The window has an exact edge

A sheet rewards the multiples of its sides found the cleanest fact about regular sheets of paper: on an equilateral triangle, the regular polygons that take more of the sheet than its inscribed circle are exactly the multiples of three, because only they can lay a side along each of the triangle’s three. The rule has a width asked how long that survives a triangle that is not quite equilateral, and measured windows: the hexagon stays ahead of the circle while the apex of an isosceles triangle runs from 44.4 to 79.1 degrees, the 24-gon from 55.7 to 64.5. The worst angle sets the price derived the first-order rate at which each polygon loses share, estimated each window as the polygon’s lead divided by that rate, found the estimate short of every measured window, and named what it had left out: the second-order terms, which “would turn the windows into closed forms as well”.

They do, and they do more than that. The windows were never second-order quantities to be approximated. On an isosceles sheet the share of the largest multiple of three is exact in one line, the circle cancels out of the comparison, and each edge of each window is a closed-form angle. Two things follow that the measurements and the estimate both missed: the opening half is wider for a reason that can be named, and the window shrinks with the number of sides about seventeen per cent more slowly than ninety degrees over the sides.

A bound that is an equality #

The earlier derivation held a regular $n$-gon at a fixed rotation and asked how large it could be. Its circumradius and its centre are three unknowns, and each side of the triangle gives one inequality: the polygon’s reach toward that side, plus how far its centre has moved toward it, may not exceed the side’s distance from the origin. Weighting each inequality by its side’s length makes the centre drop out, because a closed shape’s side lengths times their outward normals sum to zero, and leaves $R\sum_i L_i\,\text{reach}_i \le 2A$.

That was used as a bound. On a triangle it is an equality. Three inequalities in three unknowns: the largest polygon at a given rotation pushes against all three sides at once, every inequality is tight, and so the weighted sum is tight too. The size of the polygon at a rotation is not bounded by $2A/\sum L_i\,\text{reach}_i$; it is that.

The rotation is the only choice left, and on an isosceles triangle it is made for nothing. The earlier essay found the best rotation putting the polygon’s facing at the weighted middle of the three sides’ turns. On an isosceles sheet the base has not turned and the two legs have turned equal amounts in opposite directions, so the middle is the base: the largest multiple of three keeps a side flat on the base, and meets each leg at a corner turned half the apex change away from the side it used to lie along. With legs of length one, base $B = 2\sin(\theta/2)$ and $c = \pi/n$, its share of the sheet is

$$\text{share} = \frac{n\sin c\cos c \cdot 4A}{\big(B\cos c + 2\cos(c - |\Delta|/2)\big)^2},$$

where $\Delta$ is the apex’s departure from sixty degrees. Nothing in it is fitted.

The fitted hexagons sit on the formula’s curve to within a few parts in $10^{16}$ of the sheet at every apex from 46 to 82 degrees, which is to say exactly. The dial walks the same check through the nonagon, the dodecagon and up to the 24-gon, and the agreement holds at every one of them, over a range that narrows as the sides grow — a range the next sections come back to.

The circle was in both shares #

The inscribed circle has a share of the triangle too, and it is the formula’s limit: a polygon of infinitely many sides is the circle. Its radius is $2A/P$, with $P$ the perimeter, so its share is $\pi\cdot 4A/P^2$. Now write the polygon’s denominator with the perimeter pulled out. Since $P = B + 2$,

$$B\cos c + 2\cos(c - |\Delta|/2) = P\cos c + 2\big[\cos(c - |\Delta|/2) - \cos c\big],$$

and the polygon’s share becomes the circle’s times a factor:

$$\frac{\text{polygon}}{\text{circle}} = \frac{n\tan c/\pi}{\Big[1 + \dfrac{2\big(\cos(c-|\Delta|/2) - \cos c\big)}{P\cos c}\Big]^2}.$$

The factor $4A/P^2$ is in both shares, and it divides out. Everything about how the circle’s share changes as the triangle changes shape — its slow, flat-topped loss away from the equilateral — is also in the polygon’s share, and the comparison between them never sees it.

