The least area a width can hold

The rung below ends with Barbier’s theorem: every curve of constant width $w$ has perimeter exactly $\pi w$. A shape’s perimeter therefore says nothing about which member of the family it is, and the family is left with one obvious quantity that does distinguish its members.

Area does. The circle encloses the most, the Reuleaux triangle the least, and everything else lies between — which is the Blaschke–Lebesgue theorem, proved in 1915 and reproved several times since because the early proofs were hard to check.

Why area is free to vary and perimeter is not #

The two quantities behave differently and the difference is visible in the support function, which is what the rung below is about.

A convex shape is described by $h(\theta)$, the distance from the origin to the supporting line with outward normal $\theta$. Constant width says $h(\theta) + h(\theta + \pi) = w$ for every $\theta$, which is a linear condition — it holds exactly when the Fourier expansion of $h$ has only odd harmonics beyond the constant term.

Perimeter is also linear in $h$: it is the integral of $h$ around the circle. So on the family of constant-width shapes it is constant, and that single sentence is Barbier’s theorem.

Area is not linear. It is

$$A = \tfrac{1}{2}\int_0^{2\pi} \left(h^2 - h'^2\right) d\theta,$$

which is quadratic — so it genuinely varies over a linear family, and asking for its minimum is asking to minimise a quadratic on a linear space.

The formula is worth reading rather than accepting. The first term is what the area would be if the shape were a circle of varying radius; the second is a correction that subtracts whatever the boundary spends moving sideways rather than outwards. A support function that varies quickly has a large $h'$ and therefore loses area, which is the whole mechanism of the minimisation in one line. It also says immediately why the circle is the maximum: $h$ is constant there, $h'$ is zero, and the correction term vanishes.

Minimising a quadratic ought to be easy #

A quadratic on a linear space has its minimum where the derivative vanishes, and the calculation is a page. It gives the wrong answer, and the reason is a constraint that has not been mentioned.

A support function has to describe a convex shape, and that requires $h + h'' \ge 0$ everywhere — the radius of curvature must not go negative. That is an inequality, not an equation, and it turns a tidy minimisation into a constrained one.

The sweep is the whole argument in one picture. Push the third harmonic up and the area falls monotonically; the radius of curvature falls too, and hits zero first. The minimum is therefore on the boundary of the constraint, and a shape whose radius of curvature is zero somewhere has a corner there.

So the corners of the Reuleaux triangle are not a quirk of a construction with a compass. They are what a constrained minimum looks like: the objective wants to keep going and the constraint stops it, and the place it stops is a point where the curvature has run out.

Reading the sweep’s numbers #

The table and the sweep are measuring the same thing from two directions and it is worth putting the numbers side by side.

At the circle the area is $0.7854 w^2$. Raising the third harmonic to an amplitude of 4 brings it to $0.7804$; to 8, $0.7653$; and at the critical amplitude of 12.5, where the radius of curvature first vanishes, $0.7363$. Meanwhile the Reuleaux pentagon sits at $0.7585$, the heptagon at $0.7719$ and the enneagon at $0.7773$ — so the Reuleaux polygons climb towards the circle as their side count rises, and the smooth family falls towards its constraint boundary, and the two sequences pass each other.

That crossing is the useful observation. The best smooth curve the sweep reaches is better than the Reuleaux pentagon and worse than the Reuleaux triangle, so neither family contains the other’s best member and neither is the whole story. The minimiser is a Reuleaux polygon with the fewest sides a Reuleaux polygon can have, and the smooth curves approach it without belonging to that family at all.

There is one more reading. Every Reuleaux polygon has corners — three, five, seven — and its area rises with the number of them, which says that corners are what buys the saving and that fewer, sharper corners buy more. The circle is the case with none.

What the Reuleaux triangle is doing #

Read as a support function rather than as three arcs, the Reuleaux triangle is the extreme case in the most literal sense.

Its radius of curvature takes only two values: zero at each of the three corners, and $w$ along each arc. There is no intermediate behaviour anywhere. Every other member of the family spreads its curvature out; this one puts all of it in three points and none in between.

