Two sides at sixty are enough

A sheet rewards the multiples of its sides found the cleanest fact in this line of argument: on an equilateral triangle of paper, 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 three essays since asked how long that survives a triangle cut a little wrong, and all three asked it of the isosceles family, with two angles equal.

The worst angle sets the price derived the first-order loss, and it did not need the isosceles shape: a polygon loses share in proportion to the spread of the triangle’s three side directions, which is the largest of its three angle errors, whatever the triangle’s shape. The window has an exact edge then solved the isosceles edge in closed form — 15.6 degrees closing the apex, 19.1 opening it, for the hexagon — and left the general triangle as a definite question. On a triangle with angles $60 + x$, $60 + y$ and $60 - x - y$, the window is a region of the $(x, y)$ plane, and the first-order law predicts its shape: a hexagon, since the worst error is the largest of $|x|$, $|y|$ and $|x + y|$, and a level set of that is a hexagon.

The exact region is not that hexagon. It is a six-pointed, scalloped shape whose points reach half as far again as the first-order hexagon, and the points are where the triangle keeps two of its sides at sixty degrees.

The plane of two errors #

A triangle’s three angles add to 180 degrees, so two numbers describe how far any triangle is from equilateral. Writing its angles as $60 + x$, $60 + y$ and $60 - x - y$ puts every triangle in a plane, with the equilateral one at the origin. The isosceles triangles are three lines through it: $x = y$, where the first two angles agree, and its two images under relabelling the corners. On each line one direction closes the odd angle and the other opens it.

The first-order law needs only one quantity from a point of this plane, the spread of the triangle’s side directions, and that is the largest of $|x|$, $|y|$ and $|x + y|$. The points where it equals a given number form a hexagon. Its corners are not on the isosceles lines. They are on the lines between them, the triangles where one angle opens, another closes by the same amount, and the third is untouched: angles $60 + t$, $60$ and $60 - t$.

To first order a hexagonal polygon loses its lead over the circle at 13.86 degrees of worst error in every direction, since the lead it starts with on the equilateral triangle — 6.2 points of the sheet — divided by its first-order rate of loss gives that figure. The isosceles essays found the real edge further out along their lines, by 1.12 and 1.38 times. The question is what happens off them.

Fitting the polygon everywhere #

Every point of the plane is a triangle, and the largest regular hexagon in a given triangle is a linear programme. At a fixed turn, its size and two coordinates of its centre are three unknowns, and each side of the triangle is one inequality on them; the best fit has all three inequalities tight. The turn is searched over and refined. The circle’s share is a formula. So along any ray from the origin, the lead of polygon over circle is a function of the distance along the ray, and the edge is its first zero, found by stepping out until the sign changes and halving.

The figure is that edge in 72 directions. Along the isosceles lines it lands where the closed forms put it, 15.57 degrees closing the apex and 19.11 opening it. Along the six lines between them it lands at 21.85 degrees, 1.58 times the first-order hexagon and 1.40 times the closing isosceles edge. Between a corner and an isosceles side the edge scallops inward, so the region is a star with six points rather than a polygon with six sides.

What the corners have that the sides do not #

The triangles at the edge show why.

On the isosceles triangles the hexagon keeps one side flat on the base, the side whose direction is the middle of the three, and meets the other two sides at its corners. That was the configuration the isosceles closed form was built on: three sides, three unknowns, and a corner reaching out to each tilted side.

On the corner triangle, with angles $60 + t$, $60$ and $60 - t$, two of the sides still meet at exactly sixty degrees. Their directions are therefore two of the hexagon’s own side directions, and the hexagon lies flush against both, with only the third side tilted away from it. One corner of the polygon has to reach a tilted side instead of two, and the polygon’s size is fixed by one reaching corner rather than by a compromise between two.

That configuration has a closed form, and it is short. Put the sixty-degree corner of the triangle at the origin. A regular polygon with a multiple of three sides, inradius $r$, lying flush against both of its arms has its centre on the bisector at distance $2r$ from the corner. The third side of the triangle lies at distance $h$ from the corner with its normal turned by $t$ from the bisector, and the polygon touches it at whichever vertex is nearest that normal:

$$2r\cos t + \frac{r}{\cos(\pi/n)}\,\cos\psi = h,$$

where $\psi$ is the angle between the normal and that vertex. Everything else — the triangle’s area, its perimeter, the incircle — is a formula in $t$, and the polygon’s share and the circle’s tie at a single $t$. For the hexagon it is 21.848 degrees, the edge the fit found in every corner direction.

