The biggest one that can also be folded

Two questions can be asked about a regular polygon and a square sheet, and they come from different centuries and different subjects.

How much of the sheet does the largest one use? is an optimisation, answered by a maximisation over rotations, with a number for an answer.

Can it be built? is an algebraic question about whether the polygon’s angle lies in the field a given tool generates, answered yes or no, and it is the question the heptagon and the hendecagon already have essays about.

Putting them in the same table is the point of this rung.

The ranking #

Leaving out the square, which is the trivial case at 100%, the polygons rank by how much of the sheet they use like this:

The octagon at 82.84%, the twelve-sided at 80.38%, the sixteen-sided at 79.56%, the fifteen-sided at 77.32%. Then the thirteen at 76.91%, the fourteen at 76.89%, the eleven at 76.26%, the ten at 75.32%, the nine at 75.13%. And at the bottom the seven at 72.89%, the hexagon at 69.62%, the pentagon at 67.36% and the triangle at 46.41%.

The sequence is not increasing, which is the first surprise: more sides means a rounder shape and a rounder shape ought to fit a square better, and from nine to eleven it does not.

The multiples of four have a formula #

The first surprise stops being one as soon as the top three are written down properly, because they are a single expression evaluated three times.

A polygon whose side count is a multiple of four can be placed with the square’s own quarter-turn, touching all four walls, and it then has apothem $1/2$. A regular $n$-gon of apothem $r$ has area $n r^2 \tan(\pi/n)$, so the fraction of the sheet it uses is

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

At $n = 4$ that is $1$ — the square itself. At 8, 82.84%; at 12, 80.38%; at 16, 79.56%. Every one of the essay’s three best figures, from one formula with nothing fitted.

The octagon’s value is exact and pretty: $2(\sqrt2 - 1)$.

Which is decreasing, and its limit is the circle #

Now the shape of the list is settled rather than observed. $\frac{n}{4}\tan(\pi/n)$ falls monotonically as $n$ grows, and its limit is

$$\lim_{n\to\infty} \frac{n}{4}\tan\frac{\pi}{n} = \frac{\pi}{4} = 78.54\%,$$

the inscribed circle. So among the polygons that share the square’s symmetry, more sides is always worse, and the whole family is squeezed into the narrow band between the square’s 100% and the circle’s 78.5%.

That inverts the intuition the essay starts from. A rounder shape does not fit a square better; it fits it worse, converging on the circle, and the reason the octagon leads the table is simply that it is the smallest multiple of four above the square itself.

The counts that are not multiples of four cannot touch all four walls symmetrically, so they fall off this curve and interleave below it — which is why nine, ten and eleven sit under the octagon, and why the sequence looks jagged rather than being one.

The other column #

Now the tools.

Whether a polygon is constructible turns on the prime factorisation of φ(n). A compass needs it to be a power of two; a single fold allows factors of three as well; two simultaneous folds allow fives.

Of the polygons in the ranking, the compass builds three, four, five, six, eight, ten, twelve, fifteen and sixteen. A fold adds seven, nine, thirteen and fourteen. Two folds add eleven.

The agreement, and why it is not a coincidence #

Take the four best-fitting polygons: eight, twelve, sixteen, fifteen. Every one of them is compass-constructible.

The best polygon a compass cannot build is the thirteen-sided one, in fifth place. The polygon that needs two simultaneous folds — the hendecagon — ranks seventh of thirteen, behind four compass polygons and two single-fold ones.

So the extra reach that makes folding interesting buys nothing at all in this problem. Everything worth inscribing in a square was already available.

The reason is not luck and it is worth following through, because it is the same arithmetic wearing two hats.

A polygon fits a square well when it shares the square’s symmetry. The square has a quarter-turn. A polygon whose side count is a multiple of four has one too, and can be placed so that all four walls of the square do equal work. Eight, twelve and sixteen are the multiples of four in range, and they are the top three.

A polygon is compass-constructible when φ(n) is a power of two. For n a multiple of four in this range, φ(n) is 4, 4 and 8 — all powers of two.

Both properties are statements about the divisors of n. Being a multiple of four is what puts a polygon in tune with the square; and multiples of four in this range have totients built from twos, because their odd parts are small. The symmetry that makes a polygon fit is arithmetically close to the smoothness that makes it constructible, and near the small numbers they coincide.

Where the coincidence breaks #

It does break, and saying where is the honest part.

