Half the difference is a theorem

The event is an incircle gave the tree method’s designers a check they could make with a ruler. A convex quadrilateral’s universal molecule has at most one crease that touches neither the outline nor a corner, the crease that vanishes exactly when the quadrilateral has an inscribed circle. By Pitot’s theorem that happens when the two pairs of opposite sides have equal sums. On six hundred random quadrilaterals the crease was never shorter than half the difference between the sums, so a layout whose sums differ by more than twice the shortest crease a folder can place cannot have a crease that short. The essay said plainly that the bound “held on every quadrilateral drawn and is not proved here”. Six creases belong to the square found it again on 1,358 of the creases of random pentagons, and the record of that work noted it holding at exactly 0.5000 on its least case: a bound that looked like a theorem and was not one.

It is one, and the proof is short enough to fit in a paragraph. More usefully, the proof does not stop at a bound. It gives the crease’s length exactly, from four sides and two angles, and it says why the same check exists for a hexagon and an octagon and cannot exist for a pentagon.

A sum that shrinking cannot change #

The universal molecule is the straight skeleton of its polygon, the same object one straight cut folds a drawing along, and the skeleton is the record of what happens when the polygon is shrunk by moving every side inward at the same speed. Each corner slides along its bisector, the sides shorten, and the moment a side shrinks to nothing is a node where creases meet.

How fast does a side shorten? Move it inward by a small distance $t$ and each of its two ends is cut back by $t\cot(A/2)$, where $A$ is the angle at that end. So side $i$, between corners $i$ and $i+1$, has length

$$s_i(t) = s_i - t\left(\cot\frac{A_i}{2} + \cot\frac{A_{i+1}}{2}\right)$$

until something happens. Now take the alternating sum of the four sides, $s_1 - s_2 + s_3 - s_4$. Every corner’s cotangent appears in two neighbouring sides, once with each sign, and every one of them cancels. The alternating sum of the shrunk quadrilateral is the alternating sum of the original, exactly, at every stage.

The figure shows it on a quadrilateral with sides 0.965, 0.628, 0.740 and 0.777. At each of the five stages drawn the four sides have changed, the second down to almost nothing, and the alternating sum reads 0.301 every time. That number is the Pitot difference, and Pitot’s theorem is the case where it is zero. A quadrilateral with no incircle carries its distance from one unchanged through the whole shrink, and that alone forces the crease to exist.

The triangle left at the first event #

Shrink until the first side vanishes. Which one goes first is a matter of arithmetic: side $i$ vanishes at $t = s_i / (\cot\frac{A_i}{2} + \cot\frac{A_{i+1}}{2})$, and the smallest of the four wins. In the drawn quadrilateral it is the second side, at 0.330. What is left is a triangle, whose three sides are the shrunk versions of the other three, and whose alternating sum is still $d = 0.301$, with the vanished side counted as zero.

A triangle has an incircle, and the distance from any corner to where the circle touches is half of the two sides at that corner minus the third. At the new corner, where the vanished side used to be, that tangent length is half of the triangle’s sides taken with exactly the signs of the alternating sum. So it is $|d|/2$, whichever side vanished. The triangle then shrinks to its incentre, and the interior crease is the path the new corner takes there: along its bisector, from the corner to the centre. A corner whose angle is $C$ reaches the centre after travelling the tangent length divided by $\cos(C/2)$, and the new corner’s angle is what the two vanished side’s neighbours make when extended: $C = A_k + A_{k+1} - \pi$. Put together,

$$L = \frac{|a - b + c - d|}{2\,\sin\frac{A_k + A_{k+1}}{2}},$$

with $A_k$ and $A_{k+1}$ the angles at the two ends of the side that vanishes first. The sine is at most one, so $L \ge |d|/2$, and the measured bound is a theorem.

On the drawn quadrilateral the two angles at the second side are 74.9 and 118.1 degrees, which add to 193.0. The formula gives 0.1512, and the skeleton, computed by shrinking the shape event by event, gives 0.1512. On 599 random quadrilaterals the two agree to two parts in $10^{14}$, which is the rounding of the arithmetic, and every one of them sits on the curve $1/\sin(x/2)$ drawn against the angle sum $x$.

Where the bound is tight #

The formula says what the scatter in the earlier essay was. Its cloud of points rose above the line at half the difference and spread upward, and the spread was attributed to “unequal angles”. It is exactly one over the sine of half the angle sum at the vanishing side, so it depends on two angles and nothing else. A crease near the bound needs those two angles to add to a little more than a straight angle, which means the two sides beside the vanishing one are nearly parallel.

