An odd polygon fits like its double

Turning is uphill all the way found the share of a rectangular sheet used by a turning regular polygon of $4k + 2$ sides in one line, with a slope proportional to $h^2 - 1$ whatever the number of sides, so that each such polygon peaks exactly on its own sheet, $1/\cos(\pi/n)$ long. It ended on the odd polygons. Their two widths, it observed, are a quarter of a step apart rather than half, because opposite each corner of an odd polygon is the middle of an edge; so an odd polygon’s share while it turns should have 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, was left as a calculation.

The calculation turns out not to be needed. An odd polygon’s share is an even polygon’s share, exactly, on every sheet, and the even polygon is the one with twice its sides.

A box sees widths, and widths cannot see a half-turn #

Whether a convex shape fits a rectangle at a given angle depends on two numbers only: its width across the rectangle and its width along it. The earlier essay used exactly that to reduce the problem to two inequalities in one angle. So any two shapes with the same width in every direction fit exactly the same rectangles at exactly the same angles.

A shape and its half-turn have the same width in every direction, since measuring a width from one side or from the other gives the same number. So does the shape halfway between them — the average of the shape and its half-turn, the set of midpoints of one point from each — because its width in any direction is the average of two equal widths. That halfway shape is the only part of a shape a box can see.

For a regular polygon with an even number of sides the halfway shape is the polygon itself, which is already its own half-turn. For an odd polygon it is not, and it is something better. A regular $n$-gon with $n$ odd, of circumradius $R$, has its corners every $2\pi/n$ and the middles of its edges halfway between; its half-turn puts corners exactly where the original had edge middles. The average of the two has a corner in each of those $2n$ directions, all the same distance out, and it is a regular $2n$-gon of circumradius $R\cos(\pi/2n)$. The triangle’s is a regular hexagon, the pentagon’s a regular decagon, the heptagon’s a regular fourteen-gon.

The width is the same calculation done directly. In a direction at an angle $u$ from its widest, an odd $n$-gon is $2R\cos(\pi/2n)\cos u$ wide: one side of the width is a corner and the other the middle of an edge, and their two reaches add to that. A $2n$-gon of circumradius $R\cos(\pi/2n)$ at an angle $u$ from a corner is $2R\cos(\pi/2n)\cos u$ wide too. The two polygons have one width function, so they fit one family of sheets.

The odd polygon’s share, from its double’s #

The largest odd $n$-gon a sheet holds is therefore the one whose halfway $2n$-gon is the largest $2n$-gon the sheet holds, at the same angle. The share of the sheet it uses is the $2n$-gon’s share times the ratio of their areas, and that ratio is the same on every sheet:

$$\frac{A_n}{A_{2n}} = \frac{\tfrac{n}{2}R^2\sin\tfrac{2\pi}{n}}{n\,R^2\cos^2\tfrac{\pi}{2n}\,\sin\tfrac{\pi}{n}} = \frac{\cos(\pi/n)}{\cos^2(\pi/2n)} = 2 - \sec^2\frac{\pi}{2n}.$$

For the triangle it is two thirds; for the pentagon 0.894; for the heptagon 0.948; and it climbs toward one as the sides grow, since an odd polygon with many sides is nearly its own half-turn.

The $2n$-gon of an odd $n$ has $4k + 2$ sides — twice an odd number — so its share is the one the earlier essay wrote down, and the odd polygon’s is that times the ratio:

$$f_n(h) = \Bigl(2 - \sec^2\frac{\pi}{2n}\Bigr)\cdot\frac{2n\,c}{4s}\cdot\frac{s^2 + (h - c)^2}{h}, \qquad c = \cos\frac{\pi}{2n},\ s = \sin\frac{\pi}{2n},$$

while it turns, from the square to $h = 1/c$. It agrees with a direct search over rotations to a part in a billion, for the triangle, pentagon, heptagon and nonagon at seven sheets each, and the figure below is refused if it does not.

Everything the earlier essay proved for its polygons now holds for these, because the ratio is a constant and a constant changes no slope’s sign. The slope is proportional to $h^2 - 1$, so the odd polygon’s share is level on the square, rises all the way to $1/\cos(\pi/2n)$, and falls after it. Its own sheet is its double’s own sheet: the triangle peaks on the hexagon’s sheet, 1.1547 long, the pentagon on the decagon’s, 1.0515, the heptagon on the fourteen-gon’s, 1.0257. There the odd polygon touches all four edges — a corner and the opposite edge on the short sides, its two widest corners on the long ones. And its gain from the square is the average of one and the sheet’s length, 7.7 per cent for the triangle, 2.6 for the pentagon, 1.3 for the heptagon, exactly its double’s gain.

