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One of the most well known problems from ancient Greek mathematics was that of trisecting an angle by straightedge and compass, which was eventually proven impossible in 1837 by Pierre Wantzel, using methods from Galois theory.

Formally, one can set up the problem as follows. Define a configuration to be a finite collection ${{\mathcal C}}$ of points, lines, and circles in the Euclidean plane. Define a construction step to be one of the following operations to enlarge the collection ${{\mathcal C}}$:

• (Straightedge) Given two distinct points ${A, B}$ in ${{\mathcal C}}$, form the line ${\overline{AB}}$ that connects ${A}$ and ${B}$, and add it to ${{\mathcal C}}$.
• (Compass) Given two distinct points ${A, B}$ in ${{\mathcal C}}$, and given a third point ${O}$ in ${{\mathcal C}}$ (which may or may not equal ${A}$ or ${B}$), form the circle with centre ${O}$ and radius equal to the length ${|AB|}$ of the line segment joining ${A}$ and ${B}$, and add it to ${{\mathcal C}}$.
• (Intersection) Given two distinct curves ${\gamma, \gamma'}$ in ${{\mathcal C}}$ (thus ${\gamma}$ is either a line or a circle in ${{\mathcal C}}$, and similarly for ${\gamma'}$), select a point ${P}$ that is common to both ${\gamma}$ and ${\gamma'}$ (there are at most two such points), and add it to ${{\mathcal C}}$.

We say that a point, line, or circle is constructible by straightedge and compass from a configuration ${{\mathcal C}}$ if it can be obtained from ${{\mathcal C}}$ after applying a finite number of construction steps.

Problem 1 (Angle trisection) Let ${A, B, C}$ be distinct points in the plane. Is it always possible to construct by straightedge and compass from ${A,B,C}$ a line ${\ell}$ through ${A}$ that trisects the angle ${\angle BAC}$, in the sense that the angle between ${\ell}$ and ${BA}$ is one third of the angle of ${\angle BAC}$?

Thanks to Wantzel’s result, the answer to this problem is known to be “no” in general; a generic angle ${\angle BAC}$ cannot be trisected by straightedge and compass. (On the other hand, some special angles can certainly be trisected by straightedge and compass, such as a right angle. Also, one can certainly trisect generic angles using other methods than straightedge and compass; see the Wikipedia page on angle trisection for some examples of this.)

The impossibility of angle trisection stands in sharp contrast to the easy construction of angle bisection via straightedge and compass, which we briefly review as follows:

1. Start with three points ${A, B, C}$.
2. Form the circle ${c_0}$ with centre ${A}$ and radius ${AB}$, and intersect it with the line ${\overline{AC}}$. Let ${D}$ be the point in this intersection that lies on the same side of ${A}$ as ${C}$. (${D}$ may well be equal to ${C}$).
3. Form the circle ${c_1}$ with centre ${B}$ and radius ${AB}$, and the circle ${c_2}$ with centre ${D}$ and radius ${AB}$. Let ${E}$ be the point of intersection of ${c_1}$ and ${c_2}$ that is not ${A}$.
4. The line ${\ell := \overline{AE}}$ will then bisect the angle ${\angle BAC}$.

The key difference between angle trisection and angle bisection ultimately boils down to the following trivial number-theoretic fact:

Lemma 2 There is no power of ${2}$ that is evenly divisible by ${3}$.

Proof: Obvious by modular arithmetic, by induction, or by the fundamental theorem of arithmetic. $\Box$

In contrast, there are of course plenty of powers of ${2}$ that are evenly divisible by ${2}$, and this is ultimately why angle bisection is easy while angle trisection is hard.

The standard way in which Lemma 2 is used to demonstrate the impossibility of angle trisection is via Galois theory. The implication is quite short if one knows this theory, but quite opaque otherwise. We briefly sketch the proof of this implication here, though we will not need it in the rest of the discussion. Firstly, Lemma 2 implies the following fact about field extensions.

