The mark that changes what is reachable

An angle of sixty degrees cannot be cut into three equal parts with compass and straightedge. The angle that will not divide by three settles that by arithmetic: cutting it would produce a number of degree three over the rationals, and every number the two instruments reach has degree a power of two.

Now add one operation. Allow a straightedge carrying two marks a fixed distance apart, and allow it to be slid until the marked segment has its ends on two stated curves. That single addition trisects every angle.

The construction is four hundred years older than the impossibility proof and was never in doubt. What changed in 1837 was not whether it worked but what it meant.

The construction, and why it works #

Let $O$ be the centre of a circle of radius $r$ and let $A$ be a point on it, at angle $\theta$ from a chosen diameter. Extend that diameter backwards past the circle.

Lay the marked straightedge through $A$ and slide it until the segment cut off between the extended diameter and the far side of the circle is exactly $r$. Call the point where it crosses the diameter $P$, and the nearer crossing with the circle $Q$.

Then the angle at $P$ is $\theta/3$, and here is why. The segment $PQ$ has length $r$, and so does $OQ$, being a radius — so triangle $OPQ$ is isosceles and the angle at $O$ equals the angle $\varphi$ at $P$. The exterior angle of that triangle at $Q$ is therefore $2\varphi$. But $OQ$ and $OA$ are both radii, so triangle $OQA$ is isosceles too, and its angle at $A$ is also $2\varphi$. Finally, $\theta$ is the exterior angle of triangle $OPA$ at $O$, which is the sum of the two remote angles: $\varphi + 2\varphi = 3\varphi$.

Three isosceles triangles and one exterior-angle fact. There is nothing hard in it, and the figures assert both isosceles conditions by measuring the drawn distances rather than by quoting the argument.

Where the extra power comes from #

Each compass-and-straightedge step solves a linear or a quadratic equation over the numbers already built — two lines meet in a solution of a linear system, a line meets a circle or two circles meet in a solution of a quadratic. Every step is a square root is the statement that the reachable numbers form a tower of extensions each of degree at most two, so every reachable number has degree a power of two over the rationals.

A tower whose degrees multiply supplies the multiplication that makes this a proof rather than a plausibility: degrees multiply along a tower, so a number of degree three cannot appear in a tower of degree $2^k$.

A neusis step is different in kind. Sliding a fixed-length segment until both ends lie on stated curves is asking for the intersection of a curve with a conchoid — the locus of points at a fixed distance along a ray from a pole — and that intersection is governed by an equation of degree three or four rather than two.

So the reachable field grows by extensions of degree up to four rather than up to two, and a number of degree three comes into range. The trisection is precisely a degree-three problem: $\cos(\theta/3)$ satisfies $4y^3 - 3y = \cos\theta$, which for $\theta = 60°$ is $4y^3 - 3y = 1/2$, irreducible over the rationals.

Two instrument sets, and the same question asked of both #

The cleanest comparison uses angles a regular polygon hands over. A third of $360/n$ is $360/3n$, so trisecting that angle is drawing the regular $3n$-gon — and which polygons can be drawn decides that by Gauss’s criterion.

The table’s two columns are computed by routes with nothing in common — one is a factorisation into Fermat primes, the other a bisection search followed by a measurement — and they disagree in three rows. That disagreement is the whole content of adding the mark.

The conchoid, which is what the slide really draws #

There is a curve hiding in the phrase “slide until the segment has its ends on two curves”, and naming it is what turns the neusis from a physical action into a construction.

Fix a point $O$ (the pole), a line $\ell$, and a distance $k$. For each ray from $O$, mark the two points at distance $k$ from where the ray crosses $\ell$. The locus of all such points is the conchoid of Nicomedes, with Cartesian equation

$$(x - a)^2 (x^2 + y^2) = k^2 x^2$$

when $\ell$ is the line $x = a$. It is a quartic curve with a distinctive loop or cusp near the pole depending on how $k$ compares with $a$.

A neusis placement is exactly an intersection of a conchoid with the second curve. Sliding the ruler until the marked segment fits is finding where the conchoid meets the circle, and Nicomedes built a linkage that draws the curve so that the placement could be made once and reused rather than fiddled with each time.

That reformulation is what makes the degree argument available. An intersection of a quartic with a line or a circle is a system whose elimination gives an equation of degree at most four, and the quartic’s own structure reduces it to a cubic in the cases of interest. Nothing about the reach of the marked straightedge has to be argued from the sliding; it is read off the curve.

And this is where the operation set stops being an idealisation. A compass and a straightedge are idealisations of physical instruments and so is the marked ruler, but only the last of the three requires a search rather than a drawing: the position is found by moving until a condition holds, which is a different kind of act from drawing a circle through a point. That difference is why some accounts refuse it and why Pappus classified it separately.

What else the mark buys #

Doubling the cube. Nicomedes gave a neusis construction for two mean proportionals, which is what doubling the cube reduces to, in the second century BC. The same degree argument covers it: $\sqrt[3]{2}$ has degree three, out of reach of quadratics and inside the reach of cubics.

Regular polygons the classical instruments miss. The heptagon and the nonagon are both constructible by neusis, and more generally every regular $n$-gon whose $n$ has the form $2^a 3^b p_1 \cdots p_k$ with the $p_i$ distinct Pierpont primes — primes of the form $2^u 3^v + 1$. That is a strictly larger class than Gauss’s, which allows only Fermat primes and no factor of 3 beyond the first power.