The trapezoid makes that visible. Its right side vanishes first, and the two corners at its ends are a right angle and the angle the tilted top makes. Drag the tilt toward zero and the shape approaches a rectangle 1.3 by 1, whose skeleton is a horizontal crease 0.3 long between two forks. At a tilt of 0.04 the angles add to 181.8 degrees and the crease is 0.3203 against half the difference, 0.3203, equal to four places. At 0.36 they add to 195.5 and the crease is 0.509 against 0.504. At 0.68 they add to 207.6 and it is 0.745 against 0.724.

That gives a designer more than a bound. With four side lengths and two angles the crease is known exactly, before any skeleton is drawn. That is a check worth having on a layout of many molecules where one crease too short to fold is the defect being hunted. The rule for which side vanishes first is part of the check: divide each side by the sum of the cotangents of its half-angles and take the smallest.

Reading it off a layout #

The check is most useful in millimetres, on a molecule that has already been placed. Take the drawn quadrilateral as one cell of a layout on a sheet where it spans 150 mm. Its sides are then 144.8, 94.2, 111.0 and 116.5 mm, the alternating sum is 45.1 mm, and the bound alone promises a crease of at least 22.5 mm with no skeleton drawn. The angles at the second side, the one that vanishes first, add to 193.0 degrees, so the exact crease is 22.5 divided by the sine of 96.5 degrees, 22.7 mm. On a cell like this the bound is within one per cent of the truth, because the two sides beside the vanishing one are within thirteen degrees of parallel.

The case a designer actually worries about runs the other way. Suppose a quadrilateral’s opposite sides sum to 210.0 and 211.0 mm, a difference of one millimetre. The bound then guarantees only half a millimetre, which on most papers is below the shortest crease that can be placed. Whether the real crease is that short depends on the two angles. At an angle sum of 193 degrees, as here, it is 0.503 mm. At 240 degrees it is 0.577, and at 300 it is a full millimetre. A near-incircle is dangerous only when the vanishing side’s neighbours are close to parallel, and when they are not, the angles buy back some of the length the sums took away. The earlier check could only say the crease was not shorter than half the difference. This one says how much longer it is, and why.

Every even polygon keeps the bound #

The shrinking argument used one thing about the quadrilateral besides its four sides: that the first event leaves a triangle, whose tangent lengths are fixed by its sides. A hexagon’s first event leaves a pentagon, and the argument stops there. A different argument does not.

Each side of a polygon owns one face of the skeleton, the region of points nearer to it than to any other side. The face’s boundary is the side itself and a chain of creases from one end of the side to the other, so the side’s length is the chain’s projection onto the side’s own direction. Write the alternating sum of the sides as the alternating sum of those projections. A crease from a corner is a bisector, so it makes the same angle with both sides at that corner and projects onto both with the same length. In the alternating sum those two sides carry opposite signs, and the corner crease cancels. Every corner crease cancels. What survives is the interior creases, each the boundary of two faces and so projected onto two sides, by at most its own length each time:

$$|s_1 - s_2 + s_3 - \dots - s_{2k}| \;\le\; 2 \times \text{(total length of the interior creases)}.$$

On any convex polygon with an even number of sides, the interior creases add up to at least half the alternating sum. For a quadrilateral there is one interior crease and this is the bound already proved. For a hexagon and beyond it is new, and it says something the shrinking argument could not: a hexagon whose alternating sum is far from zero cannot have all its interior creases short.

On three hundred random polygons of each even size the least ratio is 1.000 for quadrilaterals, 1.036 for hexagons, 1.121 for octagons and 1.248 for ten sides. The medians run 1.02, 1.79, 2.64 and 3.12. The bound is sharp for quadrilaterals and increasingly loose beyond them, because a polygon with more sides has more interior creases than the alternating sum needs. Each projection is a cosine, and the more creases there are, the less of each one lies along the sides it is projected on. For a quadrilateral the single crease is nearly parallel to both its sides in the tight case; for a ten-sided polygon nothing makes all of its seven interior creases do that at once.

Why five sides keep nothing #

The projection argument cancels every corner crease against its neighbour. Around a polygon with an odd number of sides the signs cannot alternate all the way: going round once, the first and last sides carry the same sign, and the corner between them does not cancel. An odd polygon has no alternating sum that shrinking preserves, because one corner’s cotangent is counted twice instead of cancelling. So there is no bound to give.

