The angle that will not divide by three

Bisecting an angle is the second thing anyone learns to do with a compass. Trisecting one is the thing that has been declared solved, wrongly, more often than any other statement in mathematics.

The difference between the two operations is not effort. It is degree, and the reason bisection is easy is worth having in hand before the reason trisection is not.

Why halving is free #

An angle is given as two rays from a point. To halve it, draw a circle centred at the vertex, and circles of equal radius centred where it meets each ray; the two new circles cross at a point on the bisector. One instrument, three circles, done.

In the arithmetic of the previous rung, halving is free because it is a square root. The half-angle formula says

$$\cos\frac{\theta}{2} = \sqrt{\frac{1+\cos\theta}{2}},$$

which is exactly one square root applied to something already reachable — and one square root is precisely what a compass is for. Halving again costs another, and again another; every angle can be cut into $2^k$ equal parts, for any $k$, with $k$ circles.

Three is not a power of two, and the reader who has read this far already suspects how the rest goes.

Trisection as a cubic #

Cutting $\theta$ into three means constructing the angle $\theta/3$, which means constructing the length $\cos(\theta/3)$ from the length $\cos\theta$.

The relation between the two is the triple-angle identity, which falls straight out of the circle’s own arithmetic:

$$\cos 3\alpha = 4\cos^3\alpha - 3\cos\alpha.$$

Writing $c = \cos\theta$ and $y = \cos(\theta/3)$ turns that into

$$4y^3 - 3y - c = 0,$$

a cubic in $y$ whose coefficients involve nothing but whole numbers and the given $c$. Trisecting $\theta$ is solving that cubic by compass and straightedge.

By the degree theorem, that is possible only if the cubic has a root of degree a power of two over the field containing $c$. And a cubic has such a root exactly when it is reducible — because if it does not factor, every root has degree three.

Sixty degrees, and its four candidates #

Take $\theta = 60°$, so $c = \frac12$. The cubic is $4y^3 - 3y - \frac12 = 0$; clearing the fraction gives

$$8y^3 - 6y - 1 = 0.$$

The rational root theorem allows $p/q$ with $p$ dividing $1$ and $q$ dividing $8$: that is $±1, ±\frac12, ±\frac14, ±\frac18$. The figure evaluates the cubic at each, exactly, and none of them is a root.

So $8y^3-6y-1$ is irreducible over the rationals, $\cos 20°$ has degree three, and three is not a power of two. A sixty-degree angle cannot be trisected with straightedge and compass.

That is the entire proof. It occupies rather less space than the average claim to have refuted it.

Which is not the same as “no angle can be trisected” #

The result is almost always stated too strongly, and stating it too strongly is what keeps the false proofs coming, because a person who trisects a right angle has genuinely trisected an angle.

A right angle has $\cos\theta = 0$, the cubic becomes $4y^3-3y = y(4y^2-3)$, and $y=0$ is a root on sight. The cubic factors, the remaining quadratic gives $y = \frac{\sqrt3}{2}$, and $30°$ is constructible — which it plainly is, since it is half of the $60°$ that falls out of two circles.

A straight angle has $\cos\theta = -1$ and the cubic $4y^3-3y+1$ has the root $y = \frac12$, giving $60°$. A zero angle trisects trivially. Any angle that was built as three copies of something already drawn trisects, by construction.

So the honest statement is:

There is no construction that trisects every angle, and $60°$ is a witness.

That is what an impossibility proof about a general method looks like, and the witness is the load-bearing part.

Why the cubic is the right object to look at #

It is worth pausing on the reduction, because it does something that is easy to pass over: it converts a question about angles into a question about lengths, and the conversion is what makes the algebra applicable at all.

An angle is constructible exactly when its cosine is, and the reason is a single perpendicular. Given the angle, drop a perpendicular from a point on one ray to the other and the cosine is a ratio of two constructed lengths. Given the length, mark it along a unit segment, erect the perpendicular, and where it meets the unit circle is the angle. So angles and their cosines stand or fall together, and there is no loss in working with numbers throughout.

That move is used silently in every essay of this ladder and it is the reason a subject about drawing has an arithmetic answer. It is the same trick as putting a grid on the plane: once a picture has coordinates, every question about it becomes a question about numbers, and numbers have theorems.

How many angles do trisect #

Once the question is put that way it becomes a counting question, which is the kind this field prefers.

The angles with a rational cosine that can be trisected are exactly those whose cosine has the form $4y^3-3y$ for a rational $y$ — because a rational $y$ is a rational root, and a rational root makes the cubic reducible. So the trisectable rational cosines are the image of the rationals under one cubic map, and that image is thin.

The figure at the top of this page measures how thin, over the seventeen angles whose cosine is a multiple of an eighth. Three of them trisect: $0°$, $90°$ and $180°$, the three where nobody needed telling. The other fourteen — including $60°$, and including angles as unremarkable as the one with cosine $\frac38$ — do not.

The three figures together make a point no single one of them could. The trisectable set is not merely small at one resolution — it does not grow as the resolution improves. Every new angle admitted by a finer denominator is a new failure.

Why the claims keep coming #

Angle trisection has a small permanent literature of purported solutions, and their authors are not fools. The failures fall into a few recognisable kinds and each is instructive about what the impossibility actually forbids.

Approximations. Many proposed constructions trisect to within a fraction of a degree, and some are extremely good. Nothing in the theorem forbids that — constructible numbers are dense, so any angle can be approximated as closely as one likes by a constructible one. What is forbidden is exactness, and a construction that is exact for no angle except by accident is not a counterexample.

