Turning is uphill all the way

The sheet a polygon fits exactly found that every regular polygon of $4k + 2$ sides has one rectangle it touches on all four edges — one wide across its flats and $1/\cos(\pi/n)$ long across its corners — and that on it the polygon ranks exactly $(n - 2)/4$ among all regular polygons. The fourteen-gon, which only a fold builds, is third there; the twenty-two-gon, which needs two folds at once, fifth.

Two things in that result were measured rather than proved, and the essay said so. It followed each polygon’s rank across a sweep of sheets and found the best rank on the own sheet — but a sweep samples, and “the peak is exactly here” is a claim no sample can make. And it settled three of the four kinds of competitor by inequalities that hold for every $n$, while the fourth — a smaller polygon of the same kind, still turning on the bigger one’s sheet — was only checked up to forty-six sides. Both gaps were traced to the same missing piece: the turning polygon’s share, in closed form, for general $n$. The hexagon’s had been written down; nobody else’s had.

It can be written for all of them at once, and it turns out to be simpler than the hexagon’s looked.

Two widths, a quarter turn apart #

A regular polygon inside a rectangle fits exactly when its width across the sheet is no more than the sheet’s width and its width along the sheet is no more than the sheet’s length, because the smallest box round a convex shape held at a fixed angle has exactly those two widths as its sides. So the whole problem is the polygon’s width in two directions at right angles.

A regular $n$-gon of circumradius $R$ measured in a direction making an angle $t$ with its nearest corner is $2R\cos t$ wide, when $n$ is even: the corner in that direction and the corner opposite it both reach out $R\cos t$. As the direction turns, $t$ runs from nought, straight at a corner, to $\pi/n$, straight at the middle of an edge.

For a polygon of $4k + 2$ sides the direction at right angles is exactly half a step round the polygon from the first. A quarter turn is $\pi/2$, one step between corners is $2\pi/n$, and $\pi/2$ divided by $2\pi/n$ is $n/4 = k + \tfrac12$. So if the sheet’s width meets the polygon at $t$ from a corner, its length meets it at $\pi/n - t$ from one: one axis sees corners exactly where the other sees flats. That is the whole reason this kind of polygon has a sheet of its own, and the whole of the calculation.

On a sheet one wide and $h$ long the polygon must satisfy two conditions, $2R\cos t \le 1$ and $2R\cos(\pi/n - t) \le h$. The first allows a larger polygon as $t$ grows and the second a smaller one, so the largest polygon is the one at which both hold with equality. Dividing one by the other,

$$h\cos t = \cos(\pi/n - t), \qquad \tan t = \frac{h - \cos(\pi/n)}{\sin(\pi/n)}.$$

The share, in one line #

With $R = 1/(2\cos t)$ and the area of a regular polygon $\tfrac{n}{2}R^2\sin(2\pi/n)$, the share of the sheet is the area over $h$. Writing $c = \cos(\pi/n)$ and $s = \sin(\pi/n)$ and using $1/\cos^2 t = 1 + \tan^2 t$,

$$f(h) = \frac{n\,c}{4\,s}\cdot\frac{s^2 + (h - c)^2}{h}, \qquad 1 \le h \le \frac{1}{c}.$$

At $h = 1/c$ the tilt reaches $\pi/n$, the polygon lies flat against all four edges, and the share is $\tfrac{n}{4}\sin(\pi/n)$, which is the own-sheet share the earlier essay derived. Past it the polygon cannot use the extra length at all; it stays flat against the short edges and its share falls as $\tfrac{n}{4}\tan(\pi/n)/h$.

For the hexagon, $c = \sqrt{3}/2$ and $s = \tfrac12$, and the formula becomes $\tfrac{3\sqrt3}{8}\,(1 + (2h - \sqrt3)^2)/h$ — exactly the expression the crossing is as hard as the polygon derived for the turning hexagon and built its crossing proportions on. The general form was one substitution away.

It agrees with a direct search over rotations to a part in a billion, at seven sheets for each of the four polygons in the first figure, and that agreement is required rather than admired: the figure is refused if any sample differs.

