The shape described from outside

A shape can be described by saying where its boundary is. It can also be described by saying, for each direction, how far away the line perpendicular to that direction has to be pushed before it stops touching. The two descriptions carry the same information for a convex shape, and almost everything about curves of constant width is easy in the second and awkward in the first.

The second description is a single function of a single angle. Write $h(\theta)$ for the distance from a chosen origin to the supporting line whose outward normal points in direction $\theta$. That function is the support function, and it is the whole subject of this rung.

The condition, in one line #

The width of a convex shape in the direction $\theta$ is the distance between the two supporting lines perpendicular to $\theta$, which is

$$w(\theta) = h(\theta) + h(\theta + \pi).$$

So constant width — the property that took the previous rung a construction with a compass to produce and several pictures of calipers to verify — is the statement

$$h(\theta) + h(\theta + \pi) = w \quad \text{for all } \theta.$$

That is a functional equation, and a very mild one. It says nothing about arcs, corners, or vertices; it does not require the shape to be built from anything; and it is trivially satisfiable — take any function whatever on the half-turn from $0$ to $\pi$ and define the rest by $h(\theta + \pi) = w - h(\theta)$. Every curve of constant width arises that way and no others do.

That already answers a question the first rung left open. The Reuleaux polygons are the shapes a compass produces, and one might reasonably suspect they are the whole story, or nearly. They are not remotely the whole story; they are the members whose support functions happen to be piecewise nice, and the family is as large as the family of functions on a half-turn.

The picture is worth pausing on because it makes the constraint’s shape visible. The individual function is lumpy — three lobes, one per arc. The sum with its own half-turn shift is flat. Neither the lumpiness nor the flatness would be visible in a drawing of the shape.

Comparing the two plots gives a fact the drawings of the shapes make hard to see: the Reuleaux polygons converge to the circle rapidly as the number of sides grows. The three-sided one departs from a circle by about a seventh of its width and the seven-sided one by a few per cent, which is why the drawings of the higher members are difficult to tell from discs. The constraint does not weaken; the room inside it does.

Where the origin goes, and why it does not matter #

One thing must be checked before the function is trusted, because it looks as though a choice has been made. The support function depends on the origin: shifting the origin by a vector $v$ changes $h(\theta)$ to $h(\theta) + v \cdot u(\theta)$, where $u(\theta)$ is the unit vector in direction $\theta$.

But $u(\theta + \pi) = -u(\theta)$, so the added term cancels in the sum: $h(\theta) + h(\theta+\pi)$ is unchanged. The width function does not know where the origin is, which is exactly as it should be, since the width of a shape is not a fact about a coordinate system.

That cancellation is the same one that will do all the work below. Direction-reversal is an operation of order two, and everything on this rung is a statement about what survives it.

Barbier’s theorem falls out #

The first rung showed by a needle-counting argument that every curve of constant width $w$ has perimeter $\pi w$ — the same as the circle of that width, which is at first sight impossible for a shape with corners in it. In the language of support functions the result takes one line.

Cauchy’s formula says that the perimeter of a convex curve is the integral of its support function over a turn, and it is worth being clear that this is a theorem about all convex curves rather than a definition:

$$L = \int_0^{2\pi} h(\theta) \, d\theta.$$

The reason is a projection count: the length of the shadow the shape casts on a line in direction $\theta$ is $w(\theta)$, and integrating the shadow length over all directions counts each piece of boundary once for each direction it faces, which comes to its length times a fixed factor.

Now split the integral into two half-turns and use the constraint:

$$L = \int_0^{\pi} \big(h(\theta) + h(\theta+\pi)\big) \, d\theta = \int_0^{\pi} w \, d\theta = \pi w.$$

The shape has vanished from the calculation. It was never used — only the constraint was — so the answer cannot depend on which curve of constant width was meant.

It is worth checking the formula on the one case where the perimeter is known without it. For a circle of radius $R$ about its own centre, $h$ is the constant $R$, and $\int_0^{2\pi} R \, d\theta = 2\pi R$, which is right. For a segment of length $\ell$ — a degenerate convex set — $h(\theta) = \frac{\ell}{2}|\cos\theta|$ about its midpoint, and the integral is $2\ell$, which is also right: the boundary of a segment is the segment traversed both ways.

This is why the needle argument felt like a trick and this one does not. Buffon’s needle proof of Barbier’s theorem is correct and it works by showing that two quantities have the same expectation. The support-function proof shows that the perimeter is an integral of the width, and constant width means the integrand is constant. There is no coincidence left to explain, and the same calculation gives the perimeter of any convex shape whose width function is known.