The ratio at the equilateral is the numerator alone, $n\tan(\pi/n)/\pi$: 1.1027 for the hexagon, 1.0427 for the nonagon, 1.0235 for the dodecagon, 1.0058 for the 24-gon. That number has a much older name. A regular $n$-gon circumscribed about a circle of radius one has half-perimeter $n\tan(\pi/n)$, and Archimedes bounded $\pi$ from above with exactly that quantity, doubling the sides from six to ninety-six. On the equilateral sheet a multiple of three shares the triangle’s incircle as its own, so its area over the circle’s is Archimedes’ upper bound over $\pi$. The polygon’s lead over the circle, at the top of the window, is the error in Archimedes’ estimate of $\pi$ from that polygon — 10 per cent at six sides, half a per cent at twenty-four.

What brings each curve down is the bracket in the denominator, and it holds only two things: the corner’s extra reach, $\cos(c - |\Delta|/2) - \cos c$, and the perimeter it is divided by.

Each edge, in closed form #

A window ends where the ratio is one. Write $u$ for half the apex’s departure, so the legs have turned by $u$, and $\lambda_n = \sqrt{n\tan c/\pi} - 1$ for how far the square root of the peak ratio sits above one. The edge is where

$$\cos(c - u) - \cos c = \tfrac12\,\lambda_n\,P\cos c.$$

On the left is the corner; on the right the lead, scaled by the perimeter. The perimeter is $2 + 2\sin(30^\circ \pm u)$, plus when the apex opens and minus when it closes, and expanding both sides turns the condition into a single trigonometric equation in $u$:

$$a\cos u + b\sin u = d,\qquad a = \cos c\,(1 - \tfrac{\lambda}{2}),\quad b = \sin c \mp \tfrac{\sqrt3}{2}\lambda\cos c,\quad d = \cos c\,(1 + \lambda).$$

An equation of that form has a solution in one line: $u = \operatorname{atan2}(b, a) - \arccos\big(d/\sqrt{a^2 + b^2}\big)$, and the edge of the window is $2u$ from sixty degrees.

For the hexagon it can be done by hand. With $c = 30°$ and $\lambda = 0.0501$, the opening side has $a = 0.8443$, $b = 0.4624$ and $d = 0.9094$; the angle of $(a, b)$ is 28.71 degrees and the arc cosine 19.16, so $u = 9.55$ and the window opens to 19.11 degrees past sixty. The closing side has $b = 0.5376$, an angle of 32.48 and an arc cosine of 24.70, so $u = 7.79$ and the window closes to 15.57 degrees short. Those are the measured edges, 79.11 and 44.43 degrees, to the second decimal.

The table sets the closed form against the fit for every multiple of three to twenty-four sides, and the two agree to better than a thousandth of a degree on every edge: the nonagon from 49.16 to 72.47, the dodecagon from 51.70 to 69.23, the 24-gon from 55.73 to 64.51. The fitting is no longer evidence for the windows; it is a check on a formula that has no fitting in it.

Why opening has more room #

The earlier essays found every window wider on the opening side and offered no reason. The closed form has one, and it is the single sign in $b$.

Opening the apex lengthens the base, and the perimeter with it. The corner’s extra reach is divided by the perimeter, so the same turn of the legs costs the polygon less on a triangle with a longer base. Closing the apex shortens the base and concentrates the same cost on less perimeter. The $\mp\tfrac{\sqrt3}{2}\lambda\cos c$ in $b$ is exactly the perimeter’s slope, $\cos 30°$, times the lead.

The asymmetry is therefore proportional to the lead, and the lead fades with the sides. The ratio of the opening half to the closing half is 1.227 for the hexagon, 1.150 for the nonagon, 1.112 for the dodecagon and 1.055 for the 24-gon, falling at every step toward one. A polygon whose lead over the circle is small is barely affected by how the perimeter moves, because there is barely any lead for the perimeter to scale.

The corner bends as far as it reaches #

The first-order estimate kept the corner’s reach as a straight line, $\cos(c - u) - \cos c \approx u\sin c$, and the earlier essay found its limit to be ninety degrees over the number of sides. For many sides the closed form says something else, and the difference is not small.

