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Lindemann's proof of the transcendence of $\pi$ has settled the question, whether an arbitrary angle can be trisected, using straightedge and compass alone, to the negative.
In the following, trisectable always means with straightedge and compass alone.

I could however find nothing but some vague statements about angles, that are trisectable; a typical such statement is, that up to a few exceptions, it is impossible to trisect angles.

Question: I would therefore like to know, whether the angles, that are trisectable have been fully characterized and, if yes, who has done that first.

Edit:
in view of the valuable feedback I got, I will try to summarize things from my perspective

  • it can be proven, that an angle $\theta$ can be constructed if and only if $\cos(\theta) \in \mathbb{E},$ where $\mathbb E$ (for Euclidean), often called the "constructable numbers," is the smallest subfield of the real numbers that is closed under taking square roots of positive elements.

  • there seems to be common agreement, that an angle $\theta$ can only be trisected with straightedge and compass, if $\cos(\frac{\theta}{3}))$ is constructable.

So either being trisectable and constructable are equivalent or, there are angles that are not constructable, but can be trisected.

I suspect however, that there are certain angles (e.g. $\frac{2\pi}{7+\frac{1}{3}}$), that can be trisected despite not being constructable, namely angles $\theta$, for which

$$ \frac{\operatorname{lcm}(\theta,2\pi)}{\theta} = 3k, k\in\mathbb{N}$$

those angles, which I would like to call auto-trisecting, are then either all constructable or, there are exceptions which are counter examples to the characterisation via constructability (the notion of being auto-trisecting of course easily generalizes to being auto-$n$-secting).

Interestingly, the only "basic" (i.e. involving only a single Fermat prime) constructable angles, that can be trisected, but are not auto-trisecting, seem to be the ones of the form$$3k\frac{2\pi}{2^nF_0}, k\in\mathbb{N},n\in\mathbb{N}_0, F_0 := 2^{2^0}+1=3$$
With that observation in mind, and assuming that all trisectable angles are either constructable or auto-trisecting, a bullet-proof strategy for finding a trisection of an angle, that is known to be trisectable with straightedge and compass alone, is to simply to construct multiples of it, until a period has been completed. The so generated multiples either

  • subdivide the angle in case it is auto-trisecting and the size of a trisection is equal to the sum of $\frac{1}{3}$ of the (equal) subdivisions or,

  • do not subdivide the angle, but constitute to a constructable regular n-gon and, by determining n, it is possible to determine a constructable angle that resembles a trisection.

in both cases it is possible to find a trisection and, altogether it can be said that the topic of trisecting an angle in case that its trisectability is known, can be discussed at classroom-level.

As a remark it can be said, that there is no hope for proving the existence of further Fermat primes by demonstrating that the known Fermat primes together with auto-trisecting angles do not cover all cases of trisectable angles and thus would necessitate the existence of further Fermat primes.

Edit II:
I found the following Wiki article that elucidates the history of proving that angles are not trisectable in general:
https://en.wikipedia.org/wiki/Angle_trisection
according to that article, the proof was found by Pierre Wantzel in 1837 and is not based on Lindemann's proof of the transcendence of $\pi$, but rather on Galois theory.

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    $\begingroup$ Presumably an algebraic characterization of the angles that are trisectable is: $\theta$ is trisectable if and only if $[\mathbb Q(\sin(\theta/3)):\mathbb Q(\sin(\theta))]$ is equal to 1 or 2. This is equivalent to the existence of an angle $\psi$ such that $3\psi=\theta\bmod{2\pi}$ and $\sin\psi$ can be expressed as a rational function of $\sin\theta$ (with integer coefficients) $\endgroup$
    – Anthony Quas
    Commented Jan 19, 2014 at 18:17
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    $\begingroup$ The problem of the trisection of the angle has been settled much before Lindemann's proof, since while the angle $\pi/3$ is constructible, $cos (\pi/9)$ satisfies a degree 3 irreducible polynomial over $\mathbf{Q}$. $\endgroup$
    – godelian
    Commented Jan 26, 2014 at 12:21
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    $\begingroup$ The example I gave shows that there can be no general procedure for trisecting a given angle, whether that angle is constructible or not. It does not appeal to the transcendence of $\pi$; Lindemann's proof settles the question of the impossibility of squaring the circle, but is not needed to show the impossibility of trisecting an angle. $\endgroup$
    – godelian
    Commented Jan 26, 2014 at 12:50
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    $\begingroup$ Transcendence of $\pi$ has nothing to do with trisecting angles. $\endgroup$
    – Vladimir Dotsenko
    Commented Jan 27, 2014 at 11:56
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    $\begingroup$ Also, "it can be proven, that an angle $\theta$ can be constructed if and only if $\cos(\theta)\in\mathbb{Q}$" is a bit bizarre. Are you claiming that $\pi/4$ cannot be constructed? $\endgroup$
    – Vladimir Dotsenko
    Commented Jan 27, 2014 at 11:58

