The worst angle sets the price

The rule has a width took the cleanest result about regular sheets of paper and asked how long it survives a sheet that is not quite regular. 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 sides. On an isosceles triangle with its apex a degree away from sixty, the hexagon loses 0.45 of a percentage point, the nonagon 0.27 and the dodecagon 0.20, and every polygon’s lead over the circle runs out somewhere within twenty degrees.

Those rates were measured, by fitting each polygon into each triangle and reading the slope. The essay noticed that rate times sides came to 2.7, 2.4 and 2.3, “close to a rate inversely proportional to the number of sides, but not exactly”, and it named the theoretical successor: differentiate the fitting problem at the equilateral triangle and find out whether the product tends to a limit. It does, and the derivative is short. The rate is $2n\tan^2(\pi/n)/9\sqrt3$ of the sheet per radian. It depends on nothing about the triangle except its largest angle error, and the limit of rate times sides is $2\pi^2/9\sqrt3$.

The fitting problem, read at its corner #

Hold a regular $n$-gon at a fixed rotation. Its size and position are then three numbers, its circumradius $R$ and its centre, and every side of the triangle gives one inequality: the polygon’s reach toward that side, plus how far its centre has moved toward it, must not exceed the side’s distance from the origin. The reach is $R$ times the polygon’s support in that direction, the largest cosine between the direction and any of its corners. That is a linear programme in three unknowns with three constraints, and its optimum makes all three tight.

Weight each inequality by the length of its side and add them. A closed shape’s side lengths times their outward normals sum to zero, so the centre drops out of the weighted sum entirely, and what is left is a statement about the size alone:

$$R \sum_i L_i\,\text{reach}_i \;\le\; \sum_i L_i h_i = 2A,$$

with $L_i$ the side lengths, $h_i$ their distances from the origin and $A$ the triangle’s area. The largest polygon is the one for which the support sum is smallest.

On the equilateral sheet a multiple of three can face all three sides with sides of its own, and the support in each of those directions is the apothem, $\cos(\pi/n)$ times the circumradius. Turn a sheet side by a small angle $\delta$ away from facing a polygon side and the support stops being an apothem: the nearest thing to the turned side is now a corner, and the support grows to $\cos(\pi/n - |\delta|) \approx \cos(\pi/n) + \sin(\pi/n)\,|\delta|$. The absolute value is the corner. It is why the loss starts at first order and is the same on both sides of the equilateral, where a smooth support would give a loss only at second order.

The polygon can also turn, by some angle $\phi$, which changes all three mismatches at once. With three sides of equal length, the support sum’s first-order excess is proportional to $|\delta_1 - \phi| + |\delta_2 - \phi| + |\delta_3 - \phi|$, and the turn that makes it smallest puts $\phi$ at the middle of the three. What is left is the gap between the largest mismatch and the smallest, the range of the three sides’ rotations. Turning is uphill all the way watched a hexagon on a nearly square sheet turn as the sheet lengthened; here the turn is decided in one step, by putting the polygon’s facing halfway between the sheet’s two most different sides. For the triangle’s sides that range is exactly the largest of its three angle errors: rotate the angles by $x$, $y$ and $-(x+y)$ from sixty and the sides’ directions differ pairwise by $|x|$, $|y|$ and $|x+y|$.

Put the pieces together with the polygon’s area, $nR^2\sin(\pi/n)\cos(\pi/n)$, and use the fact that the circle’s own ratio $4A/P^2$ is stationary at the equilateral triangle. The circle’s share peaks there, so the sheet’s shape contributes nothing at first order. The result is

$$\text{share} \approx S_0\left(1 - \tfrac23\tan\frac{\pi}{n}\cdot\varepsilon\right), \qquad S_0 = \frac{n\tan(\pi/n)}{3\sqrt3},$$

with $\varepsilon$ the triangle’s largest angle error in radians. On the isosceles family that error is the apex change, and the loss rate is

$$\frac{2n\tan^2(\pi/n)}{9\sqrt3} \text{ a radian}.$$

Measured and derived #

The formula gives 2.015 percentage points a degree for the triangle itself, 0.448 for the hexagon, 0.267 for the nonagon and 0.193 for the dodecagon. The earlier essay’s 0.20 for the dodecagon was measured over a whole degree, and the curve has already begun to bend by then. Fitting each polygon two hundredths of a degree either side of the equilateral, so that the bend cannot enter, the measured rates match the formula to within 0.27 per cent at every multiple of three from three sides to forty-eight.