When the sides are many, $c$ is small, the lead $\lambda_n$ is close to $\pi^2/6n^2$ and the perimeter is close to 3, and the corner’s reach is $\cos(c - u) - \cos c \approx u(2c - u)/2$. The edge condition becomes $u(2c - u) = c^2/2$. The $u^2$ is the same size as the $2cu$, because the window is as wide as the corner is blunt: a polygon of many sides has a shallow corner, and the window it keeps over the circle is a fixed fraction of that corner’s angle. Keeping only $2cu$ gives $u = c/4$, a window of $c/2$ each way — ninety degrees over the sides. Solving the whole equation gives $u = (1 - 1/\sqrt2)\,c$, a window of

$$\Delta = (2 - \sqrt2)\,\frac{180^\circ}{n} \approx \frac{105.4^\circ}{n}$$

each way. The first-order estimate is short by a factor of $4 - 2\sqrt2 = 1.172$, at every size, and the gap does not close as the polygons multiply.

The figure measures every half-window in units of ninety degrees over the sides. Not one point is at one. The opening halves start at 1.27 for the hexagon and come down; the closing halves start at 1.04 and climb; at ninety-six sides — the polygon Archimedes stopped at — they are 1.179 and 1.164, and the limit between them is 1.1716. The two halves approach it from opposite sides because the perimeter’s help on the opening side and its hindrance on the closing side both fade with the lead, while the corner’s bend, which widens both, does not fade at all.

That settles the caveat the earlier essay left: “on every polygon measured the first-order window is the narrower one… although nothing here proves it is narrower for every polygon”. The closed form makes it the narrower one on every multiple of three to ninety-six sides and short by a fixed factor in the limit, and it says why: the first-order line overstates how fast a corner’s reach grows, and it overstates it most for the bluntest corners, which are the ones with the narrowest windows.

The two effects were one #

The earlier account of why the measured windows ran wide named two effects pulling opposite ways. The polygon’s curve bends below its straight lines further out, so it loses more than its rate says, which narrows the window; the circle loses share too, slowly and quadratically, so the finishing line moves away, which widens it; and the second won.

Both of those are the factor $4A/P^2$. The polygon’s share bends below its lines mostly because the triangle’s own shape factor is falling, and the circle’s share falls because of the same factor, at the same rate. In the ratio they cancel exactly. What is left — the bracket — curves the other way: the corner’s reach rises more slowly than its straight line, so the polygon loses less than the rate says, and the window is wider than the estimate for a reason inside the polygon rather than inside the circle. The measured windows were right, the estimate was short, and the explanation of the gap had the right sign for the wrong reason.

Where the polygon leaves its base #

The formula assumes the polygon keeps a side on the base, and the median argument guarantees that only for small turns, since it treats the corner’s reach as a straight line. Far enough from the equilateral a polygon turned off the base fits larger. Where that happens is found here by fitting rather than derived.

Found by fitting outward from the equilateral until a free polygon beats the formula, the range over which the polygon keeps its base is wide for few sides and narrows with many: the hexagon from 33.0 to 114.8 degrees of apex, the nonagon from 39.7 to 93.7, the 24-gon from 51.0 to 71.0. Every window lies inside its range. The 24-gon’s window, 55.7 to 64.5, sits within its 51.0 to 71.0 with several degrees to spare, so the closed form is the whole story of the window and the turning polygon is a story about triangles on which the polygon has already lost.

What a folder takes from it #

A folder cutting a triangle by eye wants to know how crooked it may be before the polygon being folded stops being the best use of the paper. The closed form answers that for any multiple of three, and for many sides the answer is a rule that fits on a fingertip: about 105 degrees divided by the sides, each way, a little more opening than closing. The 24-gon tolerates about four and a half degrees, the 48-gon a little over two. The first-order rule, ninety over the sides, is still a safe figure — it is narrower on every polygon checked — but it gives away a sixth of the tolerance that is actually there.