How far each multiple reaches #

The same wedge holds for every multiple of three, since every one of them has sides at sixty degrees to each other.

The closed form and the fit agree to a hundredth of a degree for every polygon from six sides to twenty-four. And the corners’ advantage grows with the sides: the hexagon’s corner reaches 1.40 times its closing edge, the nonagon’s 1.49 times, the dodecagon’s 1.53. A polygon with more sides has blunter corners, and the window has an exact edge found that blunt corners are exactly what the first-order estimate overstates, because a blunt corner’s reach bends back quickly as the side it presses against tilts away. The wedge leaves only one corner pressing, so it pays that cost once where the isosceles shape pays it twice.

Between a corner and a side #

The scallops between the points of the star have a cause of their own, and it is the same cause read the other way.

Leave the corner line by a little and the triangle’s two untilted sides no longer meet at exactly sixty degrees. The polygon cannot stay flush against both, so it peels away from one and keeps the other, and from there to the isosceles line it rests on a single side with corners reaching to the other two. Checked at a scalene triangle between the two lines, with angles 70, 64 and 46 degrees, the fitted hexagon lies flush against one side, as on the isosceles triangles. The two-flush configuration exists only exactly on the corner lines, and the window’s edge rises to meet it from both sides.

Along the way the two reaching corners share the tilt between them unequally. Near the isosceles line they share it evenly, which the window has an exact edge found is the costliest arrangement, because each corner’s reach is bent by the full amount the rule allows. Near the corner line one corner takes nearly all of the tilt and the other nearly none, and a corner pressing against a side that has hardly moved costs almost nothing. The edge therefore rises slowly from the isosceles line and steeply into the corner, and the region between two points of the star is scooped out rather than straight.

That is also why the first-order law holds everywhere at the start and nowhere at the edge. To first order a corner’s reach grows in proportion to its tilt, so two corners sharing a total tilt cost the same however they share it, and only the largest error — the total — appears in the price. The curvature of a corner’s reach is what makes an uneven share cheaper than an even one, and the curvature is second order. The first-order law is exact about the slope and silent about the bend, and the window is decided by the bend.

A tolerance with a direction #

The rule has a width first measured how far a triangle could stray before its polygons lost their place, and every measurement since has quoted a number of degrees as though a triangle’s error had a size and nothing else. In the plane of two errors it has a direction too, and the direction is worth more than a third of the size.

The practical version is a question about where a triangle’s error comes from. A triangle folded on the edge of a square has two base corners fixed by the square and an apex placed by a fold, and the largest triangle in a square is one such construction. An apex that lands a little to one side of where it should widens one base angle and narrows the other by the same amount, to first order, and leaves the apex angle alone: that is a corner line, the most forgiving direction in the plane. An apex that lands a little too high or too low moves both base angles together and is an isosceles error, and too high — closing the apex — is the least forgiving direction of all. Error is folded too found that an error repeated the same way adds up where independent errors partly cancel; here what decides the cost is not how errors add but which way the one error points, and a folder placing an apex by eye does better to err sideways than up.

It is not a large effect next to the polygon’s own tolerance. The hexagon keeps its lead to 15.6 degrees in its worst direction, far more than a hand-folded triangle errs. For the 24-gon, which keeps its lead only to 4.3 degrees closing an isosceles apex and 4.5 opening it, the corner direction stretches the tolerance to 6.9, and that is the difference between a careful hand winning and losing.

The lead falls, and comes back #

Following the corner direction further than the window’s edge turns up something the first-order picture could never show.

The hexagon’s lead falls through zero at 21.8 degrees and stays below. The dodecagon’s falls through zero at 12.7 degrees, bottoms out, and comes back above zero at about eighteen degrees, rising to 1.14 points at thirty. The eighteen-gon dips just below zero near ten degrees, returns, and peaks at twenty.