The general principle behind this is worth naming because it recurs. An extremal shape in a constrained problem is usually degenerate, and degenerate usually means “as far into the corner of the constraint set as the problem allows”. The most area a fence can hold is the opposite case — an unconstrained maximum, attained by the smoothest possible shape — and the two together make the pattern legible: maxima of area tend to be round, minima tend to be spiky, and which one a problem has decides what the answer looks like before any calculation.

The smooth family cannot reach it #

Something worth extracting from the sweep: the best smooth constant-width curve in that one-parameter family encloses 0.7363 times the square of the width. The Reuleaux triangle encloses 0.7048. The smooth family gets about three fifths of the way.

Adding more harmonics improves matters and never closes the gap, and there is a clean reason: the infimum over smooth constant-width curves is the Reuleaux triangle’s area, and it is not attained. A sequence of smooth curves can approach the triangle as closely as one likes, with the curvature concentrating into three ever-tighter regions, and the limit has corners and is not smooth.

A minimisation problem whose infimum is not attained inside the natural class is a signal to widen the class, and here widening it means allowing corners — which is the same as allowing the support function to be merely continuous rather than twice differentiable.

That widening is not a technicality and it changes what a proof has to do. Over smooth support functions the constraint $h + h'' \ge 0$ is an inequality between functions, and calculus of variations handles it in the usual way. Over merely continuous ones the second derivative is a measure rather than a function — it can have atoms, and an atom is a corner — and the constraint becomes a statement about a measure being non-negative. The Reuleaux triangle’s curvature measure is three atoms of equal size plus an absolutely continuous part on the arcs.

Stated that way the answer looks inevitable: to minimise, push as much of the curvature as possible into as few atoms as possible, and the constant-width condition is what fixes how few. Three is the smallest number of atoms compatible with the odd-harmonic condition, and three is what the answer has.

The reason the theorem is nevertheless hard is that “looks inevitable” is not an argument, and the space of non-negative measures on a circle satisfying a linear constraint has a great many extreme points.

The corner, examined #

It is worth being precise about what a corner is in this language, because the phrase “the curvature runs out” is doing a lot of work.

The boundary point with outward normal $\theta$ is at

$$\big(h\cos\theta - h'\sin\theta,\; h\sin\theta + h'\cos\theta\big),$$

and differentiating shows the boundary moves at speed $h + h''$ as $\theta$ advances. So $h + h''$ is the radius of curvature, and where it is zero the boundary point does not move at all while the normal direction sweeps on.

A stationary boundary point with a rotating normal is exactly a corner: one point of the shape supports a whole range of directions. On the Reuleaux triangle each corner supports a sixty-degree fan of them, which is the angle deficit of the arcs meeting there.

That also explains why $h + h''$ cannot be allowed to go negative. Negative speed means the boundary point runs backwards as the normal advances, which produces a curve that crosses itself and encloses a region the support function no longer describes. The constraint is not an extra requirement imposed for tidiness; it is the condition for the formula to be describing a shape at all.

And it says where the saving comes from. Area is $\frac{1}{2}\int (h^2 - h'^2)$, and the $-h'^2$ term rewards a support function that varies fast. Varying fast is exactly what drives $h + h''$ down. The objective and the constraint are pulling on the same quantity in opposite directions, which is why the answer sits precisely where one of them gives out.

Why the theorem was hard #

Blaschke and Lebesgue both proved the result in 1915, independently, and the proofs are not easy. That is surprising for a statement about a two-dimensional family with a quadratic objective, and it is worth saying where the difficulty is.

The difficulty is that the constraint set is infinite-dimensional and its boundary is complicated. A support function is a function on a circle, so the family of constant-width shapes at a given width is an infinite-dimensional convex set, and the minimum of a quadratic over such a set can be anywhere on its boundary. Showing that it is at this boundary point — the one with exactly three corners, at exactly those angles — requires ruling out every other extreme point of an infinite-dimensional set.

Several published proofs have been found to have gaps, and the currently accepted ones are recent. The history is worth a sentence for what it says about the shape of the problem: Blaschke’s and Lebesgue’s arguments were both correct in outline and both leaned on steps that later readers could not reconstruct, and the twentieth century produced half a dozen replacements, several of which were themselves found wanting. A statement this easy to believe should not have needed six proofs, and the reason it did is that believing it and constraining an infinite-dimensional set are different activities. A statement a figure can verify on any particular shape can still be a theorem nobody finds easy, and the gap between checking instances and proving the general case is the whole of the difficulty here.

