A third reached only in the limit

Every construction along this path so far has added an instrument. The mark that changes what is reachable put two marks on the straightedge and trisected every angle in one sliding step; the spiral that measures its own circle traded lengths for angles through a curve. Each was needed because compass and straightedge alone cannot trisect a general angle — Pierre Wantzel proved it in 1837, by showing that trisecting $60°$ means solving a cubic equation that no tower of square roots reaches.

Here is a way to trisect any angle with compass and straightedge alone and nothing added. Bisect the angle twice, to get a quarter of it. Bisect that quarter twice, to get a sixteenth, and add it on. Bisect again twice, add a sixty-fourth. Continue. Since

$$\frac14 + \frac1{16} + \frac1{64} + \cdots = \frac{1/4}{1 - 1/4} = \frac13,$$

the arms approach the exact third. Every stage is an honest construction. None of them trisects the angle.

The figure shows the approach as a stack of magnifications. Each row looks at a small stretch of the angle around its exact third, four times narrower than the row above, and marks where the arm stands after one more step. In every row the dot sits in the same place — an eighth of the window short of the third — because each step divides the shortfall by four and each row magnifies by four. The approach is self-similar: it looks the same at every scale, and at no scale does it arrive.

Two series, one limit #

The quarters are not the only way. A third is also

$$\frac12 - \frac14 + \frac18 - \frac1{16} + \cdots,$$

a series in which each term needs only one bisection of the last piece, alternately added and taken away. Its partial sums overshoot and undershoot the third in turn, closing in by a factor of two each step instead of four. Two steps of this series cost about what one step of the quarters costs, and gain the same factor of four, so the two schemes are the same speed measured per bisection: each bisection buys one binary digit of the answer.

The sum that fits in one square drew the sum of halves as a square cut into ever smaller pieces, with a corner always left over. The quarter series has its own picture of the same kind: cut a square into four, keep one corner, cut the opposite corner into four, keep one of those, and so on; the kept pieces fill one of three congruent L-shaped regions, and so a third of the square, exactly — in the limit, and in no finite stage.

A finite construction cannot do it #

The reason no stage is exact is arithmetic, and it is the same arithmetic that makes the stages easy.

A construction by compass and straightedge produces, from the given angle, only angles that can be reached by bisecting and adding and subtracting what has already been built — and, for a general angle with nothing special about it, the only fractions of it available are the ones with a power of two below the line: $\tfrac12$, $\tfrac34$, $\tfrac{5}{16}$, and so on. The sum after $k$ steps is

$$\frac14 + \frac1{16} + \cdots + \frac1{4^k} = \frac13\left(1 - \frac1{4^k}\right),$$

a fraction with denominator $4^k$. A third is not a fraction with a power of two below the line, so no finite number of steps reaches it.

For particular angles the story is different, because some angles have thirds that happen to be constructible.

Trisecting an angle $\theta$ is the same as constructing $\cos(\theta/3)$ from $\cos\theta$, and the triple-angle formula ties them by a cubic: $4x^3 - 3x = \cos\theta$ with $x = \cos(\theta/3)$. When the cubic has a rational root, the third is reachable; when it has none, a tower whose degrees multiply shows that no tower of square roots contains its roots, and no construction reaches the third. A right angle is trisectable, since $\cos 30° = \sqrt3/2$ is a square root away. $60°$ is not: its cubic $8x^3 - 6x - 1 = 0$ has no rational root. The bisection series trisects $60°$ as well as any other angle — in the limit, and only there.

Which angles have an exact third #

The dial settles angles with a rational cosine, and the general rule is the same idea one level up. An angle $\theta$ can be trisected exactly when the cubic $4x^3 - 3x - \cos\theta$ has a root in the field built from $\cos\theta$ itself — when it factors there, so that $\cos(\theta/3)$ is reached by square roots from what is given. For a general angle, one whose cosine is treated as an unknown with no special relations, the cubic never factors, and that is the precise sense in which “the” trisection problem is impossible.

For particular angles, the answer depends on arithmetic that has nothing to do with geometry. Any angle that is three times a constructible angle is trisectable, trivially — its third was the starting point. So the angle $3 \times 72° = 216°$ can be trisected, since $72°$ comes from the regular pentagon, and $3 \times 22.5° = 67.5°$ can, since $22.5°$ is a quarter of a right angle. But $60°$ is three times $20°$, and $20°$ is not constructible, because the regular eighteen-gon is not — its angle needs the root of the irreducible cubic $8x^3 - 6x - 1$, the same cubic that appeared above.

