The Geometry of Curves of Constant Width

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.

The Mathematics of Throwing Noodles at Paper

An Experiment

Get out a sheet of paper and draw parallel lines on it spaced two inches apart. Take a one-inch long needle and repeatedly toss it onto the sheet of paper, counting the total number of needle tosses and the number of times the needle touches one of the lines. What do you expect the ratio of number of tosses to the number of line intersections to be? In other words, what is the probability that a randomly tossed needle intersects a line? I don’t think the answer is obvious, but it turns out to be the ubiquitous number $\pi \sim 3.14159...$ You can in principle use this to experimentally calculate $\pi$, though you unfortunately need to toss the needle an impractical number of times in order to get a reasonable estimate.

This experiment is known as Buffon’s Needle Experiment; in this post I’m going to explore more general and seemingly more difficult phenomenon called Buffon’s Noodle Experiment. The setup is the same as before, only instead of tossing a needle (a line segment), we’ll toss a rigid “noodle” in the shape of any desired plane curve. We’ll find that the statistics of noodle crossings is determined just by the length, and not the specific shape, of the noodle in question. Thus the noodle experiment makes a profound connection between geometry and probability theory, a connection which helps solve difficult problems in both areas. This is encapsulated by a beautiful tool called Crofton’s Formula, the first result in an area of mathematics called “Integral Geometry” (or alternatively “Geometric Probability,” depending on whom you ask).

Part of the beauty of the integral geometry approach to Buffon’s needle experiment is that it involves almost no calculations. There are other approaches that involve writing out probability density functions and calculating double integrals, but the argument that I will give below involves only basic (but surprisingly subtle) ideas in probability theory and calculus. It’s a great example of a tricky problem that can be solved through careful abstract thought.

The Explanation

To come to grips with the needle and noodle experiments, we will try to answer the following question: what is the expected number of times that a randomly thrown noodle will cross lines on lined paper with line spacing $d$? We count with multiplicities: if the noodle intersects the same line twice, that counts as two line intersections.

Let $X$ be the random variable which represents the number of line intersections for a given noodle. Recall that the expected value of $X$ is given by:

$E(X) = \sum_{n=0}^\infty n P(X = n)$

where $P(X = n)$ represents the probability that the number of line intersections is exactly $n$. Here are a few basic observations about these expectations in the case where the noodle is actually a needle (i.e. a line segment).

• The expectation for a needle of length smaller than $d$ (the line spacing) is just $P(X = 1)$ because such a needle can cross at most one line.
• The expectation for any needle depends only on the length of the needle. Thus if $X_\ell$ denotes the random variable which represents the number of line intersections for a needle of length $\ell$, we can define a function $f$ by $f(\ell) = E(X_\ell)$.

Now, consider a noodle made up of exactly two needles of lengths $\ell_1$ and $\ell_2$ joined rigidly end-to-end. Denote the random variables representing the number of crossings for the two needles by $X_1$ and $X_2$, respectively; then the random variable representing the number of line intersections for the noodle is just $X_1 + X_2$. Note that $X_1$ and $X_2$ are not independent since the needles are joined, but it is nevertheless true that

$E(X_1 + X_2) = E(X_1) + E(X_2) = f(\ell_1) + f(\ell_2)$

A priori the expectation $E(X_1 + X_2)$ depends on the lengths $\ell_1$ and $\ell_2$ of the two needles as well as the angle at which they are joined, but the calculation above shows that the expectation is actually independent of the angle. Therefore we can calculate the expectation just by considering the case where angle is such that the two needles form a line segment, i.e. a needle of length $\ell_1 + \ell_2$. From this we conclude that

$f(\ell_1 + \ell_2) = f(\ell_1) + f(\ell_2)$

We can iterate this argument for any noodle made by chaining together $n$ needles of lengths $\ell_1, \ldots, \ell_n$ to conclude that:

$f(\sum_{i=1}^n \ell_i) = \sum_{i=1}^n f(\ell_i)$

In particular, if the needles all have the same length, we conclude that $f(n \ell) = n f(\ell)$ for any positive integer $n$ and any positive real number $\ell$. By dividing a needle into $n$ equal pieces, this also shows that $f(\frac{1}{n} \ell) = \frac{1}{n}f(\ell)$. Combining these two facts, we conclude that $f(a \ell) = a f(\ell))$ for any rational number $a$. The expected number of crossings for a longer line segment is at least as large as the expected number of crossings for a shorter one, so we also know that the function $f$ is non-decreasing. Also, we have that $f(0) = 0$ (i.e. the line segment of length zero doesn’t intersect any lines). Basic calculus tells us that the only non-decreasing function $f$ which satisfies $f(0) = 0$, $f(\ell_1 + \ell_2) = f(\ell_1) + f(\ell_2)$, and $f(a \ell) = a f(\ell)$ for every rational number $a$ is the function