The numbers, and what they mean #

The two extreme values are worth stating exactly, since the whole theorem is a comparison between them.

The circle of width $w$ has area $\pi w^2/4$, which is $0.7854 w^2$. The Reuleaux triangle has area $(\pi - \sqrt{3})w^2/2$, which is $0.7048 w^2$. The ratio is $0.8973$, so the whole family spans about a tenth.

A tenth is a small range, and that is part of why constant width is a useful property in practice. A shape of constant width is nearly determined by its width as far as its size is concerned: perimeter exactly, area to within ten per cent. A manufacturer measuring a rolling part with callipers learns the perimeter exactly and the area nearly.

It is also why the minimisation problem is delicate. The objective varies by ten per cent across an infinite-dimensional family, so the landscape is nearly flat, and nearly-flat landscapes are where uniqueness arguments are hard and where numerical searches are useless. That is the same difficulty a search without a certificate always has, in its sharpest form: a method that stops somewhere in a flat landscape has learned almost nothing about how far from the answer it stopped.

The ten per cent has a practical reading too. A coin of constant width — several countries mint them, to work in machines that measure by rolling — can be given a Reuleaux heptagon’s outline rather than a circle’s, and the metal saved is under three per cent. The reason for the shape is the rolling and not the saving, and the figures say the saving was never going to be worth having.

The three-dimensional question, and why it is different #

The same question in space is open, which is worth stating here because it makes the plane’s situation look lucky rather than typical.

In space a body of constant width has a surface area and a volume, and Blaschke’s relation ties them: $V = wS/2 - \pi w^3/3$ for every convex body of constant width $w$. So there is only one quantity to minimise, not two, and the situation looks as tidy as the plane’s.

It is not. The ball encloses $0.5236 w^3$ and the conjectured minimum is a Meissner solid at about $0.4199 w^3$ — a spread of twenty per cent rather than ten — and the conjecture has been open for a century. The candidate is known, its volume is known, and nobody can prove nothing does better.

The plane’s answer is not hard because the objective is complicated; it is hard because the feasible set is. In space the feasible set is worse, and the same shortage of extreme-point arguments that made the planar proof delicate leaves the spatial one unfinished. The next rung is about that.

What the pictures cannot show #

The sweep varies one harmonic. The full family is infinite-dimensional and a one-parameter slice through it establishes the mechanism — area falling, curvature reaching zero — and not the theorem.

The areas in the table are measured off drawn boundaries at a finite resolution and agree with the closed forms to four decimals. That is a check on the drawing rather than evidence about the extremal problem.

And the theorem itself is stated and not proved. Nothing here rules out an exotic constant-width shape with less area; what the figures show is that everything in the family they can draw obeys the bound, and that the mechanism forcing corners is real.

Where the ladder goes next #

Named here as debts. The proof itself, which is the interesting object — the modern versions go through a clever rearrangement argument and would need a rung of their own. And the same question for the perimeter of the convex hull in higher dimensions, which is where the problem becomes open.

Sideways, the linear condition on the support function is the rung below, the construction of the minimiser is the first rung, the opposite extremal problem is the isoperimetric one, and the convexity constraint that binds here is a line under every point read as a condition on curvature.

What is worth carrying away #

When a minimum is attained at a shape with corners, the corners are usually the constraint rather than the objective.

Area falls as the support function’s harmonics grow and would keep falling forever; what stops it is the requirement that the shape stay convex, and the stopping point is where the radius of curvature reaches zero. A shape with zero radius of curvature at a point has a corner there, so the answer has corners for a reason that never mentions corners.

The habit worth taking is to separate the objective from the feasible set. A surprising minimiser is nearly always the feasible set’s fault, and asking which constraint is active at the answer explains the surprise better than any amount of looking at the shape.

The corollary is about smooth approximations. A family of smooth shapes approaching the minimiser gets closer without arriving, so a numerical search restricted to smooth shapes will report a value that is wrong by a fixed amount and will not know it. When the answer is on a constraint boundary, the interior is the wrong place to look, and no amount of refinement inside it converges to the right answer.