The square is in the answer

The largest triangle in a square solves the first case and the biggest one that can also be folded crosses the whole census with the constructibility question. Between them they establish a sequence — how much of a square sheet each regular polygon can use — and the sequence has a striking feature in it.

It is not increasing. More sides is a rounder shape and a rounder shape ought to use a square better, and up to the octagon it does. Then the nine, ten and eleven-sided polygons all use less of the sheet than the eight-sided one, and the twelve-gon does not quite catch it either. The octagon’s 82.8% stands above everything but the square itself.

The ladder reads that as a fact about the octagon. It is a fact about the square.

Widening the question #

The census is a maximisation over one variable. A regular n-gon of circumradius one has a width across each direction; translation is free, so it fits in a rectangle exactly when its widths across the two axes are at most the rectangle’s sides; and the largest copy that fits is found by maximising the scale over the polygon’s rotation.

Nothing in that argument mentions the square. Replace the unit square with a w by h rectangle and the only change is that the two widths are compared against different numbers instead of the same one. The search range is unchanged — a regular n-gon is unchanged by a rotation through 2π⁄n, whatever the sheet is — and the golden-section refinement is unchanged.

So the generalisation costs nothing and the results are directly comparable, provided every polygon is judged at its own best angle on every sheet. That last point is not a formality: judging a rectangle’s polygons at the angles that were best on a square would produce a comparison of a sheet against somebody else’s optimum.

What the four sheets say #

Run it on a square, on the A-series proportion, on three-to-two and on six-to-five, taking the share of the sheet each polygon’s largest copy uses.

The square, after the square itself: the octagon at 82.8%, the twelve-gon at 80.4%, the eleven-gon at 76.3%, the ten-gon at 75.3%, the nine-gon at 75.1%, the heptagon at 72.9%, the hexagon at 69.6%, the pentagon at 67.4%, the triangle at 46.4%.

The A-series sheet: the hexagon at 61.2%, the octagon at 58.6%, the ten-gon at 57.4%, the twelve-gon at 56.8%, the eleven-gon at 54.8%, the nine-gon at 54.4%, the heptagon at 53.5%, the pentagon at 51.4%, the triangle at 40.8%.

Three-to-two and six-to-five give the same ordering as the A-series with different numbers.

Two things are different and one of them is not surprising. Every share is lower on a rectangle, which is expected: a square is the rectangle that best matches a roughly round shape, so any departure from it wastes paper.

The other is the finding. The octagon leads on the square and on nothing else. On every rectangle tried the hexagon is the best polygon after the square, and the octagon has fallen to second — and the even-sided polygons, which run 8, 12, 10, 6 on a square, run 6, 8, 10, 12 on every rectangle. That is not a disturbance of the order; it is its reverse.

Why the octagon wins on a square and not elsewhere #

The mechanism is the polygon’s symmetry group meeting the sheet’s, and it is worth setting out because it predicts the whole table rather than describing it.

A regular octagon has eightfold symmetry, which contains the square’s fourfold symmetry as a subgroup. So there is a rotation at which the octagon’s widths across the two axes are equal, and at that rotation the constraint from the horizontal side and the constraint from the vertical side bind at the same moment. Neither is wasted: the polygon is touching both pairs of edges of the square at once.

A hexagon has sixfold symmetry, which does not contain fourfold. Whatever angle it is set at, its widths across two perpendicular directions differ, so on a square one of the two constraints binds and the other has slack — and the slack is paper the polygon is not using.

Now change the sheet. On a rectangle the two sides are unequal, so a polygon whose widths are equal has no advantage at all: it is asking for a square-shaped bounding box on a sheet that is not square, and the mismatch is entirely wasted. What helps instead is a polygon whose widths across two perpendicular directions can be made to sit in the same ratio as the rectangle’s sides, and a hexagon — whose width varies more strongly with angle than an octagon’s does — can do that better.

So the octagon’s advantage is a resonance between two symmetry groups, and the resonance is a property of the pair. Change either member and it goes.