$f(\ell) = C \ell$

where $C$ is some constant. This is already a pretty strong statement about the expected number of crossings for a needle, but we can do even better. Take any noodle of total length $\ell$ made up of $n$ line segments of lengths $\ell_1, \ldots, \ell_n$ and let $X$ be the random variable representing the number of crossings for that noodle; as above, we have:

$E(X) = \sum_{i=1}^n f(\ell_i) = \sum_{i=1}^n C \ell_i = C \sum_{i=1}^n \ell_i = C \ell$

Thus the expected number of crossings for piecewise linear noodles is still just $C$ times the length of the noodle.

Now take any noodle which is in the shape of a piecewise smooth curve $\gamma$. Borrowing another fact from basic calculus, $\gamma$ is the uniform limit of a sequence of piecewise linear curves $\gamma_n$ whose lengths converge to the length of $\gamma$ (note that this second condition is not automatically implied by the first). Since the expectations for each of the piecewise linear noodles is given by the formula $C \ell$, this formula holds for any piecewise smooth noodle. Thus we have proved:

Proposition: Let $X$ denote the random variable corresponding the number of line crossings for a rigid piecewise smooth noodle of length $\ell$ tossed at lined sheet of paper. Then $E(X) = C \ell$ for some universal constant $C$

Already we have proven something that wasn’t really obvious at the outset: the expected number of intersections depends only on the length, and not on the shape, of the noodle! It remains only to calculate the constant $C$; since this constant is the same for any noodle (of any shape an any length), it suffices to work out just one example. There is one particular noodle for which this calculation is especially easy: the circle whose diameter is $d$ (the spacing of the lines). This is because no matter how you drop such a circle on a sheet of lined paper whose lines are spaced $d$ apart, the circle *must* intersect exactly two lines and therefore the expected number of intersections is simply $2$. The length of the circle of diameter $d$ is $\pi d$, so we have $2 = \pi d C$ and hence $C = \frac{2}{\pi d}$. Thus the expectation formula is $E(X) = \frac{2 \ell}{\pi d}$.

In the example at the beginning of this post, we considered a needle of length $1$ (so that $\ell = 1$) tossed at a sheet of lined paper with line spacing $d = 2$. According to our formula, this means that the expected number of line crossings is simply $\frac{1}{\pi}$. But as we observed above, the expected number of crossings for a needle which is shorter than the line spacing of the paper is simply the probability that at least one crossing will occur. Therefore this probability is $\frac{1}{\pi}$. According to the law of large numbers, this means that the ratio of the number of needle tosses to the number of crossings approaches $\pi$ as the number of tosses tends to infinity.

Crofton’s Formula

Having accounted for Buffon’s needle and noodle experiments, let us reflect on what we have done. We set out to answer a question about the statistics of tossing noodles at paper, and we found that the answer to the question is an explicit formula involving only the length of the noodle and some constants. Flipping this formula around, notice that this gives us a surprising way to calculate the length of the noodle:

$\ell = \frac{\pi d E(X)}{2}$

In other words, length is a quantity which can be measured and manipulated using statistical techniques. You might not be convinced right away that this is a useful way to think about length, but in fact there are a variety of difficult theorems in geometry which have remarkably easy proofs when they are translated into this language.

Actually, it’s useful to think about all of this in a slightly different way. Instead of throwing the noodle at the paper, we’ll imagine throwing the paper at the noodle. In other words, we’ll ask a slightly different question: what is the average number of times that a random line in the plane intersects a given plane curve? This question is conceptually a little more problematic than the old question because it is not completely clear what the phrase “random line” should mean; Bertrand’s Pardox is a good illustration of the subtleties involved.

Here is the right meaning of the phrase for our purposes. The set of all oriented lines in the plane can be parametrized by two coordinates: the (signed) distance $r$ from a line to the origin (a number from $-\infty$ to $\infty$) and the direction $\theta$ (an angle) in which it points (a number from $0$ to $2\pi$). With this parametrization, we can interpret a “random line” to simply be a random point $(r,\theta)$ in the strip $(-\infty,\infty) \times [0,2\pi]$. (To placate the highly mathematically literate members of my readership, I’ll remark that the space of lines in the plane is topologically a homogeneous space which has a unique translation invariant measure, and this space differs from the strip with Lebesgue measure by a set of measure zero.)