The slope is the same for every polygon #

Differentiate the part that depends on $h$:

$$\frac{d}{dh}\,\frac{s^2 + (h - c)^2}{h} = \frac{2(h - c)h - s^2 - (h - c)^2}{h^2} = \frac{h^2 - (s^2 + c^2)}{h^2} = \frac{h^2 - 1}{h^2}.$$

Every trace of the polygon has cancelled. The slope of every turning polygon’s share is proportional to $h^2 - 1$, whatever its number of sides. It is zero on the square, positive on every longer sheet, and so the share rises the whole way from the square to the own sheet — where it stops turning and the formula hands over to one that falls.

That settles the first question outright. The peak is exactly at the own sheet, for every polygon of $4k + 2$ sides and not merely for the ones swept, and it is a corner rather than a smooth top: the rising branch meets the falling one at $1/\cos(\pi/n)$ with the first still climbing. A sheet a hair shorter than the own sheet costs a turning polygon share; a sheet a hair longer costs it share too, in the other way.

It also says something about the square the census had only half noticed. The square is a stationary point of every one of these curves — the share is level there, with nothing gained by the first fraction of a per cent of extra length. That is the symmetry reappearing as a derivative. On a square the polygon is tilted exactly halfway, at $\pi/(2n)$, so that corners and flats meet the two axes evenly, and stretching the sheet either way is the same first move.

The gain is the average of one and the sheet #

Divide the share on the own sheet by the share on the square. At $h = 1$ the formula gives $f(1) = n\,c\,(1 - c)/(2s)$, and at $h = 1/c$ it gives $\tfrac{n}{4}s$. Their ratio is

$$\frac{s^2}{2c(1 - c)} = \frac{(1 - c)(1 + c)}{2c(1 - c)} = \frac{1 + 1/c}{2}.$$

On its own sheet a polygon of $4k + 2$ sides uses exactly the average of one and the sheet’s length times the share it used on a square. The hexagon’s own sheet is $2/\sqrt3 \approx 1.1547$ long and its gain is 7.7 per cent; the decagon’s sheet is 1.0515 long and it gains 2.6; the fourteen-gon’s, 1.0257, and 1.3; the twenty-two-gon’s, 1.0103, and half a per cent. Each gain is half the sheet’s extra length.

That is a strange thing for a maximisation to produce, and the drawings show why it is not a coincidence.

On the square the polygon’s two widths are equal, and each is the width at the halfway tilt. On its own sheet one width is across the flats and the other across the corners, in the ratio $1/c$. The area is the same function of the circumradius in both cases, and the circumradius is fixed by the narrower of the two widths — by $\cos(\pi/2n)$ on the square and by $\cos(\pi/n)$ on the own sheet. The share divides by a sheet $1/c$ long instead of one. Everything else is the half-angle identity $1 + \cos x = 2\cos^2(x/2)$, which turns the ratio of the two circumradii squared into $(1 + c)/(2c^2)$, and dividing by the extra length $1/c$ leaves $(1 + c)/(2c)$. Half the extra length is paid back as extra polygon, and half is lost as empty paper.

Where it passes the circle #

The circle is the limit every one of these polygons approaches, and the closed form says where each of them overtakes it.

A circle in a sheet one wide has diameter one whatever the length, so its share is $\pi/(4h)$ and falls from the square on. On the square every polygon of $4k + 2$ sides uses less than the circle: $f(1) = n\,c\,(1 - c)/(2s)$ is 69.6 per cent for the hexagon and 76.9 for the fourteen-gon, against the circle’s 78.5, because a polygon squeezed into the circle’s width, tilted halfway, gives up more paper at its flats than its corners gain outside the circle. On its own sheet every one of them uses more, since the earlier essay showed the ratio there is $\tan x/x$ with $x = \pi/n$. So each turning polygon passes the circle somewhere on the way.