Exact is not accurate drew a line between a construction that is exact and one that is accurate, and the windows here are a version of that line drawn on the sheet itself: an exact construction on a sheet that is only nearly regular keeps its advantage for a width of error that can now be computed rather than measured. What buys the reach costs the accuracy priced how errors grow along a construction; this is the error a construction is allowed to start with.

Where the equality stops being one #

Triangles that are not isosceles. The equality holds on any triangle, but the best rotation on a scalene triangle sits at the weighted middle of three different turns, and which side the polygon keeps flat changes with the triangle’s shape. The window becomes a region in a plane of two angle errors rather than an interval, and nothing here draws its edge.

Sheets with more sides. The equality is a triangle’s: three sides, three unknowns, every constraint tight. On a nearly square sheet or a nearly regular pentagon the programme has more constraints than unknowns, the optimum leaves some slack, and the weighted sum stays a bound. Turning is uphill all the way found closed forms on rectangles by a different route, through the sheet’s two widths, and an odd polygon fits like its double used the same widths to hand every odd polygon its double’s share; that is the box’s own simplification, and it does not carry over to a triangle.

Whether the first-order window is narrower for every polygon. It is narrower for every multiple of three checked, to ninety-six sides, and narrower by a fixed factor in the limit, which is strong evidence and not a proof for each $n$.

A perfect triangle and a perfect polygon #

The sheet is a perfect isosceles triangle with straight sides and the polygon is perfectly regular. Shares are areas; the width of a crease and the thickness of the paper are not represented, and neither is the folder’s ability to find the optimal polygon by eye, which the worst angle sets the price noted is not guaranteed on a crooked sheet.

Every edge computed twice #

The one-line share is compared with polygons fitted by a rotation search of 720 steps and a golden-section refinement, with every rotation and position free, at nine apex angles for each polygon from six sides to twenty-four. Inside the range where the polygon keeps its base the two agree to within $10^{-15}$ of the sheet, and the range itself is found by bisecting where they part.

Every window edge is computed twice: from the closed form, and by fitting polygons, bisecting where they cross the circle, and reading the apex. The two are required to agree to within five thousandths of a degree on every edge, and they agree to better than a thousandth. The ratio is required to equal one at both closed-form edges and $n\tan(\pi/n)/\pi$ at the equilateral, and both edges are required to lie inside the polygon’s range.

The limit is checked at ninety-six sides, where both halves must lie within one per cent of $4 - 2\sqrt2$, and every half on the figure must lie above the first-order estimate’s one.

Still open: the edge of the region #

The scalene case is the natural continuation, and it has a definite shape. On a triangle with angles $60 + x$, $60 + y$ and $60 - x - y$ the polygon keeps flat whichever side has the middle turn, the other two meet corners, and the same equality gives the share in one line on each of the regions where the middle side does not change. The window becomes a closed curve in the plane of $(x, y)$, made of arcs from the same kind of equation, and the worst angle sets the price predicts its first-order shape: a hexagon, since the first-order loss depends only on the largest of $|x|$, $|y|$ and $|x + y|$. Whether the exact region is that hexagon pushed outward by the same factor near 1.17, or something with the perimeter’s asymmetry folded into it, is a calculation with every piece already in hand.

Sideways from here, the Archimedes connection deserves following. The lead of every multiple of three over the circle is the excess of a circumscribed polygon’s perimeter over the circle’s, and every even polygon beats every odd one and a sheet rewards the multiples of its sides both compared polygons sharing an incircle with a sheet. On a square sheet the multiples of four share the square’s incircle, and their lead over the circle is the same Archimedean excess — unlike the polygons of four-k-plus-two sides that the crossing is as hard as the polygon followed as they turn to press all four edges of a lengthening sheet, which touch the square without sharing its incircle; whether their windows on a nearly square rhombus come out as cleanly as the triangle’s do is the obvious first test.

The habit worth carrying is about bounds. When a derivation weighs constraints into a bound, count the constraints against the unknowns before treating the bound as an estimate. Three sides and three unknowns made the bound an equality all along, and every approximation built on top of it — the rate, the estimate, the second-order terms it said were missing — was an approximation to something that had been exact from the first line.