Now, associated to any piecewise smooth curve $\gamma$ in the plane is a function $n_{\gamma}(r,\theta)$ which represents the number of times that the line determined by the values $(r,\theta)$ intersects $\gamma$. Crofton’s formula relates the average value of this function to the length of $\gamma$:

Crofton’s Formula: $\text{Length}(\gamma) = \frac{1}{4} \iint n_{\gamma}(r,\theta)\, dr\, d\theta$

This formula can be proved using more or less the same procedure that we used to calculate the expected number of crossings for a noodle thrown at a sheet of lined paper: argue that the integral on the right-hand side is additive for line segments attached end-to-end, use an approximation argument to show that it agrees with length up to a multiplicative constant, and fix the constant by calculating a single explicit example (the circle is once again a good choice).

As I alluded in the beginning of this post, Crofton’s formula is the tip of a very deep iceberg. There are analogous formulas for area, volume, and a plethora of interesting geometric quantities. In my next post I will use Crofton’s formula to deduce some facts about curves of constant width, tying this post together with post on the Coaster Roller.

The Coaster Roller

My first blog post was inspired by my visits to the Museum of Mathematics, and it appears that my second will be as well.  Like the first post this one will be suitable for a general audience, but I’ll write a follow-up which gets a little more mathematically serious.

I started volunteering at MoMath as an “integrator” this past Sunday; the primary role of the integrator is to float around the museum helping people who seem confused by an exhibit.  A secondary role is to operate the few exhibits which require staff supervision, and I was taught how to operate an exhibit called the “coaster roller”.  The coaster roller works as follows.  There is a small pit full of peculiar-looking objects and a sort of “raft” which sits on top of them (I’ll take a picture next time I visit the museum).  The objects are about the size of basketballs, but they are not spherical.  Here’s a picture of what they might look like, stolen from the internet:

A person (or several children) can sit in the raft and pull themselves along using ropes, rolling over the strange shapes at the bottom.

So what’s so special about the shape of these objects?  The important property that they each possess is that they have constant width, meaning that if one of the objects fits in a vise when it is pointing in one direction then it fits in the same vise when it is pointing in any other direction.  This is important because if the width varied then the distance between the raft and the floor would vary as it rolled along and the ride would get pretty bumpy!

The constant width property is possessed by a sphere but not by a cube: the width of a cube is larger when measured between opposite corners than when measured between opposite faces.  If you think about it for a moment, you might find it hard to convince yourself that bodies of constant width which aren’t spheres can even exist!  But in fact they exist in abundance; I will explain a procedure which allows you to construct infinitely many different ones.

To begin, note that given any plane curve of constant width one can obtain a corresponding surface of constant width by rotating the curve in three-dimensional space (you may have already noticed that each of the surfaces in the picture above is rotationally symmetric).  Not all surfaces of constant width arise this way, but in any event this shows that it is enough to look at curves of constant width if we are happy just finding a few basic examples.

To construct a curve of constant width, begin by drawing an equilateral triangle with side length $w$.  Then draw the circle of radius $w$ centered at each of the three vertices of the triangle.  The boundary of the intersection of the disks enclosed by these circles is then a curve of constant width $w$.  Can you figure out why?  Here’s a picture, stolen from Wikipedia:

In fact, the same procedure works if you start with any regular polygon with an odd number of sides (thanks to Ben Levitt from MoMath who corrected my original claim that it works for any regular polygon).  The curves of constant width obtained in this way are often called Reuleaux polygons for their discoverer (even though they aren’t technically polygons).

There are all sorts of interesting mathematical questions one can ask about bodies of constant width.  Here are some useful facts:

1. All curves of the same constant width have the same perimeter.
2. Among all curves with a given constant width, the circle encloses the largest area.  Similarly, among all surfaces with a given constant width, the sphere encloses the largest volume.
3. Among all curves with a fixed constant width, the Reuleaux triangle encloses the smallest area.
4. Among all surfaces with a fixed constant width, nobody knows which one encloses the smallest volume!

In my next blog post, I plan to discuss some of these facts in greater detail.  If you want to read more in the meantime, I recommend The Enjoyment of Math by Rademacher and Toeplitz, two great 20th century mathematicians. Actually, I recommend that book even if you don’t care to read more about bodies of constant width!