The alignment holds because the numbers are small. Twenty-eight is a multiple of four with φ(28) = 12 = 2 · 2 · 3, so it is not compass-constructible; it needs a fold. Thirty-six is the same. So “multiple of four” and “compass-constructible” part company as soon as the odd part of n stops being 1, 3 or 5.

At those sizes the fitting question is also losing its teeth. A polygon of twenty-eight sides is very nearly a circle, and the largest circle in a unit square uses π/4 = 78.54%; every high-sided polygon is converging on that from below and the differences between them are fractions of a per cent. So the ranking becomes uninteresting at roughly the point where the correspondence fails.

The honest summary is therefore bounded: over the range where the fitting question has an interesting answer, the good fits are the constructible ones, and both facts come from the divisors of the side count. Neither the correspondence nor the ranking is claimed beyond that.

The polygons that fall between #

The middle of the ranking is where the symmetry argument is easiest to see, because the polygons there are close in area and far apart in structure.

Nine, ten and eleven sides give 75.13%, 75.32% and 76.26% — three quarters of the sheet each, within about one point of each other, and all three worse than the octagon two places below them in side count. Nothing about the number of sides is doing any work here. What separates them is nothing much, because none of them has anything in common with a square.

Then twelve arrives at 80.38% and picks up five points at a stroke, because twelve is a multiple of four.

Thirteen and fourteen fall back to 76.9%, and fifteen recovers a little to 77.32% — fifteen has a five-fold and a three-fold symmetry, neither of which is the square’s, but it is odd and its tilt of exactly 3° puts a great many vertices near the walls. Sixteen returns to the multiples of four at 79.56%.

So the sequence oscillates rather than climbs, with the peaks at four, eight, twelve and sixteen and troughs between them, and the oscillation decays as the polygons approach a circle. A circle inscribed in a unit square uses π/4, which is 78.54%; the multiples of four are above that value and the others are below it, and both are converging on it from their respective sides.

That last observation is the cleanest statement of the whole effect available. A polygon in tune with the square beats the circle; a polygon out of tune with it loses to the circle. The octagon at 82.84% beats a circle by four points; the hendecagon at 76.26% loses to one by two.

The second-placed polygon shows both halves of the correspondence at once.

Whether the tool question is the right question #

There is a fair objection to this essay’s framing and it is worth putting properly.

Inscribing a polygon in a sheet is not the same as folding one. A construction that marks the polygon’s vertices with creases is what the constructibility question is about, and a construction that produces a folded object with the polygon’s shape is a different and much harder thing. This essay’s second column is about the first.

The distinction matters because the two have different answers. The vertices of a regular heptagon can be constructed by folding, in the sense that creases can be made through them; that does not mean a heptagonal sheet can be folded out of a square, and it certainly does not mean a heptagonal model is available.

What makes the marking question worth asking anyway is that it is the prerequisite. Nothing can be folded to a shape whose vertices cannot be located, so constructibility is a necessary condition on everything downstream, and a polygon that fails it fails permanently for that tool. It is a floor rather than a specification, and the table should be read as one.

Which theorem was checked, and how #

The two columns come from different kinds of computation and are checked differently.

The areas are a maximisation over rotation angles, anchored on two closed forms that the code never consults. The largest square inside a unit square must be exactly the square, area 1; the largest equilateral triangle must be 2√3 − 3, at a tilt of exactly 15°; and a corner construction solving a quadratic must agree with the maximisation to nine figures. All three hold.

The tools are Euler’s totient and its prime factorisation, both by trial division, with three predicates over the factors. What is quoted rather than computed is which degrees each tool extracts roots of — that a fold solves cubics is Beloch’s result, that two simultaneous folds solve the quintic is Alperin and Lang’s — and the essay says so rather than letting the table imply that a computation settled it.

The specific numbers the argument turns on are asserted in the site’s fold check: that the octagon uses more of the square than every polygon from nine to twelve, that eleven is the first polygon a single fold misses, that two folds reach it, that twenty-three is the first that defeats two folds, and that seven remains the first place a fold beats a compass.

Third place is where the sequence starts to lose interest, and it is worth seeing why.

What this does not say #

Three limits, and the middle one is the important one.

It is not a theorem. The correspondence between fitting well and being constructible is an observation over thirteen polygons with an explanation attached, and the explanation names the range in which it works. Nothing here proves anything about n in general.