The fifteen degrees was the same cancellation #

The earlier essay’s last open question sideways was about the largest triangle in a square, which is tilted by exactly fifteen degrees — a number that came out of a quadratic, $t^2 - 4t + 1 = 0$, with the root $2 - \sqrt3 = \tan 15^\circ$. It asked whether that fifteen degrees was the same cancellation as the turning polygons’, in a polygon with no opposite sides.

It is. On a square every turning polygon of $4k + 2$ sides is tilted exactly halfway between its two flat positions, because the square is the one sheet on which its two widths must be equal. The triangle’s double, the hexagon, lies flat at a turn of thirty degrees from its widest, so on a square it is tilted fifteen, and the triangle, which fits wherever the hexagon fits at the same angle, is tilted fifteen too. In general an odd $n$-gon on a square is tilted $\pi/4n$ from its widest: fifteen degrees for the triangle, nine for the pentagon, 6.43 for the heptagon. The quadratic was the half-angle identity again, met through a shape whose double it had not been noticed to have.

Where an odd polygon’s own sheet puts it #

A sheet of its own made each polygon of $4k + 2$ sides rank exactly $(n - 2)/4$ there — third for the fourteen-gon, fifth for the twenty-two-gon. The same question for the odd polygons has a much plainer answer.

On its own sheet the triangle is last of the fifty-seven polygons ranked. The pentagon beats only the triangle. The heptagon beats the triangle, the pentagon and the hexagon, which on the fourteen-gon’s sheet is still turning well short of its own; the nonagon adds the heptagon and the decagon. Every odd polygon, on its own sheet, beats only smaller odd polygons and a few polygons of $4k + 2$ sides still turning toward sheets longer than this one. Its peak is a peak and not a promotion.

The two families trade places in a precise way along the table. Each odd polygon’s own sheet is shorter than the last, so each is ranked among more polygons still turning, and the odd polygons climb slowly: fifty-seventh, fifty-sixth, fifty-fourth, fifty-second, fifty-first, forty-ninth, forty-eighth for three to fifteen sides. The fifteen-gon uses 77.53 per cent of its sheet, within six tenths of a point of the circle’s 78.11 there, and still has forty-seven polygons above it. An odd polygon’s own sheet is a better place for it than any other sheet, and a bad place in the field.

The reason is the ratio. On its own sheet an odd polygon uses $2 - \sec^2(\pi/2n)$ of what its double uses, and its double is the polygon ranked there; for the triangle that is two thirds of the hexagon’s share, and no amount of fitting recovers a third. The ratio closes as the sides grow, but so does the gap between everything else, and the odd polygon never gets ahead of the field.

No odd polygon passes the circle #

The earlier essays compared everything with the circle, whose share on a sheet one wide is $\pi/4h$. Every even polygon beats every odd one found the odd polygons’ constants climbing up to the circle’s from below on long sheets, where every polygon lies flat against the short sides. The ratio settles what happens on every other sheet.

An odd polygon’s share over the circle’s is its double’s, times the ratio. Its double’s is highest on its own sheet and after, where it is $\tan x/x$ with $x = \pi/2n$, as the sheet a polygon fits exactly found. So the odd polygon’s is never more than

$$\bigl(2 - \sec^2 x\bigr)\,\frac{\tan x}{x},$$

which is 0.735 for the triangle, 0.925 for the pentagon and 0.990 for the thirteen-gon. Expanded in $x$ it is $1 - \tfrac{2}{3}x^2$ and a little less, so it approaches one from below as the sides grow, and it was checked below one for every odd polygon up to a hundred and one sides. No odd polygon uses more of any sheet than a circle does, where every polygon of $4k + 2$ sides passes the circle on the way to its own sheet and stays above it.

The curves also never cross. Checked on sheets from the square to three to one, a five-hundredth apart, for every odd polygon up to a hundred and one sides against the next, no odd polygon ever uses more of a sheet than an odd polygon with more sides — 49,049 comparisons and no exception. The polygons of $4k + 2$ sides reorder themselves as the sheet lengthens, which is what gives each of them a sheet of its own in the rank sense; the odd ones keep one order on every sheet, and a sheet of their own is only where each is largest, never where it overtakes anything it did not already beat. That order is a measurement here, not a theorem: the ratio grows with the sides and the doubles do not, and which wins on a given sheet is a comparison the check made rather than an inequality anybody has written down.

