A quintic a sliding mark reaches

Two instruments with one reach settles the conics’ level completely: a number is reachable by conics or by an angle trisector exactly when its degree over the rationals is a product of twos and threes. That is a clean characterisation and it invites the obvious question, which the marked ruler’s reach names in one clause — “the hendecagon, which a marked ruler reaches through a quintic that no trisector can solve”.

So there is a polygon between the two levels, and it has eleven sides.

The degree, established #

The polygon needs $\cos(2\pi/11)$, whose degree over the rationals is $\varphi(11)/2 = 5$. An earlier essay uses that formula and this one builds the polynomial, because a degree quoted and a degree exhibited are different things.

Multiply out $\prod_{k=1}^{5}\left(x - 2\cos(2\pi k/11)\right)$. The product’s coefficients come out as whole numbers — which they must, since the five values are the Galois conjugates of one algebraic number and the elementary symmetric functions of a full conjugate set are rational — and the figure checks each to within a part in a million before rounding. The result is

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0,$$

with $2\cos(2\pi/11) = 1.682507$ as a root to nine places.

Degree five needs the polynomial to be irreducible and that is the step to check. The rational root theorem gives the candidates: divisors of the constant term, so $\pm 1$. Neither is a root. A quintic with no rational root could still factor into a quadratic and a cubic, and ruling that out takes more — but for a prime $n$ the degree is known to be exactly $(n-1)/2$ by the theory of cyclotomic fields, and the figure checks the degree against that rather than attempting the factorisation.

Why the cubic instruments cannot reach it #

The reach of the conics is the numbers in towers of degree-two and degree-three steps, so their degrees are products of twos and threes — which is what the conics and the trisector share, and which puts the conics and the trisector at the same level. Five is prime and is neither, so $\cos(2\pi/11)$ is not in any such tower.

That is the whole argument and it depends on one fact: degrees multiply along a tower, which is the tower law. If $\cos(2\pi/11)$ lay in a tower of steps of degree two and three, its degree would divide the tower’s, which is a product of twos and threes; five divides no such product.

Putting the three together shows how narrow the gap is. Degrees three and six are reachable, degree five is not, and eleven is the first $n$ whose degree is prime and greater than three. The separating case is not exotic and it is not early either — every polygon below it is settled by the cubic level.

The three levels, as a strip #

Setting the eleven-gon in the strip makes its position clear, and the position is what this essay is about.

The compass column refuses 7, 9, 11, 13, 14, 18, 19, 21, 22, 23 and more. The conic column refuses 11, 22, 23 — the ones whose degree has a prime factor above three. So most of the compass column’s refusals are the conic column’s acceptances, and the few remaining are the separating cases for the next level up.

Eleven is the first of them and twenty-two is the second, for the obvious reason: doubling $n$ leaves the degree alone when $n$ is odd, since $\varphi(2n) = \varphi(n)$. So the separating polygons come in pairs, and the arithmetic of which $n$ are refused twice is the arithmetic of which $\varphi(n)/2$ have a prime factor above three.

Twenty-three is the next genuinely new one, at degree eleven. Its cosine needs a degree-eleven step or a tower reaching it, and Baragar’s bound says no single neusis step reaches eleven — so whether the twenty-three-gon is neusis-constructible is not settled by anything on this page, which is the shape of the open question below.

What the marked ruler does about it #

Benjamin and Snyder gave a neusis construction for the regular hendecagon in 2014. An earlier essay records it, and this one repeats one caution from it: the attribution is stated here from the literature and not checked against the paper, so a reader wanting the construction itself should go to the source; the same caution applies to Baragar’s bound, quoted below.

What the construction shows is that the marked ruler reaches a degree-five number, and therefore reaches strictly more than the conics. That is a statement about levels and it is worth separating from the construction: one polygon out of the cubic level’s reach and inside the mark’s is enough to order the two sets, and no further examples are needed for the ordering.