The reason is the vertex the polygon touches with. As the third side tilts, the nearest vertex of the polygon to its normal moves further off, the polygon shrinks, and it loses. But a polygon has a vertex every $360/n$ degrees, so when the tilt reaches half that spacing a new vertex becomes the nearest, and as the tilt keeps growing the new vertex approaches the normal instead of leaving it. At a tilt of exactly $360/n$ degrees, the third side lies along one of the polygon’s own side directions, and the polygon lies flush against all three sides of the triangle — as it does on the equilateral one.

An island far from equilateral #

The thirty-sixty-ninety triangle, the half of an equilateral triangle cut along an altitude, is nowhere near equilateral: its worst angle error is thirty degrees. Its sides run at directions thirty degrees apart, which are side directions of the dodecagon, and the dodecagon lies flush against all three sides of it, beating the inscribed circle by 1.14 points against 1.42 on the equilateral triangle.

Flush against all three sides, the polygon circumscribes the triangle’s own incircle, and the comparison is then between a polygon and the circle inscribed in it: their areas stand in the ratio $n\tan(\pi/n)/\pi$, which is more than one for every polygon. So the polygon wins exactly by that ratio on every triangle whose three sides lie along its side directions, however crooked the triangle. For the dodecagon those are the triangles whose angles are all multiples of thirty: the 90-60-30 and the 120-30-30 besides the equilateral. For the nonagon they are those with angles from 20, 60, 100 and 140 that sum to 180. For a 24-gon, many.

The window round the equilateral triangle is therefore not the whole story of where a multiple of three beats the circle. It is the largest of a scatter of islands, one round every triangle whose angles the polygon’s own geometry divides, and the first-order law, which knows only distance from equilateral, cannot see any of them.

What a folder can make #

The polygons here are the largest regular polygons that fit, found by optimisation, and the question these essays have been asking is about the best use of the paper rather than about a folding sequence. How many polygons a fold reaches counted which regular polygons can be folded at all: the hexagon, the nonagon and the dodecagon all can, so every polygon drawn here is constructible in principle on the sheet that holds it. The thirty-sixty-ninety triangle is itself the cheapest crooked triangle to fold, half of an equilateral one, and the largest triangle in a square is where the equilateral triangle itself comes from.

The model is flat and exact. Polygons and circles are ideal shapes cut from an ideal sheet, and the share is area. The tolerance a real hand needs — the paper’s thickness, a crease’s width — is smaller than any window here except the largest polygons’, and the worst angle sets the price priced that hand.

How the edge and the islands were found #

Every share is a fit. For a triangle and a polygon, the largest polygon is found by a linear programme in its size and centre at each of 240 turns, refined by golden section, and checked to lie flush against the sides the essay says it does: two at the window’s corners, one at its isosceles edges, three on the islands. Every edge is a sign change, stepped toward in quarter degrees from the equilateral triangle and halved thirty times. At the corners the wedge’s closed form is required to agree with the fit to a hundredth of a degree, for every multiple of three from six to twenty-four, and the corner is required to reach further than either isosceles edge. On the islands the polygon is required to lie flush against all three sides and to lead the circle.

Still open: the whole map of islands #

The islands have not been mapped. Each is a region of the $(x, y)$ plane round a triangle whose angles the polygon divides, and its size, like the main window’s, should follow from how many sides lie flush near it — three at its centre, two along certain lines, one elsewhere. A map of the whole plane for the dodecagon, and the shrinking of its islands against the main window as the sides multiply, would say how much of the space of triangles each multiple of three wins, which is the real form of the question these essays began with.

The square sheet has the same structure waiting. An odd polygon fits like its double and every even polygon beats every odd one compared polygons on rectangles, and on a quadrilateral that is not quite square the multiples of four should have windows with corners where two sides keep their right angle — the analogue of the wedge here — and islands on quadrilaterals whose sides the octagon or the dodecagon divides.

The habit worth carrying is about first-order laws. When a first-order law says a quantity depends on one number alone, ask which configurations keep a constraint satisfied exactly along the way. The worst angle set the first-order price everywhere; whether two sides stayed at sixty decided how long the price stayed first order.