Rate times sides does tend to a limit, and the limit is $2\pi^2/9\sqrt3$ a radian, 2.21 percentage points a degree. For large $n$, $\tan(\pi/n)$ is very nearly $\pi/n$, so the rate is very nearly $2\pi^2/(9\sqrt3\,n)$: inversely proportional to the sides, as the earlier essay suspected, with the factor approached from above. The products are 2.69 for the hexagon, 2.40 for the nonagon, 2.32 for the dodecagon, 2.24 for the 24-gon and 2.22 for the 48-gon.

The derivation also says why the earlier essay’s reading of the products was right. It read the rates as “how much of its boundary the coincidence was holding against the sheet”. In the formula, that is the factor $\tan(\pi/n)$ twice over: once from the polygon’s share, $S_0$, and once from the size of the corner, $\sin(\pi/n)/\cos(\pi/n)$, which is how far a corner reaches past an edge for a given turn. A polygon with many sides has short edges and blunt corners, and a blunt corner reaches barely past the edge it replaced.

Why the triangle loses fastest #

The formula’s smallest case is the regular triangle itself, the polygon that is the sheet. With $n = 3$ the share at the equilateral is one, and $\tan^2(\pi/3) = 3$, so the rate is $2/\sqrt3$ a radian, 2.015 percentage points a degree, four and a half times the hexagon’s. The earlier essay measured 2.0 and explained it as the case where the whole of the polygon’s perimeter lies on the sheet’s. The derivation gives the mechanism behind that explanation. A triangle’s corners are the sharpest a regular polygon has, so when the sheet turns away from facing one of its sides, the corner that takes the side’s place reaches furthest past it: $\tan(\pi/3)$ is 1.73 against the hexagon’s 0.58. The sharper the corner, the more a small turn of the sheet costs, and the triangle’s corners are as sharp as regular corners get.

The same factor runs the other way at the top of the range. A 48-gon’s corners are almost straight, and a turn of the sheet barely changes what touches it. It loses 0.046 points a degree and has almost nothing to lose: its lead over the circle is a tenth of a point.

Why the non-multiples have no corner #

The polygons that are not multiples of three behaved differently in the earlier essay, and the derivation says why in one line. On the equilateral sheet a pentagon or an octagon does not face all three sides with sides of its own, so in at least one direction its support is a smooth function of the turn, with no corner. A smooth support changes at first order with a signed slope, not an absolute value. So the share of a non-multiple has a slope at the equilateral, with at most a smaller corner on top of it, where some of its sides do lie flat against the sheet. It can rise one way while it falls the other, which no multiple of three can do, and the earlier essay’s octagon gaining 0.12 points for the first degree the apex opens is that slope showing.

That is the whole difference between the two kinds of polygon, and it is visible in the support sum. A multiple of three sits at a corner of the sum in every direction, where every turn of the sheet costs it something. A non-multiple sits at least partly on a smooth part, where some turns help. The equilateral triangle is a peak for the first kind and an ordinary point for the second, and that is why the ranking inverts within twenty degrees: one kind can only lose, and the other gains on one side of sixty.

Any triangle, not only the isosceles #

The isosceles family moved two of the triangle’s angles at once and kept the base fixed. The derivation never used the isosceles shape, only the largest angle error, so it predicts that every nearly equilateral triangle, whatever its shape, costs a polygon the same amount for the same worst angle.

On 72 random triangles with every angle within 0.4 degrees of sixty, the fitted loss agrees with the prediction from the worst angle alone to within 2.3 per cent, the residue being the curve’s bend at that size. A triangle with one angle a degree out costs the hexagon the same 0.45 points as one with two angles half a degree out each way, because in both the largest error is a degree. A triangle with angles 60.5, 60.5 and 59 has its worst error at one degree and loses the same as one with angles 61, 59.5 and 59.5.

That turns the earlier essay’s tolerance into a single number a folder can check. The worst angle of a hand-cut triangle is read with a protractor, and the hexagon’s share is $66.67 - 0.448$ points a degree of it, with no need to know which way the triangle is bent.

The window is the lead over the rate #

The earlier essay’s most practical result was a set of windows: how far the apex can move before each multiple of three falls behind the circle. The circle loses only at second order, so to first order a window is the polygon’s starting lead divided by its rate. The lead is $(n\tan(\pi/n) - \pi)/3\sqrt3$, the polygon’s share less the circle’s.

For large $n$ the lead is about $\pi^3/(9\sqrt3\,n^2)$ and the rate about $2\pi^2/(9\sqrt3\,n)$, so their ratio is $\pi/2n$ radians. The window shrinks as ninety degrees over the number of sides. For the 24-gon the estimate is 3.73 degrees against the limit’s 3.75, and the measured window is 4.3 degrees closing the apex and 4.5 opening it. For the hexagon the estimate is 13.9 degrees, against measured halves of 15.6 and 19.1.

