The sheet a polygon fits exactly

Every even polygon beats every odd one ran the census of the largest regular polygon a sheet holds on seven proportions — a square, eleven by ten, six by five, the A series, three by two, two by one and three by one — and read off a verdict about tools. The best polygon only a fold can build places fourth at best; the twenty-two-gon, which needs two folds at once, places ninth on every long sheet. It ended by asking whether any polygon beyond a single fold’s reach ever does better on some other sheet.

The crossing is as hard as the polygon found that between a square and a sheet 1.1284 long the ranking moves continuously, driven by a hexagon that turns against all four edges before it can lie flat. That is a hint about where the seven sheets were not looking, and it is worth following to the end.

Every polygon with $4k + 2$ sides has a sheet of its own, and on it the polygon does better than on any sheet the census named. The fourteen-gon places third there. The twenty-two-gon places fifth. And the rank each reaches is a formula rather than a measurement: $(n - 2)/4$.

Flats one way, corners the other #

A regular polygon has a symmetry that decides how it sits in a rectangle, and it comes in three kinds.

A polygon whose number of sides is a multiple of four — the octagon, the twelve-gon — has flat edges facing all four directions of a square at once. Its width across the flats is the same both ways, so it fits a square with a flat against every edge. A longer sheet gives it nothing, and its share falls as its constant over the length.

A polygon with an odd number of sides has no pair of parallel edges at all; opposite every corner is a flat. Its least width runs from a corner to the edge across from it, and on a long sheet it is always a little worse than the circle.

A polygon with $4k + 2$ sides — the hexagon, the decagon, the fourteen-gon — is the interesting one. Its opposite edges are parallel, but a quarter turn round from a pair of flats is a pair of corners. So it is narrower across one axis than across the other, by exactly the ratio of its circumradius to its apothem:

$$\frac{R}{a} = \frac{1}{\cos(\pi/n)}.$$

There is therefore exactly one rectangle it fits against all four edges: one wide across the flats and $1/\cos(\pi/n)$ long across the corners. For the hexagon that is $2/\sqrt{3} \approx 1.1547$, the length at which the turning hexagon finally lies flat. For the decagon it is 1.0515, for the fourteen-gon 1.0257, for the twenty-two-gon 1.0103, and for the forty-six-gon 1.0023 — each sheet closer to a square, as the polygon comes closer to a circle.

Why the rank is a count #

On its own sheet a polygon of $4k + 2$ sides with apothem one half fills the width exactly. Its area is $n a^2 \tan(\pi/n)$, so its share of a sheet of width one and length $1/\cos(\pi/n)$ is

$$\frac{n}{4}\tan\frac{\pi}{n}\cos\frac{\pi}{n} = \frac{n}{4}\sin\frac{\pi}{n}.$$

That is 77.88 per cent for the fourteen-gon, and the first figure measures it to the precision of the search on all eleven polygons.

Now ask what beats it there. There are four kinds of competitor, and each can be settled in a line.

A multiple of four sits with flats on all four edges and uses its constant over the length, $\tfrac{m}{4}\tan(\pi/m)\cos(\pi/n)$. Divide both shares by $\cos(\pi/n)$ and the comparison is between $\tfrac{n}{4}\tan(\pi/n)$ and $\tfrac{m}{4}\tan(\pi/m)$ — the two polygons’ constants — and the even constants fall with the number of sides. So every multiple of four with fewer sides beats the polygon and every one with more loses to it.

Another polygon of $4k + 2$ sides with more sides has its own sheet closer to a square, so on this one it is already lying flat and falls as its constant over the length, which is smaller. One with fewer sides needs a longer sheet than this to lie flat, so here it is still turning, pressed against all four edges, and doing worse than it will later. In the census it never gets above.

An odd polygon is below the circle on every long sheet, and on this one too.

The circle itself would use $(\pi/4)\cos(\pi/n)$, and the polygon uses $\tfrac{n}{4}\sin(\pi/n)$; the ratio is $\sin x/(x\cos x) = \tan x/x$ with $x = \pi/n$, which is always above one. So no polygon approaching the circle, however many sides it has, climbs above this one — which is what makes the rank independent of where the census stops.

What is left above the polygon is the multiples of four from eight up to $n - 2$, and there are $(n - 6)/4$ of them. The rank is one more: $(n - 2)/4$. The hexagon is first on its sheet, the decagon second, the fourteen-gon third, the twenty-two-gon fifth and the forty-six-gon eleventh, and the first figure measures every one of them.