Setting $f(h) = \pi/(4h)$, the $h$ in each denominator cancels, and what is left is a quadratic in $h$ alone:

$$(h - c)^2 = \frac{\pi s}{n c} - s^2, \qquad h_\times = c + \sqrt{\frac{\pi s}{n c} - s^2}.$$

The hexagon passes the circle at a sheet 1.0947 long, sixty-one per cent of the way from the square to its own sheet at 1.1547. The decagon passes at 1.0322, sixty-three per cent of the way; the fourteen-gon at 1.0162 and the twenty-two-gon at 1.0065, both sixty-three per cent. The share of the way does not wander; it settles, and expanding both sides in $x = \pi/n$ gives its limit exactly. The own sheet is $1 + x^2/2$ long to leading order and the crossing is at $1 + (\sqrt{2/3} - \tfrac12)\,x^2$, so the share of the way tends to

$$2\sqrt{2/3} - 1 \approx 0.633.$$

A polygon of a thousand sides passes the circle 63.3 per cent of the way to a sheet five millionths longer than a square, and the hexagon passes it 61.2 per cent of the way to a sheet fifteen per cent longer. The geometry is the same at every scale; only the stretch differs, which is the tilt figure’s observation — the same curve stretched — made quantitative.

It is also a cleaner version of the fact every even polygon beats every odd one found on long sheets, where every even polygon’s constant sits above the circle’s and every odd one’s below. On a long sheet the even polygons are above the circle for good; near the square the polygons of $4k + 2$ sides are below it, and the crossing at $h_\times$ is where each one moves from the one side to the other.

The step the rank formula was missing #

The rank $(n - 2)/4$ counts the polygons above an $n$-gon on its own sheet: the multiples of four from eight to $n - 2$, and nothing else. The earlier essay proved that the multiples of four with fewer sides are above and those with more are below, that the circle and every odd polygon are below, and that a polygon of $4k + 2$ sides with more sides is below, because on this sheet it already lies flat with a smaller constant. What it could not prove was the last kind: a polygon of $4k + 2$ sides with fewer sides, whose own sheet is longer, so that on the $n$-gon’s sheet it is still turning.

The closed form proves it in two links.

First link: a turning polygon is below its own peak. The $m$-gon’s share rises all the way to its own sheet, and the $n$-gon’s sheet is shorter than that, so on it the $m$-gon uses less than $\tfrac{m}{4}\sin(\pi/m)$.

Second link: the peaks rise with the number of sides. The own-sheet share $\tfrac{m}{4}\sin(\pi/m)$ is $\tfrac{\pi}{4}$ times $\sin x/x$ with $x = \pi/m$, and $\sin x/x$ grows as $x$ shrinks, so a polygon with fewer sides has a lower peak than one with more.

Put together: on the $n$-gon’s own sheet the $m$-gon uses less than its own peak, which is less than the $n$-gon’s peak, which is what the $n$-gon uses there. No smaller polygon of the same kind passes it, for any $n$, and $(n - 2)/4$ is a theorem rather than a measurement to forty-six sides.

The figure draws the chain for the twenty-two-gon. The hexagon on the twenty-two-gon’s sheet is barely turned and uses 69.6 per cent; its own peak, on a sheet 1.1547 long, is 75. The eighteen-gon is almost at its own sheet already, 1.0154 against 1.0103, and its two dots nearly touch. All four peaks are left of the twenty-two-gon’s 78.27 per cent, which is the second link, and every turning share is left of its own peak, which is the first.

The table is the earlier essay’s, and nothing in it has moved; what has changed is its status. Every row’s rank was measured against a census that stops at sixty sides and was argued for the competitors that census could not exhaust, and the argument had one gap. With the gap closed, the right-hand column is the theorem and the measured column is its check.

What the closed form says about tools #

The rank the formula now guarantees is what the earlier census read as a verdict on tools, and the biggest one that can also be folded first set the two columns side by side on a square. A fold’s best place among regular polygons is third, taken by the fourteen-gon on its own sheet, because fourteen is the first number of $4k + 2$ sides with a heptagon in it, and the heptagon a compass cannot reach is the first polygon a fold builds that a compass does not. Two folds at once buy fifth, with the twenty-two-gon, because eleven is the first prime a single fold cannot reach. Which polygons each tool reaches at all is the count how many polygons a fold reaches makes; the rank is what the first of the right shape can do with it. Those were measured on sheets up to forty-eight sides; they are now proved on every sheet, since the proof above has no census in it.

The own sheets themselves remain the numbers the crossing essay found them to be. $1/\cos(\pi/14)$ has degree six over the rationals with a three in it, so only a fold can mark the fourteen-gon’s sheet; $1/\cos(\pi/22)$ has degree ten, with a five, and needs two folds at once. The sheet on which a tool’s best polygon does best is a sheet only that tool can measure out, and the closed form adds that it is the one sheet on which that polygon does best — not approximately, but exactly.