The width function, which is where all of it comes from #

The symmetry argument above is a story and the arithmetic under it is short, so it is worth putting down — it makes the whole table predictable instead of merely explained.

A convex shape has a support width across each direction: the distance between the two parallel lines that touch it from opposite sides. For a regular n-gon of circumradius one, rotated by θ, the width across a direction φ is the spread of the projections of its vertices onto that direction, and it is a function that oscillates with period 2π⁄n as θ runs.

The largest copy that fits a w by h rectangle is the smallest of w over the width across the horizontal and h over the width across the vertical, maximised over θ. So the whole problem is: choose the rotation that balances two oscillating functions a quarter turn apart against two fixed numbers.

Now the two cases fall out. When n is a multiple of four the two width functions are the same function — a quarter turn is a whole number of the polygon’s own symmetry steps — so they are equal at every angle, and on a square, where w and h are equal, both constraints bind everywhere. Nothing is wasted at any rotation, and the maximisation is over a function that is already as flat as it can be.

When n is not a multiple of four the two widths differ, and their ratio sweeps through a range as θ runs. On a square that is a liability: the polygon wants the two widths equal and can only approach it. On a rectangle it is an asset, because the rectangle wants the two widths in the ratio w to h, and a polygon whose width ratio sweeps widely can land on it.

So the sheet decides whether a polygon’s width variation is a cost or an asset, and four-fold symmetry is exactly the property of having none. That predicts the octagon and the twelve-gon leading on the square, the hexagon and the octagon leading on a rectangle, and the reversal of the even-sided order — which is the table.

What the amplitude of the variation buys #

The prediction can be pushed one step further and it explains why the hexagon rather than the pentagon or the decagon leads on a rectangle.

The width function’s variation shrinks as the number of sides grows: a triangle’s width across two perpendicular directions can differ enormously, a hexagon’s less, a twelve-gon’s hardly at all, and a circle’s not at all. So the sweep available to a polygon falls with n.

A rectangle asks for a width ratio away from one, and how far away depends on the proportion. A polygon can supply it only if its sweep reaches that far, so the best polygon on a rectangle is the one with the smallest number of sides whose sweep still covers the required ratio — small enough to have variation to offer, large enough to be round.

For the proportions here that is the hexagon, on all four. For a more extreme rectangle it would be a polygon with fewer sides, and for a proportion very close to square it would be one with more — which is a prediction this rung makes and does not test, and which the machinery would settle in an afternoon.

That also explains the triangle’s uniform failure. Its sweep is enormous and its area for a given circumradius is small, so it has plenty of the right kind of freedom and nothing to spend it on.

Which makes the sequence a fact about squares #

The consequence for how the ladder’s first two rungs read is direct, and it is the reason for this rung.

The sequence they establish — the share each polygon uses, non-monotone, peaking at the octagon — is not a sequence about polygons. It is a sequence about polygons against a square, and the square is doing enough of the work that the ordering does not survive changing it.

That is a stronger statement than “the numbers depend on the sheet”, which would be obvious. The numbers would depend on the sheet even if the ordering were stable, and the ordering is not: which polygon is best changes, and the even-sided ones reverse.

A result whose ordering is not stable under a change of setting is a result about the setting. The ladder’s first rung is honest about this in one place — the largest triangle in a square is tilted by fifteen degrees, and fifteen degrees is 45 − 30, which is manifestly a number about the square and the triangle together — and does not carry the observation through to the census.

What survives the change of sheet #

Not everything moves, and the parts that do not are worth separating because they are the parts the ladder can keep.

The square uses a square perfectly and nothing else does. That is the check the solver has to reproduce and it is a fact about a sheet holding a copy of itself.

The triangle is worst on every sheet, by a wide margin, at 46.4% on a square and 38-48% on the rectangles. Three sides is genuinely too few, and no proportion rescues it.

Every share falls when the sheet stops being square. Every polygon here is roughly round, a square is the rectangle closest to round, and departing from it costs every one of them — which is a different price from the one a designer packing flaps pays for a square and points the other way.

