Today I will finally fulfill my earlier promise to revisit the geometry of curves of constant width. I doubt anyone was going to hold me to this promise, but it’s generally good to keep your promises even if you only made them to your own blog. In any event, I have two goals in this post:

1. Prove that if two curves have the same constant width then they have the same length (perimeter).
2. Prove that among all curves with a given constant width the circle encloses the largest volume.

Let us begin by providing some precise definitions. Recall that a plane curve is simply a continuous function $\gamma \colon [0,1] \to \mathbb{R}^2$, and a plane curve is closed if it begins and ends at the same point, i.e. $\gamma(0) = \gamma(1)$. A closed curve is simple if it intersects itself only at the endpoints, meaning $\gamma(t_1) = \gamma(t_2)$ only if $t_1 = t_2$ or $t_1$ and $t_2$ are both endpoints of the interval $[0,1]$. The most basic fact about simple closed curves is that they divide the plane into two disconnected regions: a bounded piece (the “inside”) and an unbounded piece (the “outside”). This is called the Jordan curve theorem, and as far as I know the simplest proofs use some reasonably sophisticated ideas in algebraic topology (though only a mathematician would think it even needs to be proved!)

Given a simple closed curve $\gamma$, let $C_\gamma$ denote the image of $\gamma$, i.e. the set of all points in the plane that $\gamma$ passes through, and let $D_\gamma$ denote $C_\gamma$ together with the points “inside” $\gamma$. A line in the plane is said to be a supporting line for $D_\gamma$ if it intersects $C_\gamma$ but does not pass through any interior points of $D_\gamma$. The set $D_\gamma$ is closed and bounded, so there are exactly two distinct supporting lines for $D_\gamma$ in any given direction. The set of directions in the plane can be parametrized by an angle $\theta$ between $0$ and $\pi$ (with the understanding that $0$ and $\pi$ represent the same direction). Thus we define a “width” function $w_\gamma$ on the set of directions by letting $w_\gamma(\theta)$ denote the distance between the supporting lines for $D_\gamma$ in the direction $\theta$. Here’s what the width looks like in an example:

Finally, we say that $\gamma$ has constant width if $w_\gamma$ is constant. The goal is to prove that any two curves of constant width $w$ have the same length, and that among all curves of constant width $w$ the circle of diameter $w$ has the largest area. Before proceeding, we need to understand the geometry of constant width curves a little better.

Specifically, we want to show that every curve $\gamma$ of constant width is convex, meaning $D_\gamma$ contains the line segment between any two of its points. In fact we will prove something a bit stronger: $\gamma$ is strictly convex, meaning it is convex and $C_\gamma$ contains no line segments (so that the line segment joining any two points in $D_\gamma$ actually lies in the interior of $D_\gamma$). This requires a nice little trick that I couldn’t figure out on my own; special thanks to Ian Agol for helping me out on mathoverflow.

Proposition: Every curve of constant width is strictly convex.
Proof: Let $H_\gamma$ denote the convex hull of $D_\gamma$; this is by definition the smallest convex set which contains $D_\gamma$. According to a general fact from convex geometry, the boundary of $H_\gamma$ consists only of points in the boundary of $D_\gamma$ and possibly line segments joining points in the boundary of $D_\gamma$. So we will show that the boundary of $H_\gamma$ contains no line segments, implying that $H_\gamma = D_\gamma$ and hence that $D_\gamma$ is strictly convex.

According to another general fact from convex geometry the supporting lines for $H_\gamma$ are precisely the same as the supporting lines for $D_\gamma$, and hence $H_\gamma$ has the same constant width $w$ as $D_\gamma$. So assume that the boundary of $H_\gamma$ contains a line segment joining two points $a$ and $b$. Since $H_\gamma$ is convex, the line $\ell$ passing through $a$ and $b$ is a supporting line for $H_\gamma$. There is exactly one other supporting line for $H_\gamma$ parallel to this line; let $c$ denote a point where it intersects $H_\gamma$. Consider the triangle $abc$; its height is precisely $w$, the width of $H_\gamma$, so we have that $w$ is strictly smaller than at least one of $dist(a,c)$ or $dist(b,c)$. Assume $w < dist(a,c)$ and consider the supporting lines for $H_\gamma$ which are perpendicular to the line segment joining $a$ and $c$. The points $a$ and $c$ must lie between (or possibly on) these supporting lines, but the distance between the supporting lines is $w$ since $H_\gamma$ has constant width. We conclude that $w < dist(a,c) \leq w$, a contradiction.
QED

The reason why strict convexity is important to us is that lines intersect strictly convex curves in a very predictable way:

Lemma: Let $\gamma$ be a closed strictly convex curve and let $L$ be a line which intersects $C_\gamma$. Then $L$ intersects $C_\gamma$ exactly once if it is a supporting line or exactly twice if it is not.
Proof: Note that the intersection of two convex sets is again convex, so the intersection $I = L \cap D_\gamma$ is a convex subset of a line. Since $D_\gamma$ is closed and bounded the same must be true of the intersection, so the only possibility is that $I$ is a closed interval $[a,b]$ with $a \leq b$. Note that interior points of $[a,b]$ correspond to interior points of $D_\gamma$ and the boundary points $a$ and $b$ correspond to boundary points of $D_\gamma$, so we have that $a = b$ if and only if $L$ is a supporting line and $a < b$ otherwise. Thus supporting lines intersect $C_\gamma$ exactly once and any other line which intersects $C_\gamma$ does so exactly twice.
QED

We are now ready to calculate the length of a constant width curve. Our strategy is to use the main result of my previous post, “The Mathematics of Throwing Noodles at Paper.” There we saw that if one randomly tosses a curve of length $\ell$ at a lined sheet of paper with line spacing $d$ then the expected number of line intersections is given by $\frac{2 \ell}{\pi d}$. So let us toss our curve of constant width $w$ at a lined sheet of paper with line spacing $w$. The curve must intersect at least one line and it can’t intersect three or more lines, so it either intersects exactly one line or exactly two lines. The curve intersects exactly two lines if and only if they are supporting lines, and hence each line intersects the curve exactly once by the lemma above. If the curve intersects exactly one line then it cannot be a supporting line and thus the lemma implies that the curve intersects the line exactly twice. In either case the total number of intersections is exactly $2$, and thus the expected number of intersections is $2$. Therefore

$2 = \frac{2 \ell}{\pi w}$

and hence $\ell = \pi w$. Thus every curve of constant width $w$ has length $\pi w$, an assertion consistent at least with the circle of diameter $w$. The result is called Barbier’s Theorem, and it has a variety of different proofs; I find the argument using geometric probability to be the most beautiful.

We have now settled the length question; what about area? In fact, to place an upper bound on the area inside a constant width curve we will simply use our length calculation together with the following landmark theorem in geometry:

Theorem: Let $\ell$ be the length of a simple closed curve in the plane and let $A$ be the area that it encloses. Then:
$4\pi A \leq \ell^2$
with equality if and only if the curve is a circle.

In other words, among all curves with a given length the circle is the unique curve which encloses the largest area. This theorem is called the isoperimetric inequality, and it has many beautiful proofs, generalizations, and applications. Our claim about the area enclosed by constant width curves is an immediate corollary since they all have the same length (given a fixed width). I originally intended to prove the isoperimetric inequality in this post using geometric probability, but I would need to take some time to explain how to calculate area probabilistically and I think the post is long enough as it is. Perhaps I will revisit this in the future.