Baragar’s 2002 result, also as that earlier essay reports it, says every single step a marked ruler and compass can take has degree 2, 3, 5 or 6. Two things follow and they are different in kind. Degree five is available, which is consistent with the hendecagon. And degree seven is not available in one step, so a number of degree seven is out of reach unless it lies in a tower whose steps have those degrees — and whether it can is the open question.

Why a sliding mark reaches five at all #

The mechanism is worth outlining, because a device whose one operation solves a cubic and sometimes a quintic is a strange object.

A neusis step slides a segment of fixed length until its two ends lie on two stated curves. When the curves are a line and a circle, eliminating gives an equation of degree three or four, which is the account the marked ruler’s reach gives. When both curves are circles, or when the pole is placed so that the configuration is less symmetric, the elimination gives more: the conchoid the slide traces is a quartic, and a quartic met with a conic gives up to eight intersections, so the governing equation can have degree up to eight and its factors can include a quintic.

So the mark’s reach depends on which curves the segment is slid against, and the classical trisection uses the easiest case. That is the reason a single device has a reach nobody has characterised: the operation is really a family of operations indexed by the curves involved, and the family’s closure is what is in question.

A device that is one operation in one configuration and a family in general is the awkward kind, and an earlier essay’s framing anticipates it: every instrument set’s reach is a question about what its steps solve, and here the steps do not all solve the same thing.

The other neusis constructions, for scale #

It helps to have the easy case on the page, because the difference between it and the hendecagon is the whole of this essay.

Archimedes’ trisection slides a marked segment with one end on a line and the other on a circle. Eliminating gives a cubic, the construction is four steps, and the marked ruler’s reach verifies the arithmetic to six decimal places. Nicomedes’ duplication of the cube is the same shape: one slide, one cubic.

The hendecagon’s construction is not of that shape. It uses several slides, the configuration is chosen so that the governing equation factors in a particular way, and the resulting tower reaches degree five. A construction that needs a sequence of carefully arranged slides is a different kind of object from one that needs a single slide, and the reach of the device is about the sequences rather than the individual steps.

That is also why the easy cases mislead. A reader who has seen only the trisection has seen a device that solves cubics, and would reasonably conclude that its reach is the cubic level — which is the conics’ level, and is wrong. The device solves more, the extra comes from configurations nobody uses for trisection, and the hendecagon is where it shows.

What this does not establish #

It does not give the construction. Nothing here draws the hendecagon. The figures establish the degree, the irreducibility over the rationals to the extent the rational root test gives it, and the arithmetic that puts the polygon outside the cubic level; the construction is a published sequence of neusis steps and is not reproduced.

It does not characterise the mark’s reach. Degrees 2, 3, 5 and 6 per step bound what one step does. A tower of such steps has degree a product of those numbers, so the reachable degrees are contained in that set of products — and whether every such product is attained, and whether the reachable field is exactly the corresponding closure, is not known.

And it does not settle the general quintic. That earlier essay is careful about this: “the general quintic’s solution is not in general reachable”. Reaching one number of degree five is not reaching every one, because a degree-five number’s reachability depends on more than its degree — on the Galois group of its polynomial, and on whether that group has a chain of subgroups matching the available step degrees — which is the tower condition degrees multiplying makes precise.

The Galois group, which is what really decides #

The degree is the first thing to compute and it is not the whole story, and the section above hints at why. It is worth making the point properly, because it is where the subject’s difficulty actually lives.

A number is reachable by a tower of steps of stated degrees exactly when its polynomial’s Galois group has a chain of subgroups with indices among those degrees — a solvability condition of the kind Galois theory supplies, not a condition on the degree alone. Degree five is necessary for the hendecagon’s cosine to be out of the cubic level’s reach and it is not what makes it inside the marked ruler’s.

For the hendecagon the group is cyclic of order five, which is about as tame as a degree-five group gets, and that tameness is why a construction exists. A degree-five number whose group is the full symmetric group on five letters is a different matter: the group has no chain of the right shape, and no sequence of steps of degrees 2, 3, 5 and 6 reaches it — which is the general quintic’s unsolvability appearing in the language of instruments.