And nothing beyond degree six. A neusis step solves a cubic or a quartic, so the reachable numbers lie in towers whose degrees are products of 2s and 3s. A number of degree five is not reachable, and neither is the quintic’s general solution. The mark extends the reach and does not remove the boundary; it moves it.

The heptagon, worked #

The seven-sided polygon is the smallest one the classical instruments miss and the marked straightedge reaches, so it is the case worth following.

Constructing a regular heptagon means constructing $\cos(2\pi/7)$. That number satisfies

$$8y^3 + 4y^2 - 4y - 1 = 0,$$

which has no rational root — the candidates are $\pm 1, \pm 1/2, \pm 1/4, \pm 1/8$ and none works — so it is irreducible and $\cos(2\pi/7)$ has degree three. Out of reach of quadratic towers, and inside the reach of cubic ones.

Gauss’s criterion says the same thing from the other side: 7 is prime and is not a Fermat prime, since $7 - 1 = 6$ is not a power of two. And which polygons can be drawn decides every $n$ up to a bound by exactly that test.

The Pierpont condition, which governs the neusis case, asks instead whether $n - 1$ has the form $2^u 3^v$. For 7 that is $6 = 2 \cdot 3$, so the heptagon is constructible by neusis; for 11 it is $10 = 2 \cdot 5$, so the hendecagon is not. The move from Fermat primes to Pierpont primes is the arithmetic shadow of the move from quadratics to cubics, and it is the cleanest statement of what the extra operation is worth.

Why nobody counted the mark as cheating for two thousand years #

Greek geometry used neusis freely. Archimedes’ trisection is in the Book of Lemmas; Nicomedes built a mechanical device — the conchoid-drawer — for performing the slide; Pappus classified problems into plane (soluble by line and circle), solid (soluble by conics) and linear (needing other curves), and treated neusis as a legitimate method whose place in that hierarchy was worth arguing about.

What Pappus insisted on was not that neusis was forbidden, but that a problem should be solved by the least powerful method that suffices. Using a conic where a circle would do was the error, not using a conic at all. The restriction to compass and straightedge as the only legitimate tools is a later reading, hardened in the nineteenth century when the impossibility proofs made the restricted question interesting.

So the historical position is closer to this site’s own framing than the folklore suggests. The question was always “what does this set of operations reach?”, and the classical instruments are one set among several — which is the sentence this field is written to.

The boundary with folding, which is a different set again #

Paper folding reaches cubics too, and by a different route: a single fold can place two given points onto two given lines simultaneously, which is the sixth Huzita–Hatori axiom and is a cubic condition. So origami constructions have the same reach as neusis — the same Pierpont primes, the same trisection, the same doubling of the cube.

The two are not the same operation set and the coincidence of their reach is a theorem rather than a definition. The fold is a different operation and belongs to a different subject; what is shared is the degree, and the degree is what both essays are really about. The straightedge buys nothing makes the same kind of comparison in the other direction, between two sets whose reach coincides for a different reason.

The whole classical closure, for comparison #

It is worth looking once at what the two instruments do reach, because the picture makes the “at most quadratic” claim concrete.

Each round adds finitely many points and each new point costs at most one square root. Iterating gives a tower, the tower’s degree is a power of two, and the three classical impossibilities are three numbers of degree not a power of two — $\sqrt[3]{2}$ at degree three, $\cos 20°$ at degree three, $\pi$ of infinite degree.

The last of those is different in kind and worth separating. The circle that will not square fails not because $\pi$ has the wrong degree but because it has no degree at all — it satisfies no polynomial equation with rational coefficients. The marked straightedge does not help with it, and neither does any operation set whose steps solve polynomial equations. Two of the three classical impossibilities are removed by one scratch on a ruler and the third is untouched by any of this, which is the sharpest illustration that the three problems were never the same problem.

What the pictures cannot show #

The neusis figure solves for the sliding point by bisection to two hundred iterations, which is exact to the limits of the arithmetic and is not what a draughtsman does. A person slides a physical ruler until it looks right, and “looks right” is a tolerance rather than a solution — the construction is exact in principle and approximate in every execution, exactly as compass-and-straightedge constructions are.

The angle is reported to six decimal places and agrees with a third of the original to all of them. That is a measurement of a solved configuration, not a proof that the construction is exact; the proof is the three isosceles triangles, and the figure asserts their equal sides rather than the conclusion.

And the reach table covers seven angles. The claim that the marked straightedge trisects every angle is the geometry above, which works for any $\theta$ strictly between 0° and 180°; the table checks seven instances and the argument covers the rest.

The ladder from here #

Below: what two points can build, the closure the classical operations generate, and the angle that will not divide by three, the impossibility this operation removes. Sideways: the straightedge buys nothing, a comparison of operation sets whose reaches coincide, and the cube that will not double, the other classical problem the mark settles. Above: the conchoid of Nicomedes and its cubic, Pierpont primes and the polygons they allow, the reach of conic-assisted construction, and the fact that no finite set of these operations reaches degree five.

What is worth carrying away #

An impossibility proof is always relative to a stated set of operations, and the statement is the load-bearing part. “The angle cannot be trisected” is false; “the angle cannot be trisected with compass and straightedge” is true, and the difference is one scratch on a ruler.

The scratch does something precise. It replaces an operation whose equations are quadratic with one whose equations are cubic, and every consequence follows from that single change of degree. When a boundary moves, it is worth asking what the new operation solves that the old one did not — the answer is nearly always a change in the degree of some equation, and the geometry is downstream of it.