The bisection series does not care which kind of angle it is given. It approaches the third of $216°$ exactly as slowly as the third of $60°$, arriving at neither, although one of them has a finite construction by another route. The limit construction is uniform and blind; the finite constructions are special and depend on the arithmetic of the particular angle.

The price of each step #

The limit is approached quickly, and it is worth knowing how quickly, because it is the reason approximate trisection was never a practical problem.

A bisection costs three operations — two circles and a line — and each step needs two bisections and a transfer of the new piece onto the running total, about nine operations in all. The shortfall after $k$ steps is $\theta/(3 \cdot 4^k)$. For a $60°$ angle, five steps — forty-five operations — leave the arm less than two hundredths of a degree short, which on a drawing ten centimetres across is a gap of about three thousandths of a millimetre, far below the width of a pencil line. Ten steps put the error below a millionth of a degree.

So as a matter of drawing, trisection was never impossible. A careful draughtsman can trisect any angle to the accuracy of the paper in a few minutes. The impossibility is exactly the impossibility of finishing: the Greek problem asked for a construction that ends with the exact third, and it is the ending, not the accuracy, that no compass supplies.

One sliding step against infinitely many #

The contrast with the marked ruler is worth stating plainly. Archimedes’ verging construction, which the mark that changes what is reachable drew, trisects any angle in a single operation: a straightedge carrying two marks is slid until the marks lie on a line and a circle while the edge passes through a given point, and the angle it then makes is exactly a third. One sliding step does what infinitely many bisections approach.

Measured this way, a marked ruler is worth an infinite amount of compass work for this one task — which is the sense in which it is a genuinely stronger instrument, and not merely a more convenient one. The same is true of the conic sections, which reach every cube root and so every trisection exactly, and of the spiral, which divides angles in any ratio at once. Each of them packs into one step a limit that compass and straightedge can only chase.

Doubling the cube the same way, and faster #

The same trick attacks the other classical problem. The cube that will not double showed that no finite construction produces $\sqrt[3]{2}$, the side of a cube with twice the volume of a unit cube. But compass and straightedge can add, subtract, multiply and divide lengths, and that is enough to run Newton’s method on the equation $x^3 = 2$:

$$x_{k+1} = \frac{2x_k + 2/x_k^2}{3}.$$

Starting from $x_0 = 1$, the constructible lengths $x_1 = 1.3333$, $x_2 = 1.2639$, $x_3 = 1.25993$ close in on $\sqrt[3]{2} = 1.259921\ldots$, and from then on the number of correct digits roughly doubles at every step. Every stage is a finite construction; the limit is not.

The comparison with the bisection series is instructive. Bisection gains a fixed number of binary digits per step, two for each step of the quarter series. Newton’s method gains digits in proportion to those it already has. Both are limits, both are excluded by the same rule, and one of them is enormously faster. What Wantzel’s theorem forbids is not approximation, which is cheap and can be very fast, but arrival.

Every fraction is a binary expansion #

The bisection series is one instance of a general scheme, and the scheme shows exactly which fractions of an angle a finite construction can reach.

Any fraction $p/q$ has a binary expansion $0.b_1b_2b_3\ldots$, and it can be read as instructions: bisect the angle once and add the piece if $b_1 = 1$; bisect again and add if $b_2 = 1$; and so on. The expansion ends exactly when $q$ is a power of two — $\tfrac38 = 0.011_2$ — and then the construction is finite. For every other $q$ the expansion repeats forever: a third is $0.010101\ldots_2$, a fifth $0.00110011\ldots_2$, a seventh $0.001001\ldots_2$. The period of the repetition is the number of steps it takes $2$ to return to $1$ modulo $q$ — two for a third, four for a fifth, three for a seventh — the multiplicative order that how often two generates every remainder studied.

The same limitation appears wherever numbers are stored in binary. A computer that represents numbers as binary fractions cannot store a third or a tenth exactly, which is why adding a tenth to itself ten times on most machines does not give exactly one; the error is in the last binary digit, exactly as the bisection series stops one piece short of the third at every stage.