The pentagon in the figure is drawn round a circle, so its molecule is a single node where all five creases meet, with nothing in between. Its alternating sums, one for each side the count starts from, are all well away from zero, and each is exactly twice the tangent length at the corner where the first and last signs collide. That is the same fact the pentagon essay reached from the other side. Five equations for five tangent lengths always have a solution, so the sides alone never rule an incircle out or in. The quadrilateral’s ruler check is a property of evenness, not of the number four, and the pentagon’s lack of one is a property of oddness rather than of difficulty.

The pentagon essay’s 1,358 creases fall into place too. Each was a crease the pentagon shares with the quadrilateral its four side lines make, so it is that quadrilateral’s interior crease, and the theorem applies to it. The 242 that fell below the bound were creases the quadrilateral draws differently, where a different side vanishes first, and the theorem never applied to them. What was found by sampling was this theorem applied to the cases it covers.

What the picture cannot show #

Whether the crease can be folded. The bound says how short the crease can be from the sides, not whether a crease of that length is foldable in a given paper. That threshold belongs to the paper’s thickness, as the last free parameter and the incircle essay both set it, and the theorem only turns a ruler reading into the crease length that threshold is compared with.

The molecule’s other creases. The universal molecule is the straight skeleton and the perpendiculars dropped from its nodes. The perpendiculars, which carry the hinges between flaps, are not interior creases in this sense and have no bound here. A designer’s shortest crease can be one of them.

The tree method’s other ways to fill a polygon. Lang’s universal molecule is one filling, and the theorem is about it. Gusset molecules and other fillings place their creases by other rules, and where they share the straight skeleton they share the bound; where they do not, nothing here applies.

The idealisations underneath #

The polygon is convex and its molecule is its straight skeleton, shrunk with every side moving at the same speed. A reflex corner splits the shrink and the molecule that does not exist is the result, so the argument is for convex polygons only, as the molecule is. Lengths are in the polygon’s own units, and the crease is a line with no width.

How the claims were checked #

The invariant is checked on the drawn quadrilateral by shrinking it to five stages, computing every side from the shrunk polygon itself, and requiring the alternating sum to agree with the original to $10^{-9}$ at each.

The exact length is checked against the skeleton, which is computed by an event-by-event shrink that never forms a cotangent sum or a sine. The two must agree to $10^{-9}$ on every one of 599 random convex quadrilaterals, and they agree to $2 \times 10^{-14}$. The trapezoid on the dial is required to have its right side vanish first and its crease match the formula at every tilt the dial offers.

The even bound is checked on three hundred random polygons of four, six, eight and ten sides: total interior crease at least half the alternating sum on every one. The odd counterexample is required to have no interior crease and five alternating sums each equal to twice the tangent length at its colliding corner.

Still open: the hexagon’s exact length #

The quadrilateral’s crease came out exactly because its first event leaves a triangle and a triangle has no freedom left. A hexagon’s first event leaves a pentagon, whose shrink is fixed by its sides and angles but carries no invariant. Whether a hexagon’s interior creases have an exact total in terms of its sides and angles, as a quadrilateral’s one crease does, is the natural next question, and the projection argument says what such a formula would have to reduce to: the alternating sum, divided by a weighted average of the cosines between each interior crease and its two sides.

The other direction is the designer’s. The check is now a formula, and the skeleton changes its mind found that a molecule’s crease count jumps at flips, where four side lines touch one circle. A flip in a quadrilateral is the shape with an incircle, where $d = 0$ and the formula gives no crease. In a pentagon a flip is a zero of one of five such quadrilateral sums, one for each side dropped. A pentagon designer could read all five off a layout and know which flips are near without computing a skeleton, which is the use the incircle essay hoped the ruler check would have. Sideways, a tree cannot argue found that molecules almost never contradict their own letters because their skeletons are trees, and the corner that splits the shrink found the event a reflex corner adds. The invariant here is a third property that shrinking keeps, and it is the only one of the three that depends on parity.

The habit worth carrying is about measured bounds. When a bound holds on every case drawn with its least case exactly on the line, look for the quantity the process keeps. The bound was tight at exactly 0.5000 because the alternating sum was being carried through the shrink unchanged. Once that was seen, the proof took a paragraph and handed over the exact length as well.