The estimate falls short of both halves, and by more for the hexagon, because of two effects the first-order argument leaves out on purpose, and they pull in opposite directions. The polygon’s own curve bends below its straight lines further out, as the cone figure shows, so it loses a little more than the rate says, which would narrow the window. The circle loses share too, slowly and quadratically, and the finishing line moves away. At eight degrees from the equilateral the hexagon has lost half a point more than its lines predict and the circle two thirds of a point, so the second effect wins. Both grow with the distance travelled, and the hexagon travels furthest. On every polygon measured the first-order window is the narrower one, so it is the safe figure to design to, although nothing here proves it is narrower for every polygon.

What this says about cutting a triangle #

A folder cutting an equilateral triangle from a square by eye is off by a degree or two in the worst angle. The formula says what that costs before anything is measured. A worst angle of two degrees takes 0.9 points from the hexagon’s 66.7 per cent and leaves it ahead of the circle by more than five. It takes 0.19 points from the 24-gon, which started only 0.35 ahead, so more than half of the 24-gon’s lead is gone, and at four degrees all of it is. For a polygon a folder would actually make, the regular-sheet rule holds at any accuracy a careful hand reaches. For the polygons that nearly tie with the circle anyway, it fails at accuracies a careful hand can easily miss.

Exact is not accurate made the same kind of point about divisions: an exact construction and an accurate one are different things, and the second is the one a folder holds. Here the exact sheet makes a sharp ranking and the accurate sheet keeps most of it, and the formula says which parts survive. The surviving parts are the polygons with few sides, whose large leads are spent slowly.

What the picture cannot show #

The second-order terms. Everything here is the first derivative at the equilateral triangle. The bend of each polygon’s curve and the circle’s own loss are measured in the figures and not derived, and they are what makes the measured windows wider than the guarantee.

Sheets that are not triangles. The argument used one fact peculiar to a triangle: any small perturbation of its three sides is a rotation of each, and the support-sum weights are equal. A nearly square sheet has four sides and two families of polygons that match it, the multiples of four and those of the crossing that is as hard as the polygon. Its rate would come from the same linear programme, with the range of four rotations in place of three, and it has not been derived.

Whether a hand can find the polygon. The share is for the largest polygon placed by computation. A folder placing a hexagon by eye on a slightly crooked triangle does not find the optimal turn, and a hexagon laid flat against the base is not the best one on a scalene sheet.

The idealisations underneath #

The sheet is a perfect triangle with straight sides, and the polygon is a perfect regular polygon, fitted by an optimisation that searches every turn and solves for the largest size at each. Shares are areas, and paper’s thickness and the width of a crease are not represented. The derivation is first-order in the angle error.

How the claims were checked #

The rate is checked by fitting each polygon, with a rotation search 720 steps fine and a golden-section refinement, at the equilateral triangle and two hundredths of a degree either side, and comparing the slope with $2n\tan^2(\pi/n)/9\sqrt3$. The two must agree to within half a per cent at every multiple of three from three sides to forty-eight, and they agree to 0.27 per cent. The cone figure also checks the derived lines at a tenth of a degree either side for every polygon its dial offers.

The worst-angle law is checked on 72 random triangles, 24 for each of three polygons, with every angle within 0.4 degrees of sixty. The fitted loss must match the prediction from the largest angle error to within four per cent.

The windows are found by stepping the apex outward and bisecting where the polygon’s share meets the circle’s, and each must contain its first-order estimate on both sides. The estimate for the 24-gon must lie within two per cent of ninety degrees over twenty-four.

Still open: the sheet that is nearly square #

The derivation is general in everything but its last step. On any convex sheet the largest regular polygon solves a linear programme whose dual weights are the sheet’s side lengths, and on a sheet that matches the polygon’s directions the support sum has a corner in every matched direction. On a square, the same argument would give the hexagon’s and the octagon’s rates on a nearly square rectangle in closed form, and the square is a choice already set up the family. The open part is that a rectangle’s perturbation is not a rotation of its sides but a stretch, which moves the sides’ distances rather than their directions. What the support sum does under a stretch is a different derivative, and it is the one an odd polygon fits like its double would want.

The other open part is second order. The windows run wider than the guarantee because of two curvatures, the circle’s and the polygon’s, and both are derivatives of the same programme one order up. Computing them would turn the windows into closed forms as well. They are where the earlier essay’s asymmetry, more room above sixty degrees than below, would come from. Every even polygon beats every odd one and a sheet rewards the multiples of its sides are both statements about directions matching, and this essay prices the first degree of mismatch for one of them.

The habit worth carrying is about measured slopes. When a measured rate times a parameter comes out nearly constant, look for the programme it is the derivative of. The products 2.7, 2.4 and 2.3 were a curve approaching a limit, and the curve and its limit were written down in one line once the corner in the support function was seen.