So dividing an angle into $q$ equal parts by bisection alone is computing $1/q$ in binary, one digit per bisection, and it ends exactly when binary arithmetic can represent $1/q$ exactly. The limitation of the compass on a general angle is the limitation of a binary computer on a third: both can get as close as desired and neither can finish.

What a construction is made of #

It is worth being precise about what a construction produces at each stage, since the argument turns on it.

Start with two points. One round of construction draws every line through two known points and every circle centred at one known point through another, and adds the points where they cross. Another round does the same with the enlarged set. Every construction, however elaborate, is contained in some finite number of rounds — that is what it means to be a construction. The points reached after any finite number of rounds have coordinates built from the starting ones by the four arithmetic operations and square roots, a finite number of each.

The union over all rounds is the set of constructible numbers, and it is dense: between any two lengths there is a constructible one. So every length, and every angle, can be approached as closely as desired. What is special about the constructible numbers is not that they are everywhere — they are — but that each is reached at a finite round. The third of $60°$ is approached by the bisection series from round to round and belongs to none of them.

Why the limit is not allowed #

A reasonable objection is that mathematics uses limits everywhere, and a construction that converges might as well count. The answer is that it changes what the question is.

If limits are allowed, every real number is constructible, because every real number is a limit of dyadic fractions of a unit, each reachable by bisection. The distinctions this path has drawn — square roots but not cube roots for compass and straightedge; cube roots but not fifth roots for conics; the heptagon but not the hendecagon for a marked ruler, as a quintic a sliding mark reaches complicated — all collapse. The ancient problems were interesting precisely because they asked for exact, finite constructions, and the answers measure the algebraic strength of each instrument. Allowing the limit erases the instruments’ differences, and with them the theory.

That is also why the transcendental curves of the last two constructions are suspect in the same way. The quadratrix delivered $\pi$ at its foot, a limit point the defining motions never reach; the spiral delivered it through a tangent, the limit of secants. Each smuggled a limit into a single step. The bisection series makes the limit explicit and infinite, and in doing so shows what the other constructions were hiding.

What the figures can and cannot show #

Every stage is computed exactly. The partial sums are checked against the closed form $\tfrac13(1 - 4^{-k})$, the stack’s rows against the fixed fraction of their windows, and the binary expansions digit by digit against the fractions they represent.

The limit is not drawn. A figure shows finitely many stages. That the arms converge to the exact third is the sum of a geometric series; that no stage equals it is the statement about denominators; and that no other finite construction equals it is Wantzel’s theorem, which the dial of rational-cosine angles illustrates for a handful of cases and the degree argument proves for all.

The operation counts are a convention. Counting nine circles and lines per step assumes a particular way of transferring the new piece onto the running total; a cleverer bookkeeping might save a few, and would change the slope of nothing.

Still open: whether a length of e + π can be drawn #

A length is constructible from a unit exactly when it is algebraic of a special kind — a root of a polynomial with whole-number coefficients whose degree, over the rationals, is a power of two and whose field satisfies a further condition. Proving that a length is not constructible therefore needs knowing something about its algebraic nature. For $\pi$ that knowledge exists: Ferdinand von Lindemann proved in 1882 that it is transcendental, and so not constructible, which ended the squaring of the circle.

For many natural numbers the knowledge does not exist. It is known that at least one of $e + \pi$ and $e\pi$ is transcendental, but not which. Nobody can currently prove that $e + \pi$ is irrational, let alone transcendental, and so nobody can prove that a segment of length $e + \pi$ cannot be constructed with compass and straightedge from a unit. No one believes it can be. But the question of which numbers can be drawn in finitely many steps runs directly into the question of which numbers are algebraic, and for $e + \pi$, Euler’s constant $\gamma$, and many others, that question is open.

Close enough is not the same as there #

The habit worth keeping is to separate reaching a thing from approaching it.

The bisection series gets as close to a third as anyone could want, in a few steps, with the plainest instruments. For a draughtsman it solves the problem. For a mathematician it solves a different problem, and the difference is the whole content of Wantzel’s theorem: the constructible numbers are dense, so everything can be approached, and they are countable and algebraic, so almost nothing can be reached. A third of a general angle lies in the first set and not the second, and an infinite process is exactly what it takes to cross from one to the other. Every instrument on this path, from the marked ruler to the spiral, is a way of hiding that process inside a single step.