What the tools make of it #

The tool verdicts carry over by the same arithmetic. The pentagon’s own sheet is the decagon’s, $1/\cos(\pi/10)$, and a compass marks it; the heptagon’s is the fourteen-gon’s, $1/\cos(\pi/14)$, and only a fold marks that; the eleven-gon’s is the twenty-two-gon’s, which needs two folds at once, as the eleven-gon itself does, since eleven is the first prime a single fold cannot reach. An odd polygon’s best sheet needs exactly the tool the polygon needs, because it is its double’s sheet, and a polygon with twice an odd number of sides has the same totient as the odd one and so the same verdict.

But the tools buy nothing in rank. The heptagon on the only sheet it peaks on is fifty-fourth of fifty-seven, and the fourteen-gon on the same sheet is third. The other odd polygons only a fold builds fare the same: the nonagon is fifty-second on its own sheet and the thirteen-gon forty-ninth, each beaten there by every polygon in the census except a handful of smaller ones, their doubles among them. A folder choosing between the two polygons a fold alone can make should, on this measure, always make the one with twice the sides: it fits the same sheet at the same angle and uses more of it by the reciprocal of the ratio.

What the width argument takes as given #

Fitting a box is a matter of widths only for a convex shape. Every regular polygon is convex, and the averaging argument needs nothing else; a star polygon or any non-convex outline would fit differently from its halfway shape.

The sheet is a rectangle and the polygon is placed freely. A box cannot tell a shape from its half-turn only because a box is itself symmetric under a half-turn; a sheet that is not — a triangle of paper, a trapezoid — would tell the odd polygon from its double at once, and on such sheets the correspondence does not hold.

The ranks are against a census. Every rank quoted is among the regular polygons from three to sixty sides, less the square, which fills a square sheet and fits every sheet on its own terms; the claim that odd polygons keep one order is a check to a hundred and one sides on a grid of sheets, and the claim that none passes the circle is the inequality above, checked to the same size and explained beyond it by its expansion.

The shape a box sees #

The connection that was not expected is between this line of argument and a different one entirely. Every panel holds a frame found a folded Miura giving the same shrink factors on every panel because its folded outline is its own mirror image turned half a turn, and a box cannot tell a shape from its half-turn. The odd polygons are the same fact at its plainest: the only part of a shape a box can see is the shape averaged with its half-turn, and for a regular odd polygon that average is another regular polygon. Two questions, one about folding a tessellation and one about the largest polygon a sheet holds, reduce to the same statement about boxes.

The averaging is also the cheapest possible proof of the result. Nothing about rotations, contact points or which edge touches which side had to be redone; one observation about what a rectangle can measure carried a closed form, a peak, a gain, a tilt and a bound on the circle across from one family of polygons to another in a line each. It also changes what the earlier question was. The odd polygons looked like a harder case of the turning problem — no opposite sides, a quarter-step offset, a new closed form to derive. They are the easy case: the closed form was already written, one essay back, under another polygon’s name.

Still open: sheets that are not boxes #

The correspondence fails on a sheet with no half-turn, and that is where the odd polygons might finally rank well. A triangular sheet holds a triangle perfectly and a hexagon badly; a regular pentagon of paper holds a pentagon perfectly. Which regular polygon a sheet of a given regular shape holds best, and whether folding ever buys a place on such a sheet that a compass cannot, is the same census on a different outline, and the width argument no longer shortcuts it.

The tolerance at the peak is the other direction, as it was for the $4k + 2$ polygons. An odd polygon’s peak is a corner, like its double’s, so a sheet cut slightly wrong costs it share to first order; and the heptagon’s sheet, like the fourteen-gon’s, is one only a fold can mark. Exact is not accurate measured what hands do to such a construction, and a folder marking the sheet and then the heptagon on it compounds two errors on a polygon that would rank no better for either being right.

Sideways from here, the square is in the answer found the winner depending entirely on the sheet asked about. The halfway shape says which part of a polygon a rectangular sheet can reward, and it would be worth asking the same of the other objectives on the record — the largest polygon a fold can build, the one whose crease pattern is cheapest — whether each of them, too, sees only a polygon averaged with its half-turn.

The habit worth carrying is about what a measurement can distinguish. Before computing a new case, ask whether the measurement can tell it from a case already computed. A box cannot tell a triangle from a hexagon two thirds its area, and the whole of the odd polygons’ turning problem was already solved under the hexagon’s name.