Every even polygon beats every odd one

The square is in the answer runs the census of the largest regular polygon a sheet holds on rectangles as well as on a square, and finds the familiar result is a fact about the square. On a square the octagon uses 82.8 per cent of the paper and beats every polygon but the square itself; on every rectangle tried the hexagon leads, and the even-sided polygons reverse their order.

It ends with a check it did not run. The biggest one that can also be folded had crossed the square’s census with which polygons a fold can construct, and found that the best polygon and the best foldable polygon agree. If the octagon’s lead is a fact about the square, the agreement might be too, and on a rectangle a polygon only a fold can build might rise to the top where a compass polygon was.

It does not. On every proportion from a square to three to one, the best polygon is one a compass already builds. But running the crossing across proportions turns up the thing that makes the agreement unsurprising, and it is a cleaner result than either census found: past a proportion of about 1.128, the ranking of every regular polygon stops moving, and what ranks it is whether the polygon has an even or an odd number of sides.

Three tools, one verdict #

Which regular polygons a tool builds is a condition on the totient of the number of sides — how many whole numbers below it share no factor with it. A compass and straightedge build the n-gon when the totient is a power of two, which is the condition Gauss found and Wantzel proved necessary. A fold, which also solves cubics, builds it when the totient has no prime factor above three, which is what the heptagon a compass cannot reach worked through. Between three and twenty-four sides, the polygons only a fold builds are those with 7, 9, 13, 14, 18, 19 and 21 sides; the ones no single fold builds are 11, 22 and 23, the first of them the eleven-sided one nobody can fold.

The first figure puts those verdicts beside the census on seven sheets. On the square the best polygon after the square is the octagon, at 82.8 per cent; on an eleven-by-ten sheet it is still the octagon, at 75.3; on six by five, the A series, three by two, two by one and three by one it is the hexagon. The octagon and the hexagon both have totients that are powers of two, so a compass builds both, and the best polygon a fold builds on each sheet is the best polygon.

The agreement is not a coincidence of the square. It holds on every sheet, and it holds for the simplest reason available: the polygons that win are six- and eight-sided, and a compass reaches them.

Folding’s extra reach — four times as many polygons as a compass below a thousand sides — shows further down. The best polygon only a fold can build is the fourteen-gon on every rectangle, fourth on eleven by ten and fifth everywhere from six by five to three by one. On the square it is the twenty-one-gon, seventh, at 77.9 per cent against the octagon’s 82.8. The hendecagon family — the polygons no single fold builds — does worse still: the twenty-two-gon ninth on every rectangle, the twenty-three-gon sixth on the square.

Where the octagon’s lead ends #

The census on six by five, the A series and three by two was already enough to show the hexagon leading off the square. The eleven-by-ten sheet in the first figure says the change does not happen immediately. The second figure follows the shares continuously from a square to a sheet 1.4 times as long as it is wide.

The octagon’s share falls from the first moment the sheet lengthens, as $0.8284/h$ for a sheet one wide and $h$ long: an octagon fits a square exactly with its flat sides against all four edges, and extra length gives it nothing to use. The hexagon’s share rises at first, because on a square it is held by the sheet’s height as well as its width, and extra length lets it grow. The two meet at a proportion of 1.1284, where each uses 73.4 per cent of the sheet. Past that point the hexagon leads.

So the earlier census’s finding needs a boundary. It is true that on the rectangles it tried the hexagon leads and the even polygons reverse their order; it is not true of every rectangle. A rectangle within about an eighth of a square still ranks the octagon first, and the eleven-by-ten sheet is one.

The ranking stops moving #

What happens past 1.1284 is the more surprising part. The census at 1.1284, at 1.5, at 2 and at 3 gives not merely the same best polygon but the same order of all twenty-one polygons from three sides to twenty-four: six, eight, ten, twelve and so on up to twenty-four, then twenty-three, twenty-one, nineteen and so on down to three.

The reason is that on a long enough sheet a polygon touches only the two long sides. The sheet’s length stops mattering, the largest polygon is the one whose least width equals the sheet’s width, and its share of the sheet is its area divided by the sheet’s area. With the width fixed at one, that is a constant of the polygon divided by the length. Every polygon’s share falls as one over the length at the same rate, so nothing overtakes anything, and the order is set once and for all by the constants.