Corollary 3 Let ${F}$ be a field, and let ${E}$ be an extension of ${F}$ that can be constructed out of ${F}$ by a finite sequence of quadratic extensions. Then ${E}$ does not contain any cubic extensions ${K}$ of ${F}$.

Proof: If $E$ contained a cubic extension $K$ of $F$, then the dimension of $E$ over $F$ would be a multiple of three. On the other hand, if $E$ is obtained from $F$ by a tower of quadratic extensions, then the dimension of $E$ over $F$ is a power of two. The claim then follows from Lemma 2. $\Box$

To conclude the proof, one then notes that any point, line, or circle that can be constructed from a configuration ${{\mathcal C}}$ is definable in a field obtained from the coefficients of all the objects in ${{\mathcal C}}$ after taking a finite number of quadratic extensions, whereas a trisection of an angle ${\angle ABC}$ will generically only be definable in a cubic extension of the field generated by the coordinates of ${A,B,C}$.

The Galois theory method also allows one to obtain many other impossibility results of this type, most famously the Abel-Ruffini theorem on the insolvability of the quintic equation by radicals. For this reason (and also because of the many applications of Galois theory to number theory and other branches of mathematics), the Galois theory argument is the “right” way to prove the impossibility of angle trisection within the broader framework of modern mathematics. However, this argument has the drawback that it requires one to first understand Galois theory (or at least field theory), which is usually not presented until an advanced undergraduate algebra or number theory course, whilst the angle trisection problem requires only high-school level mathematics to formulate. Even if one is allowed to “cheat” and sweep several technicalities under the rug, one still needs to possess a fair amount of solid intuition about advanced algebra in order to appreciate the proof. (This was undoubtedly one reason why, even after Wantzel’s impossibility result was published, a large amount of effort was still expended by amateur mathematicians to try to trisect a general angle.)

In this post I would therefore like to present a different proof (or perhaps more accurately, a disguised version of the standard proof) of the impossibility of angle trisection by straightedge and compass, that avoids explicit mention of Galois theory (though it is never far beneath the surface). With “cheats”, the proof is actually quite simple and geometric (except for Lemma 2, which is still used at a crucial juncture), based on the basic geometric concept of monodromy; unfortunately, some technical work is needed however to remove these cheats.

To describe the intuitive idea of the proof, let us return to the angle bisection construction, that takes a triple ${A, B, C}$ of points as input and returns a bisecting line ${\ell}$ as output. We iterate the construction to create a quadrisecting line ${m}$, via the following sequence of steps that extend the original bisection construction:

1. Start with three points ${A, B, C}$.
2. Form the circle ${c_0}$ with centre ${A}$ and radius ${AB}$, and intersect it with the line ${\overline{AC}}$. Let ${D}$ be the point in this intersection that lies on the same side of ${A}$ as ${C}$. (${D}$ may well be equal to ${C}$).
3. Form the circle ${c_1}$ with centre ${B}$ and radius ${AB}$, and the circle ${c_2}$ with centre ${D}$ and radius ${AB}$. Let ${E}$ be the point of intersection of ${c_1}$ and ${c_2}$ that is not ${A}$.
4. Let ${F}$ be the point on the line ${\ell := \overline{AE}}$ which lies on ${c_0}$, and is on the same side of ${A}$ as ${E}$.
5. Form the circle ${c_3}$ with centre ${F}$ and radius ${AB}$. Let ${G}$ be the point of intersection of ${c_1}$ and ${c_3}$ that is not ${A}$.
6. The line ${m := \overline{AG}}$ will then quadrisect the angle ${\angle BAC}$.

Let us fix the points ${A}$ and ${B}$, but not ${C}$, and view ${m}$ (as well as intermediate objects such as ${D}$, ${c_2}$, ${E}$, ${\ell}$, ${F}$, ${c_3}$, ${G}$) as a function of ${C}$.

