The longest fence of diameter one

The largest polygon of diameter one asked which polygon with a given number of sides, and no two points more than one unit apart, encloses the most area. For an odd number of sides the answer is the regular polygon, as Karl Reinhardt proved in 1922. For an even number it never is: the regular hexagon spends its width on three diameters through its centre, and Ron Graham’s lopsided hexagon does better. That essay was about area. Reinhardt’s same paper of 1922 asked the other natural question — which polygon of diameter one has the longest perimeter? — and its answer is different in an instructive way.

The bound is easy to state. A convex polygon with $n$ sides and diameter one has perimeter at most

$$2n \sin\frac{\pi}{2n},$$

which is 3 for a triangle, $8 \sin(\pi/8) = 3.06147$ for a quadrilateral, and creeps up towards $\pi$ as $n$ grows. The surprise is in when it is reached: not only for odd $n$, as with area, but for every $n$ that has an odd factor greater than one — six, ten, twelve, fourteen and so on — by polygons that are mostly not regular. Only the powers of two fall short.

The bound, and why the sides are short #

A first, weaker bound takes one line. Every set of diameter one can be enlarged to a set of constant width one — a classical theorem about completing convex sets — and a convex polygon inside a convex curve has a shorter perimeter than the curve. Every curve of constant width one has perimeter exactly $\pi$, so every polygon of diameter one has perimeter less than $\pi$. Reinhardt’s bound is the sharp version for a fixed number of sides. Its proof studies the diameter graph of an extremal polygon — the pairs of vertices exactly one unit apart — and shows that in an optimal polygon every side is a chord of a unit circle centred at a vertex across from it. A chord of a unit circle subtending an angle $\theta$ has length $2\sin(\theta/2)$; the angles subtended by the $n$ sides, each from its vertex across the polygon, add to at most $\pi$; and since the sine is concave the total length is largest when the angles are all equal, to $\pi/n$. That gives $2n\sin(\pi/2n)$.

The same argument says exactly what a polygon must do to reach the bound: every side must subtend exactly $\pi/n$ from a vertex exactly one unit away. For odd $n$ the regular polygon does that — each side is seen from the opposite vertex at the angle $\pi/n$, by the inscribed angle theorem, and its long diagonals are its diameters. For even $n$ the regular polygon’s diameters run between opposite vertices, through the centre, and its sides subtend from the opposite vertex an angle that wastes the width: the regular hexagon of diameter one has perimeter 3, against the bound of 3.10583.

Polygons cut from curves of constant width #

The polygons that do reach the bound for even $n$ come from curves of constant width. A Reuleaux polygon of width one is made from a regular polygon with an odd number of corners $m$ by replacing each side with an arc of radius one centred at the opposite corner; each arc turns through $\pi/m$, and the curve has the same width, one, in every direction.

Cut each of its $m$ arcs into $k$ equal pieces. The $m$ corners and the cut points are $n = mk$ points on the curve, and the polygon through them has a striking property: each of its sides is a chord of one of the arcs, seen from that arc’s centre — a corner of the curve, exactly one unit away — at the angle $\pi/(mk) = \pi/n$. So every side has length $2\sin(\pi/2n)$, the polygon’s diameter is one because every point of the curve is within one of every other, and the perimeter is exactly Reinhardt’s bound.

The hexagon from the Reuleaux triangle has all six sides equal but its angles alternate, so it is not regular: it is the three corners of an equilateral triangle of side one, with the three midpoints of the bulging arcs between them. Its perimeter, 3.10583, beats the regular hexagon’s 3 by more than three per cent, which is a large margin for polygons that look so alike. The figure checks every side of every such construction against $2\sin(\pi/2n)$ and every diameter against one, for every $n$ up to 16 with an odd factor.

The same construction works for any odd factor. Ten is twice five, so the Reuleaux pentagon with each arc halved gives a decagon at the bound; twelve is four times three, so the Reuleaux triangle with each arc cut in four gives a dodecagon. Every $n$ with an odd factor $m > 1$ has at least one such polygon, built from the Reuleaux $m$-gon with $n/m$ pieces per arc. For a power of two there is no odd factor, no Reuleaux polygon to cut, and — this is the other half of Reinhardt’s theorem — no polygon of any kind reaches the bound.

The maximum is shared #

For area, the maximiser for odd $n$ is unique: only the regular polygon has the most area. For perimeter the bound is reached by many shapes at once.