Curves with no corners #

The Reuleaux polygons all have corners, and a reasonable reader of the first rung might conclude that corners are somehow necessary — that a shape of constant width other than a circle has to be pointy. Writing down a support function settles it in a sentence.

Take

$$h(\theta) = \tfrac{w}{2} + a\cos 3\theta + b\cos 5\theta.$$

Every harmonic present is odd, and an odd harmonic changes sign under $\theta \mapsto \theta + \pi$, so the sum $h(\theta) + h(\theta+\pi)$ is $w$ identically. The shape is of constant width by construction, and it is smooth, because the boundary point with outward normal $\theta$ is

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

which is as differentiable as $h$ is.

There is one condition to respect and it is where the interest is. The radius of curvature of the curve is $h + h''$, and convexity needs that positive everywhere. For the harmonic $a\cos k\theta$ the contribution to $h + h''$ is $a(1 - k^2)\cos k\theta$, which is negative wherever the cosine is positive and grows as $k^2$. So a high harmonic must have a small coefficient, and the whole family is a ball of finite radius around the circle in each odd frequency.

That is the honest statement of how much room there is. The curves of constant width are a large family and a bounded one: an amplitude of eight per cent of the radius at the third harmonic is close to the limit, which is why every smooth example ever drawn looks nearly circular, and why the Reuleaux triangle — sitting at the boundary of what convexity permits, with corners where the curvature degenerates — is the extreme member rather than a typical one.

Two examples with different harmonics make the pattern plain. The frequency chooses the symmetry and the convexity condition chooses the amplitude, and there is nothing else to decide. A shape of constant width with $k$-fold symmetry exists for every odd $k$, with a maximum departure from circularity falling off like $1/k^2$, and the whole family is parametrised by a sequence of coefficients rather than by any geometric construction.

What the language is really doing #

The support function is a change of coordinates on the set of convex shapes, and the two properties that make it worth having are worth naming separately.

It linearises. The support function of a sum of two shapes, in the sense of adding every point of one to every point of the other, is the sum of their support functions. So the set of convex bodies, which has no obvious arithmetic, inherits one; the mixture of a Reuleaux triangle and a circle is again of constant width, because the constraint is linear and the mixture of two solutions of a linear constraint is a solution. That single observation produces uncountably many curves of constant width from any two, which is a stronger statement than the previous rung’s there are infinitely many and is obtained without drawing anything.

The linearity is worth one more sentence, because it explains why the support function is the right coordinate rather than merely a convenient one. Adding two shapes point by point — the operation the previous paragraph describes — is the natural way of combining convex bodies, and it is the operation under which almost every classical inequality of the subject is stated. A coordinate in which that operation becomes ordinary addition is a coordinate in which those inequalities become statements about functions, and that is what convexity buys everywhere it appears: a local, additive description of a global condition.

It turns a shape into a function on a circle, which is the natural home for Fourier series. Read that way, the constant-width condition says the even harmonics of $h$ all vanish except the constant term — the constant survives because it is the width and $\cos 0 = 1$ does not change sign — and the odd ones are unconstrained. A geometric restriction has become a statement about which frequencies are allowed, which is exactly the kind of statement that is easy to work with and impossible to see in a drawing. It is the same move as reading a rotating point as a wave, applied to the family of shapes rather than to one point.

Both properties are used throughout convex geometry, and constant width is the smallest interesting example of each.

Where the idea came from #

Barbier proved the perimeter result in 1860, in a paper about Buffon’s needle rather than about convex geometry, and his argument is the probabilistic one the first rung gives. The support function is older in substance — it is implicit in Cauchy’s work on convex bodies from the 1830s, and Cauchy’s projection formula is exactly the ingredient this rung uses — but it was not yet a tool anybody reached for.

It became one through Minkowski, around the turn of the twentieth century, who used it to define the addition of convex bodies and to prove the inequalities that carry his name. Once shapes could be added, the set of convex bodies became an object with structure rather than a collection of examples, and the support function was the coordinate that made the structure visible.

The lesson in the chronology is the same one the perimeter proof teaches. Barbier’s theorem was proved thirty years before the language that makes it a line existed, and it was proved by a beautiful argument that generalises to almost nothing. The support-function proof generalises immediately: any linear functional of $h$ is constant across the family, and the perimeter happens to be one. What the later language bought was not the theorem but the ability to say which theorems of that shape there are.