The constant is the polygon’s area over the square of its least width, and it has a closed form. A regular polygon with an even number of sides has parallel opposite sides, its least width is the distance between two of them, and the constant is

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

which is $\sqrt{3}/2 \approx 0.866$ for the hexagon and $0.828$ for the octagon, and which falls toward $\pi/4$ as $n$ grows. A regular polygon with an odd number of sides has no parallel sides; its least width runs from a vertex to the opposite side, and the constant is

$$\frac{\frac{n}{2}\sin\frac{2\pi}{n}}{\left(1+\cos\frac{\pi}{n}\right)^2}$$

which is $0.577$ for the triangle and $0.727$ for the pentagon, and which rises toward $\pi/4$.

The fourth figure measures those constants on a sheet three times as long as it is wide and compares them with the two formulas; they agree to the precision of the search. And the two families sit on opposite sides of the same line. Every even polygon’s constant is above $\pi/4$ and every odd polygon’s is below it, so on a long sheet every even polygon beats every odd one — the twenty-four-gon beats the twenty-three-gon, the fourteen-gon beats the heptagon, and the circle, which is what both families approach, is beaten by every even polygon and beats every odd one.

That is why the even-sided polygons reversed their order off the square. On a square an even polygon uses both widths, and those with fourfold symmetry fit best; on a long sheet only one width matters and the formula for even polygons, which decreases with $n$, orders them from six upward.

Why the two families sit on opposite sides #

The two constants can be derived in a few lines, and the derivation says more than the formulas: it says how fast each family approaches the circle, and that the even polygons approach twice as fast.

Take a regular polygon with circumradius $R$ and write $x = \pi/n$ for half the angle each side subtends at the centre. Its area is $n$ triangles from the centre, $\tfrac{n}{2}R^2\sin 2x$. An even polygon’s least width is twice its apothem, the distance from the centre to a side, which is $2R\cos x$. So its constant is

$\frac{\tfrac{n}{2}R^2\sin 2x}{4R^2\cos^2 x} = \frac{n}{4}\tan x = \frac{\pi}{4}\cdot\frac{\tan x}{x}.$

An odd polygon’s least width runs from a vertex to the side opposite it, which is the circumradius plus the apothem, $R(1+\cos x)$, and its constant is

$\frac{\tfrac{n}{2}\sin 2x}{(1+\cos x)^2} = \frac{\pi}{4}\cdot\frac{\sin 2x}{2x}\cdot\frac{4}{(1+\cos x)^2}.$

Both are $\pi/4$ times a correction that tends to one as the polygon gains sides. For the even family the correction is $\tan x/x$, which is always greater than one, so every even polygon sits above the circle; for small $x$ it is about $1 + x^2/3$. For the odd family the two factors pull in opposite directions — $\sin 2x/2x$ is below one and $4/(1+\cos x)^2$ is above it — and their product, for small $x$, is about $1 - x^2/6$: below one, so every odd polygon sits below the circle.

The two corrections have the same shape and different sizes. An even polygon is above the circle by a third of $x^2$ and an odd polygon below it by a sixth, so the even family closes on $\pi/4$ from above twice as fast as the odd family closes from below. An even polygon with $n$ sides is as far above the circle as an odd one with about $n/\sqrt{2}$ sides is below it: the twenty-gon’s 0.7919 is 0.0065 above $\pi/4$, and the fifteen-gon’s 0.7796 is 0.0058 below.

That is also why the fourteen-gon, the first even polygon only a fold builds, cannot do better than fifth. By fourteen sides the even correction has fallen to within two per cent of one, and the four even polygons with fewer sides — six, eight, ten, twelve — are each further above the circle than it is. A fold’s first even polygon arrives after most of the even family’s advantage has been spent.

Where folding’s polygons land #

The ranking puts the polygons only a fold builds in fixed places, and the places are unflattering.

Among even polygons the first that a compass cannot build is the fourteen-gon: its totient is six, which has a factor of three. It ranks fifth, after six, eight, ten and twelve, all of which a compass builds. The eighteen-gon, also fold-only, is seventh. The odd polygons only a fold builds — twenty-one, nineteen, thirteen, nine and seven — sit in the bottom half, below every even polygon, because they are odd.

So the answer to the question the earlier census posed is sharper than yes or no. The agreement between the best polygon and the best foldable polygon holds on every sheet; and the reason it holds is the parity result. The polygons that fit a strip best are the even ones with few sides, and the even polygons with few sides are exactly the ones a compass already reaches. The first totient with a factor of three among even polygons belongs to the fourteen-gon, and by fourteen sides the constant has come most of the way down to $\pi/4$.