It does not devalue folding’s extra degree. The one-degree advantage settles two of the three classical problems and is exactly the reason the heptagon is foldable. What this essay establishes is narrower and is about a particular optimisation: for inscribing a regular polygon in a square, the extra reach is not needed, because the polygons it unlocks are not the ones worth inscribing.

It is about one polygon in one sheet. Packing several shapes is the problem origami design actually faces, and it has none of this structure — the answers there are not known in closed form for almost any count and no symmetry argument settles them.

The operation that separates the two tool columns is one fold and one alignment.

What the coincidence is really about #

There is a way of stating the finding that makes it less surprising and more useful.

Both questions are asking about the factor structure of n, and it is the same structure both times.

Fitting a square asks whether the polygon’s rotational symmetry has anything in common with the square’s, which is a question about the greatest common divisor of n and four.

Being constructible asks whether φ(n) is smooth, and φ(n) is smooth when n’s prime factors are Fermat or Pierpont primes and appear to low powers — which is another way of saying n is built from small, well-behaved factors.

Numbers with small, well-behaved factors are numbers that share divisors with four. So the two conditions are not the same, and they are not independent either, and near the beginning of the integers they select nearly the same set.

That is a satisfying place for two questions with three centuries between them to meet.

The same table for a different sheet #

One way to test whether the correspondence is really about arithmetic is to imagine changing the sheet, and it is worth doing even though the computation is not made here.

Inscribe the polygons in an equilateral triangle instead. The symmetry argument then says that polygons with three-fold symmetry should do best: three, six, nine, twelve. The constructibility column would not move at all, because it depends only on the side count and not on what the polygon is being fitted into.

And the two would part company immediately. Nine is three-fold and is not compass-constructible — φ(9) is 6, which has a three in it — so a triangular sheet would put a fold-only polygon high in its ranking. The correspondence found here would break, and it would break for a reason that identifies exactly what was producing it: the square’s four-fold symmetry selects side counts divisible by four, and side counts divisible by four happen to have power-of-two totients in this range.

That is the clearest available statement of how much the finding depends on the square. It is not a fact about polygons or about tools. It is a fact about four — that the number the sheet contributes is a power of two, which is the same number the compass condition is built from.

A hexagonal sheet would be the interesting middle case, since six-fold symmetry selects counts divisible by six, and φ(6k) picks up threes rather than staying a power of two. A fold would earn something there that it does not earn on a square. None of that is computed here and all of it follows from the same arithmetic, which is the sense in which the correspondence has an explanation rather than merely a demonstration.

Folding’s extra degree earns its reputation elsewhere, and the elsewhere is worth pointing at.

Who asked each half, and when #

The constructibility half is Gauss and Wantzel for the compass, in the early nineteenth century, and Pierpont at its end for the sequence a fold would later turn out to need. That folding realises the Pierpont condition is a twentieth-century result, and it sat in the wrong language for fifty years before anybody in the folding world noticed.

The inscribed-polygon half is recreational geometry, mostly anonymous, mostly rediscovered. The largest equilateral triangle in a square appears in puzzle collections repeatedly; the general question for arbitrary n is the kind of thing that gets answered numerically and left.

Putting them together is this repository’s, and the joint table is the object worth having: an area column that is a search, a set of tool columns that are arithmetic, and a correspondence between them that neither column knew about.

The number this essay is really about #

Strip everything away and the finding is about a single integer appearing in two roles.

The sheet is a square, so the symmetry it can share with a polygon is four-fold. Four is a power of two.

The compass builds a polygon when the totient of its side count is a power of two. The same power of two.

Those are not the same statement and they are not independent either. A polygon whose side count is divisible by four is in tune with a square; a side count divisible by four and by little else has a totient made of twos. In the range where the fitting question is interesting the two conditions select nearly the same polygons, and the coincidence is the fact that both are built out of the number two.

Which is why the correspondence is expected to fail on any other sheet. A triangular sheet contributes a three, a hexagonal one contributes a two and a three, and neither matches the compass’s power-of-two condition — so on either of them a fold would earn something it does not earn here.

The result is therefore about the square, not about folding, and the essay’s title question has a slightly disappointing answer: on a square sheet, the biggest one that can also be folded is simply the biggest one.

Where the ladder goes next #

Back down to the construction itself. The largest triangle is reachable by two folds and two axioms, and its being a quadratic is what made this essay’s question worth asking at all.

And sideways to the tool. What a second simultaneous crease buys is a whole level of the constructibility hierarchy, and the finding here is that on this particular problem it buys a polygon that ranks seventh of thirteen.