Following one polygon through the sheets #

A rank measured on a handful of sheets is a sample of a curve, and the third figure draws the curve. On a square the fourteen-gon is buried: every multiple of four up to the census limit fits a square better than it does. As the sheet lengthens it climbs past them, one crossing at a time, reaches third place at 1.0257, and then falls back as the multiples of four with more sides stop being beaten and the evens settle into their long-sheet order, with the fourteen-gon fifth.

The twenty-two-gon traces the same shape nearer the square: fifth at 1.0103, ninth on every long sheet. The eighteen-gon, between them, peaks at fourth on 1.0154. Each polygon has one proportion where it does best, and it is the one it fits exactly. Nothing about that is visible on a sheet of eleven by ten, which is already too long for all three.

What the seven sheets said, and what every sheet says #

The fourth figure sets the two answers side by side. On the seven sheets, the best polygon only a fold builds is the fourteen-gon, fourth; over every sheet it is still the fourteen-gon, and it is third. The best polygon needing two folds at once is the twenty-two-gon at ninth, and over every sheet it is fifth. The best polygon that no two folds build is the forty-seven-gon at twelfth on the named sheets, and the forty-six-gon at eleventh over all of them.

So each tool can buy a place, and the place is set by its first polygon of $4k + 2$ sides. The polygons only a fold builds first include a $4k + 2$ at fourteen, because fourteen is twice seven and the heptagon’s three is in its totient. The polygons needing two folds first include one at twenty-two, twice eleven, with a five in its totient. The first $4k + 2$ polygon beyond both is forty-six, twice twenty-three, whose totient is twenty-two and carries an eleven. Put into the formula, those three numbers are the whole table: third, fifth, eleventh.

It is a pleasingly blunt answer to the question of what a better tool buys in this contest. A compass already takes first place, on every sheet longer than 1.1284 with the hexagon and on every shorter one with the octagon. A single fold can buy third. Two folds at once can buy fifth, and anything stronger than that, eleventh. Each extra power the tool gains is paid for in places, because the first polygon needing it has more sides, and more sides means more multiples of four in front of it.

The square, and a rank that depends on the census #

The census on a square behaved differently from the start, and the own-sheet argument explains why it should be treated with care.

On a square, the multiples of four sit with flats against all four edges and use their constants, all of which are above the circle’s $\pi/4$. Every other polygon uses less than $\pi/4$ of a square. So on a square every multiple of four beats every polygon that is not one — and there are infinitely many multiples of four. A census that stops at twenty-four sides ranks the twenty-one-gon seventh, because it has counted only five multiples of four; a census that stops at forty-eight ranks the same polygon thirty-first. The number measured on a square is a statement about where the census ended.

The stable answer on a square is the smallest multiple of four each tool needs. The first multiple of four only a fold builds is the twenty-eight-gon, sixth on a square whatever the census, because only the octagon, twelve-, sixteen-, twenty- and twenty-four-gon come before it. The earlier reading of the twenty-one-gon at seventh was a real measurement of a census that happened to stop at twenty-four, and the own-sheet ranks are the version of the same question that does not depend on where the counting stops.

A rank is a count of what lies above, and it is only a property of the polygon when that count is finite. On a square and on every long sheet the polygons that are not the favoured kind have infinitely many competitors above them. On a polygon’s own sheet the count is finite, and it is $(n - 6)/4$.

A sheet near the square, drawn #

The twenty-two-gon’s sheet is only a per cent longer than a square, and drawn at this size the two polygons are hard to tell from circles. The difference between them is the point. The twenty-gon has flats against all four edges and is as large as a square allows; it uses 78.4 per cent. The twenty-two-gon touches the short edges with flats and the long edges with corners, and uses 78.3. Any polygon with more sides than twenty-two, on this sheet, uses less — the twenty-four-gon because its constant is smaller, the twenty-three-gon because it is odd, and a polygon of a thousand sides because it is almost a circle, and the circle loses to every polygon on its own sheet by the factor $\tan x/x$.

The margins are small: the twenty-two-gon’s lead over the twenty-four-gon on this sheet is under a tenth of a per cent of the paper. The ranking is exact and the stakes are not, which is worth saying about every result in this census. What makes the ranks interesting is not that a folder would notice the difference but that they sort, cleanly, by the arithmetic of the number of sides.

The same contest, read from the triangle #

The census began with a single shape. The largest triangle in a square found it tilted by fifteen degrees and pressed against the square at three corners and a side, and the biggest one that can also be folded then ran the same maximisation for every number of sides and set the verdicts beside it. In both, the shape that wins is the one whose symmetry is spent against the edges of the paper, and the own sheet is the purest case of that: a polygon of $4k + 2$ sides spends all four edges, and on no other rectangle can it spend more than three.