The one sheet where a fold-only polygon places fourth is eleven by ten, inside the octagon’s region: there the order is still the square’s, only partly converted, and the fourteen-gon slips in above the hexagon, which is still held back by the sheet’s height.

The square is the exception to the rule it taught #

The square was where all of this started, and the square is the one sheet on which the parity rule does not apply — one more way the square is a choice with a price.

On a square an even polygon can use both of the sheet’s widths at once only if its own symmetry lines up with the square’s, and the census on the square ranks the polygons by that alignment instead: the octagon, the twelve-gon, the sixteen-gon, the twenty-gon, the twenty-four-gon — the multiples of four — lead, and the hexagon, which leads every long sheet, is nineteenth of twenty-one.

Folding’s best polygon on the square is the twenty-one-gon, seventh at 77.9 per cent: odd, so it touches the square’s edges at vertices and sides in turn, and with enough sides that it is nearly a circle, which uses 78.5 per cent of a square. It beats the fourteen-gon here only because the square’s own ranking rewards near-circles over even polygons out of step with its symmetry.

The census on rectangles in the last figure shows the two regimes side by side: the square ranking the multiples of four, and three long sheets ranking the even polygons in increasing order of sides.

Coins that roll #

The difference between even and odd polygons has a practical expression that has nothing to do with paper, and it runs in the opposite direction.

The British twenty- and fifty-pence coins are heptagons with their sides curved outward, each arc centred on the opposite vertex. That makes them shapes of constant width: measured across in any direction, they are the same size, so they roll through a vending machine’s gauge like a circle and cannot jam at an angle. The construction needs an odd number of sides, because it pairs each vertex with an opposite side, which only an odd polygon has — the same pairing that makes an odd polygon’s least width run from a vertex to a side.

An even polygon has the opposite property. Its opposite sides are parallel, so its width across flats is smaller than its width across corners, and it can be laid flat between two parallel edges with nothing wasted — which is what a long sheet asks of it. The feature that lets an odd polygon roll like a circle is the feature that costs it room between two parallel lines, and a strip of paper is two parallel lines.

What the census assumes #

A polygon is regular and inscribed. Each share is the largest regular n-gon inside the sheet, maximised over its rotation and position, not the best polygon of any shape with n sides and not a polygon allowed to overhang.

The square is left out of every ranking, as the census that started with the triangle left it out, because on a square it is the sheet itself. On a rectangle it is an ordinary polygon, and on every sheet long enough it would lead the ranking outright, since its constant is one.

A tool builds a polygon when the totient allows it. The verdicts are the classical conditions applied to the arithmetic of n; they say a construction exists, not that it is short, well conditioned, or foldable within a sheet of the given proportion — which is a different question, and on a rectangle a construction written for a square can quietly return a different number, as a construction written for a square found.

What the census cannot show #

It cannot say what a folder actually wants. The largest polygon is one objective among several a design might have, and a polygon chosen for its symmetry, its number of points or its relation to a base will be chosen whatever share it uses.

It stops at twenty-four sides. The parity rule holds for every n, since both closed forms are monotone and lie on opposite sides of $\pi/4$, but the verdicts of each tool are computed only as far as the census goes, and past twenty-four the polygons only a fold builds include even ones with larger totients whose ranks are not drawn.

And the crossover is found by search. The proportion 1.1284 is where the searched shares of the octagon and hexagon agree; it has no closed form here, because the hexagon’s best rotation on a sheet shorter than about 1.155 is not the obvious one, and the simple formula that assumes it gives a proportion a thousandth too large.

Still open: a closed form for the crossover #

The octagon’s share has a formula at every proportion, and the hexagon’s has one on every sheet longer than $2/\sqrt{3}$. Between a square and that length the hexagon rotates away from the obvious position to use the extra height, and the proportion at which its share equals the octagon’s is the root of an equation this census never writes down. Finding it would turn 1.1284 from a measurement into a number, and would say whether it is algebraic of a degree a fold can reach — which would be a small, pleasing symmetry for a question about which tool builds the winner.

The other continuation is the tool that builds more. Two folds made at once build the hendecagon, and what two creases at once buy found the polygons that need them. On a long sheet the twenty-two-gon, which a pair of simultaneous folds reaches, sits ninth; whether any polygon beyond a single fold’s reach ever places higher on any sheet is a finite question once the census is extended.

The habit worth carrying is a question to ask of any ranking that changes with a parameter. Ask whether there is a regime where the parameter drops out. When it does, the ranking freezes, the order is set by constants, and the constants often have a closed form that explains the whole order at once.