Fifteen is three times five, so the Reuleaux triangle cut into fives, the Reuleaux pentagon cut into threes, and the regular 15-gon — itself a Reuleaux 15-gon cut into ones — all reach the bound, and they are visibly different shapes. They are not the only ones. Reuleaux polygons need not be regular: any curve of constant width made of circular arcs centred at its own corners will do, and if every arc turns through a whole multiple of $\pi/n$, cutting each into pieces of exactly $\pi/n$ gives another polygon at the bound. Counting these Reinhardt polygons for each $n$ is a combinatorial problem in its own right, studied by Michael Mossinghoff and others, and their number grows with the number of odd factors $n$ has.

That multiplicity is the sharpest contrast with the area problem. Area rewards evenness in a way perimeter does not: a polygon of diameter one gains area by being round, and the roundest polygon of odd order is the regular one. Perimeter only asks that every side be seen at the same angle from a point one unit away, and many polygons can arrange that.

Perimeter and area pull in different directions #

The two problems’ winners are not the same polygons, and for six sides the difference can be computed by hand. The perimeter-optimal hexagon is an equilateral triangle of side one with a low triangular cap on each side: each cap has base one and apex on the arc of radius one about the opposite corner, at height $1 - \cos 30° = 0.134$ above the base, so each adds $0.067$ to the triangle’s $0.433$, for a total area of $0.634$. The regular hexagon of diameter one has area $0.650$, and Graham’s hexagon, the area-optimal one, has $0.675$. So the hexagon with the longest fence encloses the least area of the three, and the one with the most area has a fence shorter than Reinhardt’s bound.

The reason is in what each problem rewards. Area at fixed diameter rewards roundness, which is why the circle wins among all shapes and why the answers drift towards the circle as $n$ grows. Perimeter at fixed diameter rewards something more like evenness of angle: every side seen at the same angle from a point one unit away. Those coincide for odd regular polygons, which are both as round as a polygon of their order can be and evenly seen, and they part company for even $n$, where roundness and evenness ask for different shapes.

The proofs part company too. Sliding every chord to the middle, Steiner’s symmetrisation, keeps area fixed, never increases the diameter, and never lengthens the boundary; it is the engine of the classical proofs that the circle holds the most area for its fence and the most area for its diameter. For the longest fence at a fixed diameter it is useless, since it can only shorten the fence, and Reinhardt’s proof works instead with the combinatorics of the diameters. That is also why the triangles and quadrilaterals that settle the area problem’s small cases have no counterpart here: the perimeter problem’s small cases are settled by Reinhardt’s bound for three sides and by Tamvakis’s separate analysis for four.

How many ways to reach the bound #

The constructions drawn here give one Reinhardt polygon for each odd divisor $m > 1$ of $n$: cut the regular Reuleaux $m$-gon into $n/m$ pieces per arc. So $n = 6$ has one, from the triangle; $n = 15$ has three, from the triangle, the pentagon and the 15-gon itself; $n = 45$ has five, from the Reuleaux polygons with 3, 5, 9, 15 and 45 corners. For a prime $p$ there is exactly one such construction, and it is the regular $p$-gon.

Irregular Reuleaux polygons add more. A Reuleaux polygon need not have equal arcs, only arcs centred at its own corners and fitting together, and whenever every arc turns through a whole multiple of $\pi/n$ the cutting works again. The arcs’ turning angles must add up to $\pi$ and the arcs must meet in corners that are each the centre of the arc opposite, so not every list of multiples of $\pi/n$ can be realised; but many can, and each one realised gives a Reinhardt polygon that is neither regular nor cut from a regular curve, with sides all equal and angles that follow the pattern of the arcs. Which arc patterns close up is a question about sequences of signs and turns, and Mossinghoff’s enumerations show the number of Reinhardt polygons growing quickly with the number and size of $n$’s odd factors.

Four sides, by search #

For $n = 4$ there is no Reuleaux polygon to cut, and the bound $8 \sin(\pi/8) = 3.06147$ is out of reach. The best quadrilateral was found by Nikolaos Tamvakis in 1987, and the figure finds it again by brute force: two hundred random placements of four points, each improved by small moves that increase the perimeter of their convex hull divided by their diameter, keeping the best.