Let us now do the following: we begin rotating ${C}$ counterclockwise around ${A}$, which drags around the other objects ${D}$, ${c_2}$, ${E}$, ${\ell}$, ${F}$, ${c_3}$, ${G}$ that were constructed by ${C}$ accordingly. For instance, here is an early stage of this rotation process, when the angle ${\angle BAC}$ has become obtuse:

Now for the slightly tricky bit. We are going to keep rotating ${C}$ beyond a half-rotation of ${180^\circ}$, so that ${\angle BAC}$ now becomes a reflex angle. At this point, a singularity occurs; the point ${E}$ collides into ${A}$, and so there is an instant in which the line ${\ell = \overline{AE}}$ is not well-defined. However, this turns out to be a removable singularity (and the easiest way to demonstrate this will be to tap the power of complex analysis, as complex numbers can easily route around such a singularity), and we can blast through it to the other side, giving a picture like this:

Note that we have now deviated from the original construction in that ${F}$ and ${E}$ are no longer on the same side of ${A}$; we are thus now working in a continuation of that construction rather than with the construction itself. Nevertheless, we can still work with this continuation (much as, say, one works with analytic continuations of infinite series such as ${\sum_{n=1}^\infty \frac{1}{n^s}}$ beyond their original domain of definition).

We now keep rotating ${C}$ around ${A}$. Here, ${\angle BAC}$ is approaching a full rotation of ${360^\circ}$:

When ${\angle BAC}$ reaches a full rotation, a different singularity occurs: ${c_1}$ and ${c_2}$ coincide. Nevertheless, this is also a removable singularity, and we blast through to beyond a full rotation:

And now ${C}$ is back where it started, as are ${D}$, ${c_2}$, ${E}$, and ${\ell}$… but the point ${F}$ has moved, from one intersection point of ${\ell \cap c_3}$ to the other. As a consequence, ${c_3}$, ${G}$, and ${m}$ have also changed, with ${m}$ being at right angles to where it was before. (In the jargon of modern mathematics, the quadrisection construction has a non-trivial monodromy.)

But nothing stops us from rotating ${C}$ some more. If we continue this procedure, we see that after two full rotations of ${C}$ around ${A}$, all points, lines, and circles constructed from ${A, B, C}$ have returned to their original positions. Because of this, we shall say that the quadrisection construction described above is periodic with period ${2}$.

Similarly, if one performs an octisection of the angle ${\angle BAC}$ by bisecting the quadrisection, one can verify that this octisection is periodic with period ${4}$; it takes four full rotations of ${C}$ around ${A}$ before the configuration returns to where it started. More generally, one can show

Proposition 4 Any construction of straightedge and compass from the points ${A,B,C}$ is periodic with period equal to a power of ${2}$.

The reason for this, ultimately, is because any two circles or lines will intersect each other in at most two points, and so at each step of a straightedge-and-compass construction there is an ambiguity of at most ${2! = 2}$. Each rotation of ${C}$ around ${A}$ can potentially flip one of these points to the other, but then if one rotates again, the point returns to its original position, and then one can analyse the next point in the construction in the same fashion until one obtains the proposition.

But now consider a putative trisection operation, that starts with an arbitrary angle ${\angle BAC}$ and somehow uses some sequence of straightedge and compass constructions to end up with a trisecting line ${\ell}$:

What is the period of this construction? If we continuously rotate ${C}$ around ${A}$, we observe that a full rotations of ${C}$ only causes the trisecting line ${\ell}$ to rotate by a third of a full rotation (i.e. by ${120^\circ}$):

Because of this, we see that the period of any construction that contains ${\ell}$ must be a multiple of ${3}$. But this contradicts Proposition 4 and Lemma 2.

Below the fold, I will make the above proof rigorous. Unfortunately, in doing so, I had to again leave the world of high-school mathematics, as one needs a little bit of algebraic geometry and complex analysis to resolve the issues with singularities that we saw in the above sketch. Still, I feel that at an intuitive level at least, this argument is more geometric and accessible than the Galois-theoretic argument (though anyone familiar with Galois theory will note that there is really not that much difference between the proofs, ultimately, as one has simply replaced the Galois group with a closely related monodromy group instead).