What the formula cannot say #

It is about polygons of $4k + 2$ sides only. For a multiple of four the two axes see the same thing, corners or flats together, so the polygon is best flat on a square and falls as $1/h$ from the start; that is the easy case and the earlier essay settled it. For an odd polygon the two sides of each width are not symmetric — opposite a corner is a flat — and the width is $2R\cos(\pi/2n)\cos(t - \pi/2n)$ rather than $2R\cos t$, with the axis at right angles offset by a quarter of a step rather than a half. The same method gives an odd polygon’s share in closed form, and it is not done here.

It assumes the polygon is regular and the sheet exact. A sheet cut to 1.0257 by any real means is near the fourteen-gon’s own sheet and not on it. Because the peak is a corner rather than a rounded top, the loss from missing it is first order in the error: a sheet a thousandth too long costs the fourteen-gon about a thousandth of its share, where a smooth maximum would cost a millionth.

And a larger polygon is not a better model. The share of the sheet is one objective among several a designer might have, and the square is in the answer already showed how completely the winner depends on the sheet asked about; this essay settles one contest exactly and says nothing about whether it is the contest anybody should be running.

The sheet the calculation stands on #

The sheet is a rectangle one wide and $h$ long, $h$ at least one, and every polygon is regular, placed with its centre wherever it fits and turned to whatever angle makes it largest. Holding the width at one loses nothing, since a sheet twice the size holds a polygon twice the size and the share is unchanged.

A polygon fits when both its widths fit. This is exact for a convex shape at a fixed angle, and it is what lets the whole problem reduce to two inequalities in one angle; for a shape that is not convex it would not be enough.

And the tilt is measured from a corner. A tilt of nought puts a corner against the short edges and a flat against the long ones; $\pi/n$ reverses them. Turning past $\pi/n$ repeats the same shapes mirrored, since the polygon is unchanged by a step of $2\pi/n$.

How the formula was checked #

The closed-form share is required to agree with a direct search over rotations to a part in a billion, at seven sheets from the square to 1.2 and at the own sheet, for the hexagon, decagon, fourteen-gon and twenty-two-gon, and the figure is refused if any sample does not.

The gain from the square to the own sheet is required to equal $(1 + 1/c)/2$ for each of the four polygons, to twelve decimal places, which checks the algebra above rather than a reading of the curve. The tilt the search settles on is required to be the tilt the formula gives, on four sheets for each polygon. And the two links of the rank argument are checked on the twenty-two-gon’s sheet, for every smaller polygon of its kind, before the chain is drawn.

Still open: the odd polygons’ sheets #

The formula answers the question the earlier essay left for polygons of $4k + 2$ sides, and it makes the next question sharp. An odd polygon’s two widths are a quarter of a step apart rather than half, so its share while it turns has the same kind of closed form with a different offset — and whether an odd polygon has a sheet on which its share peaks, and what it ranks there, is now a calculation rather than a sweep. The circle bound suggests the rank will be poor, since an odd polygon’s constant is below the circle’s; whether some odd polygon has a sheet on which it beats every odd polygon with more sides, and by how much, is what the calculation would say.

The other direction is the corner at the peak. A share whose maximum is a corner is one whose best sheet is a tolerance problem as much as an optimum, and exact is not accurate measured what a folder’s hands do to an exact construction. Marking a polygon’s own sheet by folding and then folding the polygon on it compounds two errors, and whether the rank survives the first is a question with a millimetre in it.

Sideways from here, the cancellation is worth recognising elsewhere. A family of optima whose slopes all reduce to the same expression is a family with a hidden common structure, and here it was the half-angle identity, surfacing as the gain $(1 + L)/2$. The largest triangle in a square found its tilt of fifteen degrees from a quadratic; whether that fifteen degrees is the same cancellation in a polygon with no opposite sides is a question the formula above invites.

The habit worth carrying is about sweeps that find a peak. When a sweep finds the best value at a special setting, look for a derivative that vanishes or changes sign there for a reason. Here the reason was a slope proportional to $h^2 - 1$, the same for every polygon, and finding it turned a peak located to the resolution of a sweep into one located exactly, and a rank checked to forty-six sides into one proved for all of them.