It also recasts what the verdict column was measuring. The heptagon a compass cannot reach is ranked near the bottom of every long sheet because it is odd, and its doubled form, the fourteen-gon, is ranked third on its own sheet because it is not. The tool is the same — a fold builds either, since doubling a polygon’s sides costs one bisection — and the rank is not, which is the clearest single statement that the rank and the tool are separate facts that happen to meet. Gauss’s polygon makes the same point from the compass side: the thirty-four-gon is its doubled form, a compass builds it, and on its own sheet, 1.0043 long, it ranks eighth.

What this assumes #

The polygons are regular and the sheet exact. A sheet cut to 1.0257 by any real means is near the fourteen-gon’s sheet, and on a sheet a hair longer or shorter the fourteen-gon is still third only while no multiple of four above it has overtaken; the third figure shows that band is narrow.

The tool verdicts are the classical ones, read from the totient: a compass when it is a power of two, one fold when its primes are twos and threes, two folds at once when fives are allowed, which is the result two creases at once works through. They say a construction exists and nothing about its length or its conditioning.

The square is left out of the ranking, as every census here has left it out, because it is the sheet itself on a square and the first polygon on any sheet long enough.

What the census cannot show #

It proves the rank for the polygons it measured and argues it for the rest. The multiples of four and the circle are settled by inequalities that hold for every $n$. That a turning $4k + 2$-gon with fewer sides never climbs above is measured for every polygon to forty-six rather than proved, and a proof would need the turning polygon’s share in closed form, which the hexagon’s crossing supplies only for six sides.

It does not say which sheet a folder has. Nobody cuts paper to $1/\cos(\pi/14)$, and the ranks here are about what each polygon can reach, not about what anybody will meet.

And it keeps the objective fixed. The share of the sheet is one thing to maximise; the square is in the answer already showed how much the answer depends on which sheet is asked, and a different objective — the polygon’s perimeter, its number of points, its relation to a base — would give a different contest.

Where tools and symmetry meet #

The result is a coincidence of two quite different kinds of fact, and it is worth separating them.

One is about symmetry. Which polygons have a sheet of their own is a question about how their edges face: $4k + 2$ sides give flats on one axis and corners on the other, and nothing else does. That fact knows nothing about tools.

The other is about arithmetic. Which tool builds a polygon is a question about the prime factors of its totient, which how many polygons a fold reaches counts and the eleven-sided one nobody can fold meets at its first obstruction. That fact knows nothing about edges.

The best rank a tool can buy is where the two meet: the first polygon that has both the right symmetry and the tool’s own prime. Seven gives three, and fourteen has the right symmetry, so a fold buys third. Eleven gives five, and twenty-two has the right symmetry, so two folds buy fifth. The smallest prime $p$ whose $p - 1$ carries a prime above five is twenty-three, so the next tool’s first polygon of the right kind is forty-six.

Still open: whether a polygon can do better off its own sheet #

The own sheet is the best place the census finds for each polygon, and the third figure is consistent with that being true everywhere. It is not proved. A $4k + 2$-gon on a sheet slightly shorter than its own is turning, and turning cost the hexagon share; it plausibly costs every such polygon share. Writing the turning share in closed form for general $n$, as the hexagon’s was written, would settle whether the peak is exactly at the own sheet or merely near it for every polygon measured.

The other direction is the multiples of four, which get the square for their own sheet and the same kind of formula: on a square the $4k$-gon ranks $k - 1$, and a tool’s best rank on a square is set by its first multiple of four — the twenty-eight-gon for a fold, sixth, and the forty-four-gon for two folds, tenth. Whether the odd polygons have any sheet on which their rank is finite and small is the remaining case, and the circle bound suggests not: an odd polygon’s constant is below the circle’s, and so is its share on every sheet long enough to hold it.

Sideways from here, the sheets themselves are numbers. $1/\cos(\pi/14)$ is a number of degree six, with a three in it, so a compass cannot mark the fourteen-gon’s own sheet, exactly as the crossing proportions could not be marked for the polygons they involve. The sheet on which a fold’s best polygon does best is a sheet only a fold can measure out.

The habit worth carrying is about samples of a parameter. Before reading a best case off a handful of settings, ask whether the object has a setting of its own. A polygon, a mechanism or a material with a symmetry usually has one parameter value where the symmetry is used exactly, and a sample of round numbers will step over it every time.