What it does not settle #

The support function makes the constraint easy and leaves the hard questions exactly where they were.

Area is not linear. The area enclosed is $\frac{1}{2}\int (h^2 - h'^2)\,d\theta$, which is quadratic in $h$, so the mixture argument above says nothing about it. Minimising area over the curves of constant width is the Blaschke–Lebesgue problem, whose answer is the Reuleaux triangle — proved, but not by anything so cheap as the perimeter calculation, and every known proof is long. Perimeter is constant across the family and area is not, and that asymmetry is a direct consequence of one being a linear functional of $h$ and the other not.

There is a second thing the language does not do, and it is the more surprising of the two: it gives no obvious way to recognise a support function. Not every function on the circle is one — the condition is $h + h'' \geq 0$, which is convexity written in this coordinate system, and it is a differential inequality rather than a bound. So the set of convex bodies sits inside the set of functions as a convex cone with a curved boundary, and questions that are easy for functions can become hard the moment the answer is required to stay inside it. Blaschke–Lebesgue is exactly such a question: minimise a quadratic over a linear constraint, subject to staying in the cone, and the minimiser is on the cone’s boundary — which is what having corners means.

Three dimensions are worse than they look. The support function of a convex body in space is a function on the sphere, and the constant-width condition is again $h(u) + h(-u) = w$. So the family is again enormous. But the analogue of Barbier’s theorem is a statement about mean width rather than about surface area, and the surface areas of bodies of constant width genuinely differ. The rotated Reuleaux triangle is not the minimiser; the Meissner bodies are conjectured to be, and after a century that is still a conjecture.

What the pictures cannot show #

The support function is drawn as a graph and it is not a shape. The plotted curve is a function of an angle, and reading it as an outline is the standing hazard of the whole rung. The shape it describes is the envelope of the lines it specifies, which is a construction no single picture performs.

A supporting line is drawn as a segment. It is an infinite line, and the finite stroke on the page invites the reading that the shape touches the middle of something. What matters is only the perpendicular distance from the origin, which the arrow shows and the line does not.

The width in a direction is a difference of two distances and only their sum is drawn. The hero marks $h(\theta)$ and $h(\theta+\pi)$ as two arrows from an origin, which is the correct picture of the algebra and a misleading picture of the geometry: the width is the gap between the two lines and has nothing to do with the origin, which the arrows both start from. Moving the origin lengthens one arrow and shortens the other by the same amount, and no picture shows that without being drawn twice.

And the smooth curve is drawn as a polygon. It is sampled at seven hundred and twenty directions and joined up, so the corners it does not have are, at the level of the file, corners of the sampling. The width is measured on those sampled points, which is the honest thing to do, and the assertion is therefore about the drawn approximation rather than about the ideal curve. The ideal curve’s own constant width is the identity in the algebra above, which no picture checks.

Where the ladder goes next #

The rungs above are about what happens when the family is asked a harder question than what is the perimeter. Blaschke–Lebesgue in full, which asks for the smallest area and gets the corner-heavy extreme member. Bodies of constant width in three dimensions, where the Meissner solids repair the naive tetrahedron and the minimisation is open. And curves of constant brightness, where a different projection is held fixed and the answer changes.

Also unwritten, and named here as a debt: the width function of a non-constant-width shape is an object in its own right, and the shapes whose width function is a single harmonic are as interesting as the ones whose width function is constant. Nothing above touches them, and the circle’s own claim to be extreme is a statement about that larger family rather than about this one.

Sideways, the failure of area to be linear is the reason the isoperimetric problem cannot be solved this way, and the constraint that the odd harmonics survive is the same parity bookkeeping as a sum over a group’s characters. The linearisation of shapes into functions is the same move that turns a circle into a wave and, one dimension up, a triangulation into a hull.

What is worth carrying away #

Choosing the right description turns a construction into an equation, and the equation is usually the thing that was interesting.

Constant width described by boundaries is a family of compass constructions with a surprising property. Described by support functions it is a linear constraint on a function on a circle, and the surprising property is one line of integration. Nothing was proved by the change of language — the theorem was already true — but the argument became short enough that it stopped looking like a coincidence.

The habit worth taking is to notice which description makes the constraint linear. Width is a linear functional of the support function and a horrible functional of the boundary. Everything easy on this rung is a consequence of that single fact, and everything still open — area, three dimensions, the Meissner bodies — is a place where the quantity in question is not linear.