Extra tools, unnoticed. A construction that at some point says “mark this length on the straightedge and slide it until the marks touch both curves” has left the operation set. Archimedes’ trisection does exactly this, deliberately and correctly; it is a fine construction and not a compass-and-straightedge one. The slide is a neusis, and a marked ruler reaches every cubic.

Infinitely many steps. An angle can be trisected by bisecting repeatedly and taking the limit: $\frac13 = \frac14 + \frac1{16} + \frac1{64} + \cdots$, so trisecting is an infinite alternating sequence of bisections. Every finite stage is constructible and the limit is the answer. The rules require finitely many steps, and this is where that clause earns its place.

A different angle. The most common and the hardest to argue with: the construction trisects some particular angle correctly, and the author has not noticed that the general claim is what was at issue.

There is a fifth kind, rarer and more interesting, which is a construction correct for a family of angles. Any angle of the form $3\alpha$ where $\alpha$ is itself constructible trisects, obviously, and the constructible angles are dense, so a construction that happens to work on a dense set of inputs can look convincing on every case anybody tries. The test that separates it from a real trisection is not “does it work on this angle” but “does it work on $60°$”, and that is why the witness matters so much more than the theorem. An impossibility with a named witness can be checked by anyone in an afternoon; an impossibility without one is a claim about an infinite family and cannot be tested at all.

The same asymmetry runs through every negative result in this field. Proving something can be done means producing it. Proving it cannot means producing the one instance where it fails and a reason — and the reason is almost always an argument about whole numbers wearing geometric clothes.

The approximation that is very nearly good enough #

The first failure mode deserves more than a line, because it is the one that is mathematically interesting rather than merely mistaken.

Constructible numbers are dense on the line: between any two of them there is another, and any real number has constructible numbers as close to it as one likes. So for any angle there is a constructible angle within a millionth of a degree of a third of it, and a construction producing one is not wrong about anything — it is simply not answering the question asked.

The relationship between exactness and approximation here is the same one that runs through continued fractions, where a number’s best rational approximations can be listed and none of them is the number. A trisection accurate to eight decimal places and a trisection are different objects, and the difference is not one of degree. It is the difference between a statement about a construction and a statement about a limit.

The clean version is the infinite bisection above. Writing $\frac13$ in binary gives $0.010101\ldots$, so

$$\frac{\theta}{3} = \frac{\theta}{4} + \frac{\theta}{16} + \frac{\theta}{64} + \cdots,$$

and each partial sum is reached by bisections and additions, both free. The sequence converges to the answer and never arrives, which is exactly what the finiteness clause in the rules forbids. That clause is not a technicality; it is where the whole subject’s content lives, and the point set the operations generate is defined by finitely many rounds for the same reason.

Trisectors that do something else entirely #

The angle trisectors have one famous theorem to their name, and it is worth knowing because it shows the trisected angles are perfectly good objects — it is only their construction that is barred.

Take any triangle at all. Trisect each of its three angles, and consider the six trisecting rays. The two adjacent to each side meet in a point; those three points are the corners of an equilateral triangle, always, whatever the original triangle was.

That is Morley’s theorem, found in 1899, and it is startling in the same way the inscribed angle theorem is startling — a quantity that has no business being constant turns out to be constant. The Morley triangle exists for every triangle, has the same shape every time, and cannot be drawn by the two instruments, because drawing it would require trisecting an arbitrary angle.

There is no tension there. The theorem is about angles that exist; the impossibility is about which of them a particular pair of instruments can reach. It is a reminder that the boundary being mapped in this field is a boundary of a method and never of the mathematics.

The step the argument does not take #

One caution belongs on this page, because the argument as given proves slightly less than it appears to.

The reduction says trisection is possible only if the cubic is reducible over the field containing $\cos\theta$. When $\cos\theta$ is rational, that field is $\mathbb{Q}$ and reducibility means a rational root, which is the finite search the figures perform. When $\cos\theta$ is not rational — say it is $\frac{\sqrt2}{2}$, for $45°$ — the search must be run over $\mathbb{Q}(\sqrt2)$ instead, and a table of rational candidates says nothing.

The figures on this page therefore decide exactly the angles they claim to decide: those with a rational cosine. That is enough for the theorem, because $60°$ is among them, and it is enough for the counting claim, because the counting claim is stated about rational cosines. It is not a decision procedure for every angle, and no figure here pretends to be one.

$45°$, for the record, trisects if and only if $15°$ is constructible, which it is, being $45° - 30°$. The general question for constructible $\cos\theta$ is decidable, and deciding it needs the same machinery at one level up.

What it costs to be able to trisect #

There is a converse worth stating, because it puts a value on the thing the compass lacks.

Any instrument that solves cubics trisects every angle. The conic sections do it, as Menaechmus and Pappus knew; the marked ruler does it; folding paper does it. All of them reach exactly the numbers of degree three and four, and once cubics are available both the cube and the angle fall on the same day.

That is the pattern this whole field is built to show. An operation set has a reach, the reach has a description, and two problems that look unrelated turn out to sit on the same side of the same boundary. Doubling a cube and trisecting an angle are the same problem in the only sense that matters: both need a cubic, and neither needs anything more.

Where this ladder goes #

The next rung takes the same machinery to a question Gauss answered before Wantzel proved the theorem that explains it: which regular polygons can be drawn.

It is the most interesting of the four, because the answer is neither “all” nor “almost none”. It is a precise and strange list — $3, 4, 5, 6, 8, 10, 12, 15, 16, 17$ and on — and the reason it looks like that has nothing to do with geometry and everything to do with which numbers are one less than a power of two.