And the non-monotonicity survives. On every sheet the sequence rises, falls and rises again rather than increasing with the number of sides — which is the ladder’s real finding, and it is about the interaction of two symmetries in general rather than about the octagon in particular.

So the rung is a correction rather than a demolition. The phenomenon is real; the exemplar is local.

Which theorem was checked and how #

The unit square’s answers are unchanged. The generalised solver is required to reproduce the existing one exactly at w = h = 1, for every polygon, so the wider census cannot be quietly solving a different problem.

The square uses all of a square. That is the closed form a solver with a lost factor fails, and it is asserted rather than inspected.

The octagon must lead on the square, or the ladder’s own established result has not been reproduced and the comparison has no baseline.

No rectangle may agree with the square about which polygon is best, and the even-sided ordering must differ. Both are stated as conditions on the figure existing, so a version in which the ranking were stable would refuse to draw rather than being drawn and misread.

Where the model stops #

Four proportions is four. The claim is that the octagon leads on the square and not on the rectangles tried, and it is checked on those four; the arithmetic behind it — symmetry resonance — predicts it generally and is not a proof.

Only rectangles are compared. A sheet with a different shape entirely, or with a hole in it, is a different optimisation, and this collection’s sheet-shape ladder measures those for a different objective.

Translation is free and rotation is the only variable. That is exact for a convex shape in a convex box and it is the whole of why the problem is a one-dimensional maximisation.

And nothing here says which of these polygons a fold can construct — that is a condition on a totient and has no sheet in it. That crossing is the ladder’s own second rung, it is a question about arithmetic rather than about geometry, and it is unchanged by the sheet — a polygon’s constructibility does not depend on what it is being fitted into.

What the picture cannot show #

A table of shares cannot show the arrangements, and the arrangements are where the symmetry argument lives. Whether a polygon is touching two pairs of edges or one is visible in a drawing and invisible in a percentage, and the drawing is available for one case at a time.

Nor can it show what happens between the proportions. The census is computed at four ratios and the ordering presumably changes somewhere between the square and the next one along; where, and whether the change is a crossing or a jump, is a computation this rung does not make and the machinery would.

The deepest thing it cannot show is whether the reversal is general. Four rectangles agreeing is suggestive and the symmetry argument explains why they would, and neither is a proof that no proportion puts the octagon back on top.

The idealisation, named #

A polygon is a plane region, a sheet is a rectangle, and fitting means containment. Nothing here is folded and nothing is constructed: the census is a packing question, and the ladder’s second rung is what crosses it with the folding question.

That separation is worth being clear about, because the two are easy to run together. A polygon that uses a sheet best is not thereby foldable, and a foldable polygon is not thereby efficient. The ladder’s second rung finds the two questions agreeing over sixteen polygons and gives the reason — both are questions about the arithmetic of the same number — and this rung changes the first of the two without touching the second.

So the crossing that rung makes should be re-run on a rectangle, and the interesting possibility is that the agreement it found is another thing that does not survive: the efficiency ordering has moved and the constructibility verdicts have not, so the two lists that matched on a square need not match on an A-series sheet.

Where the ladder goes next #

That re-run is the rung this anchor now owes, and it is a small computation with a definite answer. Cross the rectangle census with the constructibility verdicts and see whether the two still agree. On a square the best foldable polygon and the best polygon are near neighbours; if the hexagon’s rise on a rectangle puts a foldable polygon at the top where an unfoldable one was, the agreement was a coincidence of the square and the ladder’s second rung has been reading it as a theorem.

Sideways from here, the finding belongs with the sheet’s own effect on what a folder can reach, where the square is again the extreme case and again for its symmetry — folds that would be distinct on a rectangle coincide on a square. Two ladders in this field have now found the square costing something through the same mechanism, and a third has found it costing a designer paper. The square is a choice with a price, and the price keeps being its own symmetry.

The habit worth carrying is a test to run on any optimisation result before generalising it. Change the constraint’s shape and see whether the ordering survives. Magnitudes always move and orderings usually do not, and when an ordering does move, the result was about the constraint rather than about the things being ordered.