The search settles on a quadrilateral with two sides of length one and two of length $0.5176 = 2\sin 15°$, and a perimeter of 3.03527, which is $2 + 4 \sin 15° = 3.035276$ to the five decimals the search reaches. Its shape explains the number. It is an equilateral triangle of side one with a fourth corner placed on the arc of radius one about one of its corners, halfway between the other two: a Reuleaux triangle with one arc cut in half and the other two arcs replaced by their straight chords. The cut arc contributes two sides seen at $30°$ each, and the two chords contribute sides of length one seen at $60°$. The angles add to $\pi$ as the bound requires, but they are not all equal, and that inequality is exactly the gap between 3.03528 and 3.06147.

For eight sides the best octagon is not regular, does not reach the bound, and was found by Charles Audet, Pierre Hansen and Frédéric Messine in 2007 by a global optimisation that proved its own optimality: its perimeter is 3.121147, just below the bound of 3.121445. For sixteen sides the optimum is not known exactly.

The limit is π, for every curve #

As $n$ grows the bound $2n\sin(\pi/2n)$ approaches $\pi$, and that limit is not a coincidence.

Every curve of constant width one has perimeter exactly $\pi$. That is Barbier’s theorem, from 1860, and it holds for the circle of diameter one, for every Reuleaux polygon, and for every smooth curve of constant width alike. The Reinhardt polygons are inscribed in such curves with their vertices spaced evenly in angle, so their perimeters are the inscribed polygon’s approximations to $\pi$, short by about $\pi^3/(24n^2)$ — 0.0057 for fifteen sides, 0.00064 for forty-five. The perimeter problem for polygons is the discrete shadow of Barbier’s theorem: a curve of diameter one can have perimeter $\pi$ but no more, and a polygon gets as close to $\pi$ as its number of sides allows exactly when it can be inscribed evenly in a curve of constant width.

Barbier’s theorem itself has a one-line proof through the average width. Cauchy’s formula says that the perimeter of any convex curve is $\pi$ times its average width, taken over all directions — the same fact that lets a length be counted by the lines that cross it. A curve of constant width one has average width one, so its perimeter is $\pi$. Applied to a polygon, the same formula says that the perimeter is $\pi$ times the average width, so the longest perimeter at diameter one belongs to the polygon whose width stays as close to one as possible in every direction. That is a different reading of the Reinhardt polygons: they are the polygons of diameter one that are as nearly of constant width as $n$ sides allow, and when $n$ has an odd factor they manage it exactly at $n$ evenly spaced directions.

That also explains the powers of two from the other side. A curve of constant width made of circular arcs has an odd number of arcs, because each arc’s centre is the corner opposite it and the corners and arcs alternate round the curve in a pattern that only closes up for an odd count. An even polygon inscribed evenly in such a curve must have a number of vertices divisible by the odd number of arcs. When $n$ has no odd factor, no such curve fits it, and the polygon must waste some of its angle.

Reinhardt’s theorem does the proving #

Optimality. The figures construct polygons that reach the bound and check them exactly; that they are the best is Reinhardt’s theorem, and that the powers of two cannot reach it is the other half of his proof. The quadrilateral search finds the optimum to five decimals and is consistent with Tamvakis’s proof, but a search is not a proof that nothing better exists.

All Reinhardt polygons. Only regular Reuleaux polygons, evenly cut, are drawn. Irregular Reuleaux polygons give more Reinhardt polygons for most $n$, and the figures neither draw nor count them.

Equal sides do not follow. Every Reinhardt polygon has all its sides equal, but the converse fails: an equilateral polygon of diameter one need not be a Reinhardt polygon, as the regular hexagon shows.

Still open: the powers of two #

What is the longest perimeter of a polygon with $2^k$ sides and diameter one? It is known for four sides, by Tamvakis, and for eight, by Audet, Hansen and Messine. For sixteen and beyond, only bounds and good constructions are known; the best polygons found are close to the bound, by an amount that shrinks rapidly with the number of sides, and their exact shape and the exact deficit are open. The same asymmetry between odd and even that runs through the area problem appears here in a purer form: the obstruction is not evenness but the absence of any odd factor, and every number except the powers of two escapes it.

There is a natural companion question in which the roles of perimeter and width swap. The least area a width can hold is Blaschke and Lebesgue’s Reuleaux triangle, and the most area a fence can hold is the circle. The Reinhardt polygons sit between the two families of results: polygons inscribed in curves of constant width, maximising the length of their fence rather than the area inside it, and settled completely except at the powers of two.