The crossing is as hard as the polygon

Every even polygon beats every odd one found that the ranking of regular polygons on a rectangular sheet freezes once the sheet is 1.1284 times as long as it is wide, and that everything past that point is decided by a constant of each polygon. It left one number unexplained, and it said so. The proportion 1.1284 came out of a search, the formula that seemed to give it was a thousandth out, and the question of which tool could mark it on paper was left open.

The number has a closed form, and so does every crossing of its kind. More interesting than the form is what it is made of. The proportion at which the hexagon overtakes a polygon is exactly as hard to mark as that polygon is to build. The octagon’s crossing is a compass number. The heptagon’s crossing needs a fold. The hendecagon’s needs two folds made at once. An optimisation that has nothing to do with construction hands back numbers that sort themselves by tool.

Why the obvious formula was a thousandth out #

On a long sheet a hexagon lies with two corners pointing along the length and two flat sides against the long edges. Its width across the flats is $\sqrt{3}R$ for a circumradius $R$, so the largest one has $R = 1/\sqrt{3}$ on a sheet of width one, it needs a length of $2R = 2/\sqrt{3} \approx 1.1547$, and its share of a sheet $h$ long is $(\sqrt{3}/2)/h$. That formula is exact on every sheet at least 1.1547 long.

The octagon on the same sheets sits with four flat sides against all four edges of a square and gets nothing from extra length, so its share is $0.8284/h$. Setting the two equal gives no crossing at all, because both fall as one over the length. The crossing must therefore happen on a sheet shorter than 1.1547, where the hexagon is too long to lie flat and has to do something else.

What it does is turn. On a sheet a little longer than a square, the largest hexagon is pressed against all four edges at once. Write $\delta$ for the angle between one of its corners and the sheet’s width. Across the width it measures $2R\cos\delta$, and across the length, a quarter turn away, it measures $2R\cos(30^\circ - \delta)$. Both edges bind, so

$$2R\cos\delta = 1, \qquad 2R\cos(30^\circ - \delta) = h.$$

Expanding the second cosine and using the first to replace $2R\cos\delta$ gives

$$h = \frac{\sqrt{3}}{2} + \frac{1}{2}\sqrt{4R^2 - 1},$$

which is a sheet length as a function of the hexagon’s size. It meets the lying-flat case exactly where it should: at $\delta = 30^\circ$ the first condition gives $R = 1/\sqrt{3}$ and the length becomes $2/\sqrt{3}$. Inverting it, the hexagon’s share of a sheet shorter than that is

$$\frac{3\sqrt{3}}{8}\cdot\frac{1 + (2h - \sqrt{3})^2}{h}.$$

The obvious formula was a thousandth out because it assumed the hexagon lying flat on a sheet too short for it.

The second figure checks that against the search. At twenty-one proportions from a square to 1.2 the searched share of the hexagon agrees with the closed form to the precision of the arithmetic, on both sides of 1.1547, so the turning hexagon is the right picture and not merely a picture that fits.

Every crossing of the hexagon has the same form #

The polygons the hexagon overtakes are all held by the short side alone by the time it reaches them, which is the regime where the constant of each polygon decides everything. A polygon with $n$ sides uses $K_n/h$ of the sheet, with

$$K_n = \frac{n}{4}\tan\frac{\pi}{n}$$

for even $n$ and $\tfrac{n}{2}\sin\tfrac{2\pi}{n}\,/\,(1+\cos\tfrac{\pi}{n})^2$ for odd. Setting the turning hexagon’s share equal to $K_n/h$ cancels the length from the denominators and leaves a quadratic, whose root is

$$h_n = \frac{\sqrt{3}}{2} + \frac{1}{2}\sqrt{\frac{8K_n}{3\sqrt{3}} - 1}.$$

For the octagon $K_8 = 2(\sqrt{2} - 1)$, and the crossing is

$$h_8 = \frac{\sqrt{3}}{2} + \frac{1}{2}\sqrt{\frac{16(\sqrt{2}-1)}{3\sqrt{3}} - 1} = 1.1284406\ldots$$

which is the number that was searched for. The measurement was a number all along, and it is built from $\sqrt{2}$, $\sqrt{3}$ and one further square root.

The same formula covers every polygon the hexagon passes. On the first figure the searched crossing and the closed form agree to a ten-millionth for all ten polygons from seven sides to sixteen, and the table carries on without trouble to twenty-three. That is eighteen of the fifty-five crossings that happen anywhere among the polygons up to twenty-four sides between a square and a sheet 1.1284 long, all in one line of algebra.

What each number is made of #

A closed form says what a number is built from, and that decides which tools can mark it — the same question the numbers a fold reaches asks of lengths and folding beats the compass answers for single folds. A straightedge and compass reach numbers built by square roots, so a compass number has a degree over the rationals that is a power of two. A single fold also solves cubics, so its numbers have degrees made of twos and threes. Two folds made at once reach the quintic, which lets fives in.