4 Answers 4

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There is an aspect of the word "construct" that comes off a bit sour here. There are infinitely many angles such that $\theta$ and $\theta/3$ are both constructable. However, the set of constructable angles $\theta$ for which $\theta/3$ is not constructable is dense. By continuity, it follows that there is no procedure for beginning with $\theta$ and using a finite number of compass and straightedge operations to actually find $\theta/3,$ even when it is constructable.

I wrote an article on the hyperbolic plane, constant curvature $-1.$ As pointed out by Bolyai, there are a countably infinite set of pairs, one circle and one square, such that the radius of the circle and the edge of the square are both constructable, and the two figures have the same area. However, there is no way to begin with one and find the other. The phrase "square the circle" is misleading, the reverse phrase "circling the square" would also be misleading if anyone ever used it.

I need to backtrack a little on that last phrase. If you have some sort of political plot going on in Moscow, Russia, and the hero is going around the perimeter of Red Square looking for the bad guys, it makes a good deal of sense to say that the hero is circling the square. The same would work with Times Square in New York.

Back to the Euclidean plane. Given an angle $\theta $ but not told what it is (in radians, say) there is no procedure for finding $\theta/3.$ Ever.

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    $\begingroup$ your last sentence is a very good point; constructing a trisection must not make use of values, I want however a characterization of the values, whose knowledge allows one to triangulate an angle with such a known value. There are other issues related to my question, which I will address in an edit. $\endgroup$
    – Manfred Weis
    Commented Jan 22, 2014 at 5:18
  • $\begingroup$ oops, of course I meant "to trisect an angle" and not to "triangulate an angle" $\endgroup$
    – Manfred Weis
    Commented Jan 23, 2014 at 5:28
  • $\begingroup$ @Will Jagy: I do not agree. If you are given an angle $\theta$, you know enough about it to be able to compute the minimal polynomial of $\cos(\theta/3))$ over some field containing $\cos(\theta)$. According to what this polynomial is (essentially, degree $1$ or $2$, vs. degree $3$), you will say how to perform the construction, or know it cannot be constructed. $\endgroup$
    – ACL
    Commented Jan 27, 2014 at 17:23
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    $\begingroup$ @ACL, my picture of being given an angle is, essentially, a pair of rays on a piece of paper, with no method for deciding the angle measure to arbitrary accuracy. $\endgroup$
    – Will Jagy
    Commented Jan 27, 2014 at 19:15
  • $\begingroup$ it is more a philosophical problem whether an angle or a circle drawn on paper actually are mathematical angles or circles. The problem of trisecting an angle can also be interpreted as a practical one, where imprecision plays a role. $\endgroup$
    – Manfred Weis
    Commented Jan 29, 2014 at 7:49
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In a comment you wrote: ... I was hoping for a characterization via fractions of $2 \pi$ which could eventually be presented in a classroom

For this you may find the following papers useful:

Joe Dan Austin and Kathleen Ann Austin, Constructing and trisecting angles with integer angle measures, Mathematics Teacher 72 #4 (April 1979), 290-293.

Keith Robin McLean, Constructing rational angles (1), Mathematical Gazette 67 #440 (June 1983), 127-128.

David Harold Armitage, Constructing rational angles (2), Mathematical Gazette 67 #440 (June 1983), 128-129.

