduce this to the impossibility of squaring a circle, but there are not really any constructions relating the area of circle sectors. –  Paŭlo Ebermann May 28 '11 at 22:58
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If $\alpha$ is constructible, then $n\alpha$ is constructible, so you can restrict the problem to $\mathbb{Z}\setminus\{0\}$ –  Thomas Andrews May 28 '11 at 23:01
    
My guess is that $e^i=\sin 1 + i \cos 1$ is transcendental. Then $\sin n$ is transcendental for all $n$. –  Thomas Andrews May 28 '11 at 23:09
    
@Thomas: in fact, to the positive elements in that set :) –  Mariano Suárez-Alvarez May 28 '11 at 23:16
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@Thomas: I think this is solved this way ... do you want to make this into a real answer so I can accept it? (If not (and nobody else comes first), I'll write this up instead.) –  Paŭlo Ebermann May 28 '11 at 23:26
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1 Answer

up vote 9 down vote accepted

An angle is constructible if and only if the cosine of the angle is a constructible distance. In particular, if $\theta$ is constructible then $\cos\theta$ must be an algebraic number (or more specifically a constructible number). By the Lindemann-Weierstrass Theorem, this can only occur if $\theta$ is transcendental. Thus no nonzero algebraic angle is constructible.

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Nice theorem. We learn something everyday. –  Patrick Da Silva May 29 '11 at 6:01
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