The crossing $h_n$ lives in a field built from $\sqrt{3}$, from the constant $K_n$, and from one more square root. The square roots can only contribute factors of two. So every odd prime in the degree of $h_n$ comes from $K_n$, and $K_n$ is a trigonometric number of the angle $\pi/n$ — the same angle whose arithmetic decides whether a regular $n$-gon can be built at all.

The degree column of the first figure counts it. For each polygon, the number $h_n$ has a finite set of conjugates, obtained by replacing $\pi/n$ with the other angles $k\pi/n$ that the cyclotomic arithmetic allows and taking both signs of each square root, and the degree is how many different values that produces. Counted that way, and checked independently against exact minimal polynomials for every polygon from seven to twenty sides:

• the octagon’s crossing has degree 8, with no odd prime in it;
• the heptagon’s has degree 24, which is eight times three;
• the nonagon’s has degree 12, which is four times three;
• the hendecagon’s has degree 40, which is eight times five;
• the fifteen-gon’s has degree 16, and the seventeen-gon’s 64, both powers of two.

And the odd primes are, on every row, exactly the odd primes of the polygon’s totient — the count of whole numbers below $n$ sharing no factor with it, which is the number the heptagon a compass cannot reach found decides every tool’s verdict. Seven has a totient of six, with a three in it, and its crossing has a three in its degree. Eleven has a totient of ten, with a five, and so does its crossing. Fifteen and seventeen have totients of eight and sixteen, and their crossings have no odd prime at all, although both have far more sides than the heptagon.

The heptagon’s proportion cannot be drawn with a compass #

The consequence is best seen on one sheet. A sheet 1.069643 times as long as it is wide is the one on which the largest hexagon and the largest heptagon use the same share of the paper — both about 70.8 per cent. The degree of that proportion is twenty-four, and twenty-four has a factor of three.

No straightedge and compass can mark that length. A compass reaches only numbers of degree a power of two, and a factor of three cannot be removed. So the sheet on which the hexagon and the heptagon tie cannot be cut to its exact proportion by the classical tools, for the same reason that the heptagon itself cannot be drawn by them — and a fold, which reaches cube roots, can in principle mark it, as it can build the heptagon. The two impossibilities are one impossibility met twice.

Nothing in the optimisation asked for that. The share curves are two maximisations over rotation and position, and the crossing is the root of a quadratic in the sheet’s length. What carries the heptagon’s arithmetic into the root is the constant $K_7$, which records how much of a strip a heptagon can use — and that is a statement about the heptagon’s shape, which is a statement about $\cos(\pi/7)$, which is where the three was all along.

The hendecagon’s needs two folds #

The same reasoning one prime further up gives a sheet no single fold can mark. The hexagon and the hendecagon tie at a proportion of 1.085296, whose degree is forty. Forty is eight times five, and a five is what the eleven-sided one nobody can fold found no arrangement of cubics produces.

So a folder with the seven single-fold operations cannot mark this sheet exactly, and one allowed two creases at once, which solves the quintic, can. The twenty-two-gon’s crossing, at 1.099203, has the same degree and the same prime, as the first figure’s extension to twenty-three sides shows; the twenty-three-gon’s, at 1.092640, has degree eighty-eight, with an eleven in it, beyond two folds as well.

The order is a sequence of tools laid along a strip of proportions between 1.07 and 1.13, and it is not an order of difficulty. As the sheet lengthens the hexagon passes the heptagon, which needs a fold, then the nonagon, which needs a fold, then the hendecagon, which needs two, then the thirteen-gon, which needs a fold again, then the fifteen-gon, which a compass reaches — and last of all the octagon, which a compass reaches too. Where a crossing falls says nothing about how hard it is to mark; what the polygon is says everything.

Twenty-four sides, to see the pattern hold #

The larger polygons repeat the pattern and add the one case it would be easy to guess wrong. The seventeen-gon has seventeen sides and a crossing of degree sixty-four — a large number with no odd prime in it — because its totient is sixteen and Gauss’s polygon is a compass polygon. Its crossing proportion, 1.090881, can be marked by a compass however many square roots it takes. The twenty-three-gon has only six more sides and a crossing of degree eighty-eight, with an eleven that neither a compass, a single fold nor a pair of simultaneous folds can supply.

Degree is not difficulty in the sense of size. Sixty-four is a larger degree than twenty-four, and the seventeen-gon’s crossing is easier to mark than the heptagon’s, because what a tool cannot do is extract a root of the wrong prime order, and sixty-four is made only of twos. The same point was made about polygons themselves in twos and threes run out, where the reachable degrees are the lattice of numbers built from two and three; this is that lattice arriving by a different road.

Why the prime survives the optimisation #

It would be reasonable to expect an optimisation to scramble arithmetic. A maximisation over a rotation, followed by a comparison of two maxima, followed by a root of a quadratic, has plenty of room to produce a number unrelated to anything.