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    $\begingroup$ are there any online copies of the articles available? I live in Germany and therefore it will be hard to get access to the print versions. $\endgroup$
    – Manfred Weis
    Commented Jan 22, 2014 at 5:11
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    $\begingroup$ All three papers are in the JSTOR archive. I don't have access to JSTOR, but if you have access or if you know someone who has access and would be willing to help (most anyone at a sufficiently large university will have access), you should be able to get digital copies of the papers. A couple of minutes ago I embedded the JSTOR URL's of the papers in my answer. (The URL's can be obtained without a subscription to JSTOR.) $\endgroup$
    – Dave L Renfro
    Commented Jan 22, 2014 at 15:47
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    $\begingroup$ In addition to the 3 references I gave in my answer, some of the references in "7. References: General and Historical" (pp. 36-41) of my manuscript A Detailed and Elementary Solution to $x^{17} = 1$ might also be useful. $\endgroup$
    – Dave L Renfro
    Commented Jan 23, 2014 at 15:58
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    $\begingroup$ @Manfred Weis: This answer was upvoted in the last roughly 12 hours, which led me to glance at it (I didn't recall it from the title alone, and apparently I haven't looked at it in many years) and see your comment. If you still don't have copies of (or even access to) one or more of the papers, then I can make .pdf scans for you from my photocopies. I didn't offer before because then I did not have the means to make .pdf scans (aside from a local photocopy store, which charged an unusually high fee -- $1.00 per page? -- due to it not being one of their normal services), but now I can. $\endgroup$
    – Dave L Renfro
    Commented Nov 29, 2023 at 8:20
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Ruler and Compass construction is discussed in length in 'Galois Theory' - Ian Stewart.

Essentially if a field extension of a point we wish to construct is of degree 2^k (over the rational numbers with k being a natural number) this point can be constructed by ruler and compass. When it comes to trisecting angles it reduces down to the case of whether or not the angle (x/3) can be constructed by constructing the point b=2*cos(x/3). Taking cos(x/3)=cos(y) we can use the trigonometric identity cos(3y)=4*cos^3(y)- 3*cos(y) which can be rearranged to obtain the polynomial f=b^3-3*b-2*cos(x). Which is of degree 3 over the rational numbers (hence not degree 2^k). This polynomial may be reducible to a degree 2 polynomial multiplied by a degree 1 polynomial. If this is the case the angle you wish to trisect is therefore constructible by ruler and compass.

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An angle $\theta$ is trisectable if and only if $\cos (\theta/3)$ is constructable number and this happens just when the degree $[\mathbb{Q}(\cos (\theta/3)):\mathbb{Q}]$ is a power of 2.

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  • $\begingroup$ Note that my answer and the answer of Antony Quas are essentially the same. $\endgroup$
    – Sh.M1972
    Commented Jan 19, 2014 at 18:28
  • $\begingroup$ thank you for your answer; while it certainly answers the question, I was hoping for a characterization via fractions of $2\pi$ which could eventually be presented in a classroom. $\endgroup$
    – Manfred Weis
    Commented Jan 19, 2014 at 18:41
  • $\begingroup$ Since $\mathbf{Q}(e^{i \alpha})$ has degree 2 over $\mathbf{Q}(\cos(\alpha))$ for $\alpha\not\in \mathbf{Z}\pi$, necessarily $e^{i\theta/3}$ has 2-power degree over $\mathbf{Q}$. But $e^{i z} = e^{(z/\pi)(\pi i)}$ with $\pi i$ a complex log of the algebraic $-1 \ne 0, 1$, so by Gelfond-Schneider it's algebraic iff $z/\pi$ is rational. You seek rational $\alpha$ so that $e^{i \pi \alpha}$ has 2-power degree over $\mathbf{Q}$; the $3\alpha$'s are the $\theta$'s. So the answer is $\theta = a/b$ in reduced form with $b = 2^r \prod p_i$ for distinct Fermat primes $p_i > 3$ (!) (avoiding 6...). $\endgroup$
    – user76758
    Commented Jan 19, 2014 at 19:41
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    $\begingroup$ The above is only addressing when the trisected angle can be built by straightedge and compass starting from nothing but unit length. This is sort of the "wrong" viewpoint, as Quas identifies the correct viewpoint: one seeks trisectability given the angle $\theta$, so it becomes an algebraicity question over $\mathbf{Q}(\cos \theta)$, which in turn looks basically hopeless for a nice answer. $\endgroup$
    – user76758
    Commented Jan 19, 2014 at 19:44
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    $\begingroup$ Is the same true for bisectable? An angle $\theta$ is bisectable if and only if $\cos(\theta/2)$ is constructible? I thought all angles were bisectable. $\endgroup$
    – bof
    Commented Jan 22, 2014 at 19:42

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