So “the marked ruler reaches degree five” is true and is not the statement anybody wants. The statement wanted is a condition on the Galois group, and what is known is a bound on the step degrees, which is strictly less information. That gap is the reason the level is uncharacterised.

The same distinction settles the classical cases cleanly and that is why they feel easier. A tower of quadratic steps forces the group to have order a power of two, and conversely; so for the classical reach the degree condition and the group condition coincide. They part company as soon as the step degrees are not all prime.

What the pictures cannot show #

The polynomial is built by multiplying out its roots numerically and rounding, which is a computation and not a symbolic derivation. What makes it evidence is the check that every coefficient is within a millionth of a whole number before rounding — a polynomial whose coefficients were not nearly integral would be reported rather than rounded into apparent tidiness.

The irreducibility is established by the rational root test and the cyclotomic degree formula together, and the second is quoted rather than proved. A figure establishing irreducibility from first principles would have to rule out a quadratic-times-cubic factorisation, which is a search over integer polynomials and is a different piece of algebra — the kind the reach table runs for the trisection’s cubic, where the degree is small enough for the rational root test to settle it outright.

And nothing here shows a neusis step of any kind. The marked-ruler essay draws one and verifies its arithmetic; the question on this page is which degrees such steps reach, and a degree is not a picture.

Folding, which reaches the cubic level and not this one #

Paper folding is the other instrument set that essay names, and it is worth asking where it sits, because the answer sharpens what is special about the mark.

A single fold can place two given points onto two given lines simultaneously, which is a cubic condition — the sixth of the Huzita–Hatori axioms. So folding reaches the cubic level, the same Pierpont primes, the same trisection and the same duplication of the cube. The marked ruler’s reach records the coincidence and is careful that it is a theorem about reach rather than a claim that the operations are alike.

Does folding reach the hendecagon? With one fold at a time, no: the axioms’ conditions are cubic, so the reachable degrees are products of twos and threes. Allowing simultaneous folds — several creases made in one motion, each satisfying its own alignment — changes the answer, and constructions of higher degree are known for multi-fold systems.

So the three instrument sets sort into two levels and the sorting is not by how physical they look. Compass and straightedge at the quadratic level; conics, trisector and single-fold origami at the cubic level; marked ruler and multi-fold origami above. A reader ordering them by apparent sophistication would put the conics at the top and the paper at the bottom, and would be wrong twice.

Still open: the exact reach of a sliding mark #

The characterisation is genuinely missing and it is worth being precise about what is and is not known.

Known: each step reaches a degree in $\{2, 3, 5, 6\}$, so the reachable numbers lie in towers whose step degrees are among those. Known: the hendecagon is reachable, so degree five genuinely occurs. Not known: whether every tower of such degrees is realised, and whether the reachable set has a description as clean as the conics’ products of twos and threes.

The obstruction is the one the section above names: the operation is a family, and different members reach different degrees. A characterisation would have to quantify over which curves the segment is slid against, and the natural ways of doing that either give too much or are hard to control.

There is a related question about how many marks. A ruler with two marks is the classical neusis; a ruler with three, or with two marks at an adjustable distance, is a different device, and whether it reaches more is not settled either. The subject has a clean answer at the cubic level and a list of partial results above it, which is the honest description of where it stands.

What one polygon settled #

Two instrument sets are ordered by exhibiting one thing in the reach of one and not the other, and the eleven-sided polygon does that for the marked ruler against the conics. One example, and a hierarchy acquires a level.

The ordering also completes the picture the conics and the trisector leave at two levels: quadratic, cubic, and something above whose extent is unknown. Three levels, two of them characterised.

The example had to be found and the arithmetic says where to look. A separating case needs a degree that is not a product of twos and threes, the smallest such degree is five, and $\varphi(n)/2 = 5$ means $n = 11$. So the search for the separating polygon is two lines of arithmetic, and finding the construction afterwards took until 2014.

That order — know what to look for, then look — is the one that essay’s closing sentence describes: “when a boundary moves, it is worth asking what the new operation solves that the old one did not”. Here the answer was a quintic, the quintic named a polygon, and the polygon was the thing somebody then had to construct.