It does not, for a structural reason that is worth stating plainly. Every step after the constant is a square root or a rational operation. The turning hexagon introduces $\sqrt{3}$ and one square root; the comparison introduces nothing; the quadratic introduces one more square root. Square roots can only multiply a degree by two. So the odd part of the degree is fixed the moment $K_n$ is written down, and $K_n$ is the polygon’s area divided by the square of its least width, both of which are polynomials in the cosine and sine of $\pi/n$.

That is also why the pattern is special to crossings with the hexagon. The hexagon’s own contribution is $\sqrt{3}$, which is a compass number, so it adds nothing odd. A crossing between the heptagon and the nonagon would carry threes from both, and one between the heptagon and the hendecagon would carry a three and a five; a crossing between two polygons that are both turning would pass through a different quadratic, or a quartic. The hexagon is a clean instrument because it is a compass polygon, and it reads out the other polygon’s arithmetic unmixed.

What is being assumed #

The sheet and the polygons are exact. The proportions are real numbers and the polygons are regular; a sheet cut to 1.0696 by any real process is a sheet near the crossing, not at it, and at the crossing the two shares agree only to the tolerance of the cutting.

The tools are ideal. A compass number is one a perfect compass reaches in finitely many steps and a fold number one a perfect fold reaches, which says nothing about how many steps it takes or how well they are conditioned — what buys the reach costs the accuracy is the reminder that the operation reaching further is also the one crossing its creases most shallowly.

The degree test is used in its classical sense. A degree with an odd prime in it rules a tool out. A degree built only of twos and threes does not by itself rule one in, but for numbers built as these are, from cyclotomic arithmetic by square roots, the degree is the whole story, which is the same reasoning Gauss and Wantzel applied to the polygons.

What the table cannot show #

It does not give a construction. A compass can mark the octagon’s crossing, and nothing here supplies the sequence; it would begin by constructing $\sqrt{2}$ and $\sqrt{3}$ and continue through two nested square roots, and it would be long.

It covers the crossings with the hexagon and no others. Fifty-five crossings happen among the polygons up to twenty-four sides between a square and 1.1284, and eighteen of them involve the hexagon. The decagon’s crossings, between 1.02 and 1.05, follow from the turning decagon in the same way, with a different quadratic; the rest are not derived here, and the claim that their degrees also carry both polygons’ primes is a prediction rather than a finding.

The degrees are counted, not proved. The count of distinct conjugates matches exact minimal polynomials on every polygon checked, and a count can in principle be twice the true degree when a square root already lies in the field below it; that would change a power of two and never an odd prime, which is the only part of the degree any argument above uses.

And the ranking is still a ranking of largest polygons. None of this says a folder should prefer a sheet 1.0696 long, only that on that sheet two particular maxima coincide.

A number from a different kind of question #

The largest triangle in a square found the optimum tilted by exactly fifteen degrees and its area a quadratic number, and concluded that a compass reaches that optimum and folding is not needed. The biggest one that can also be folded found the best polygon and the best foldable polygon agreeing on a square. Both kept the two kinds of question apart: what is largest, and what can be built.

The crossings do not keep them apart. An optimisation over sheets produces boundaries, and the boundaries are constructible numbers of exactly the difficulty of the shapes on either side. That is a stronger statement than agreement between two lists; it says that a geometric extremum inherits its polygon’s field, and it says it in a form that can be checked on every polygon by counting conjugates.

It also answers the question the earlier census asked about 1.1284 in the most ordinary way. The proportion is algebraic, of degree eight, and a compass reaches it. The pleasing symmetry it hoped for is there, but not where it was looked for: it is not that the winner’s crossing needs the winner’s tool, but that every crossing needs the tool of the polygon being passed.

Still open: whether a rank can be bought #

The crossings say where polygons change places, and they point at a question about the places themselves. On the seven sheets tabulated earlier the best polygon only a fold can build places fourth at best, and the twenty-two-gon, which needs two folds, places ninth on every long sheet.

Between those sheets the ranking moves, and a polygon that moves up somewhere between them might reach a better place than any named sheet shows. The turning hexagon suggests where to look: a polygon of $4k + 2$ sides has corners along one axis and flats along the other, so a sheet exactly $1/\cos(\pi/n)$ long fits it against all four edges at once, as the hexagon is fitted at $2/\sqrt{3}$. On that sheet it should do unusually well, and the rank it reaches there is a count that could be made.

Sideways from here, the degrees suggest a second use for the same instrument. A sheet proportion that a compass cannot mark is a proportion no amount of halving, doubling and square-root sheets — including the rectangle that keeps its shape — will ever produce, and the crossings give an explicit list of such proportions, each attached to a polygon. Whether any sheet a folder actually buys sits at one of them is a question with an easy answer and an interesting reason.

The habit worth carrying is about where arithmetic hides. When an optimisation returns a number, ask which operations produced it and which of them could have changed its degree. If only square roots and rational steps lie between a constant and the answer, the answer carries the constant’s odd primes intact, and a question about what is largest turns out to have been a question about what can be built.