Round is not the only way to be the same width

Set a shape between the jaws of a caliper and close them until they touch. Read off the gap. Turn the shape and read again.

For almost anything, the two readings differ — that is what it means for a shape to be longer one way than another. For a circle they do not, and the usual conclusion is that a constant reading identifies a circle.

It does not, and the counterexample is easy enough to draw with a compass.

How it is built #

Draw an equilateral triangle. Put the compass point on one corner, open it to the length of a side, and draw the arc joining the other two corners. Do that at each corner in turn.

The result is a curved triangle with three arcs and three corners. Every arc is centred at the corner opposite it and has radius equal to the side length $w$.

Now take a caliper reading. One jaw rests on a corner; the other rests somewhere along the arc centred at that corner. Every point of that arc is exactly $w$ from its centre — that is what an arc is — so the gap is $w$. Turn the shape and the same thing happens with a different corner and a different arc.

That is the whole argument. It is not an approximation and it does not depend on the angle: at every orientation, one jaw is on a centre and the other is on that centre’s arc, so the reading is always the radius.

The transitions deserve a moment, since they are where a construction like this usually breaks. As the shape turns, the jaw eventually reaches the end of one arc and the start of the next — and at that instant the corner and the arc swap roles, which is exactly the moment the other jaw arrives at a corner. The handover is simultaneous at both ends, and the reading never flickers.

Rolling without rising #

The property has a consequence that is more striking than the property.

Put the shape on a flat floor and lay a plank across the top. Roll it. The plank rests on the top of the shape and the floor supports the bottom, so the plank’s height is the distance between two parallel supporting lines — which is the width, which never changes.

So a plank carried on Reuleaux rollers travels perfectly level, on rollers that are not round. This is occasionally offered as a practical technique and it is worth being clear about why it is not one. The rollers work; an axle through them does not. The shape’s centre moves up and down as it rolls, so anything mounted on a fixed axle bounces even though the plank above does not. Rollers under a load are fine; wheels on a cart are not, and the difference is whether the thing is carried on top of the shape or hung from its middle.

There are infinitely many #

The construction works on any regular polygon with an odd number of sides, and oddness is the requirement.

Each arc has to be centred on the vertex opposite it, and a polygon has a vertex opposite each edge only when the number of sides is odd. On a square, edges face edges and vertices face vertices, so there is nothing to put the compass point on.

That is a parity obstruction, and it has the same shape as the one that settles whether a walk across Königsberg exists: the construction is not merely difficult on an even polygon, it is unavailable, and one line of counting says so. Impossibility arguments of this kind are cheap when they work, and they are the only arguments that finish a search — the same service the angle budget performs for the list of regular solids.

And the regular ones are only the tidy cases. Any Reuleaux polygon can be built from an irregular arrangement of arcs, the corners can be rounded off by running a compass around the outside, and there are smooth curves of constant width with no corners at all and no circular symmetry. The set of them is large — infinite-dimensional, in the sense that a constant-width curve can be specified by choosing a function fairly freely and the constraint uses up much less freedom than one might expect.

Every one of them has the same perimeter #

Here is the fact that turns this from a curiosity into a theorem.

Every curve of constant width $w$ has perimeter exactly $\pi w$. Not approximately, and not only for the circle — the Reuleaux triangle, the Reuleaux heptagon, every lopsided smooth one, all of them.

For a circle that is the familiar $\pi d$. For the Reuleaux triangle it can be checked directly: three arcs, each of radius $w$, each subtending $60°$, so the total is $3 \cdot \tfrac{1}{6} \cdot 2\pi w = \pi w$. The pentagon: five arcs of radius $w$ subtending $36°$ each, which is $5 \cdot \tfrac{1}{10} \cdot 2\pi w = \pi w$ again.

This is Barbier’s theorem, and the cleanest proof of it is one already told on this site from the other end. Dropping a bent needle on a lined floor establishes that the expected number of crossings is proportional to the curve’s length, with one constant for every curve there is. Drop a curve of constant width $w$ onto a floor ruled at spacing $w$, and it crosses a line exactly twice however it lands — because its extent perpendicular to the lines is always exactly $w$. So the expected crossing count is $2$ for every such curve, the lengths must therefore all be equal, and the circle fixes the value at $\pi w$.

One argument, run in two directions: the needle problem uses the circle to get $\pi$, and this uses $\pi$ to get every other constant-width curve’s perimeter. That the two are the same argument is not a coincidence and is the reason integral geometry exists as a subject.

It is worth pausing on how little the theorem uses. It never asks what the curve looks like, whether it has corners, whether it is symmetric, or how it was constructed. It asks one thing — that the extent between parallel supporting lines is the same in every direction — and returns the perimeter exactly. Results with that shape, where a single constraint determines a quantity outright, are rare enough to be worth collecting; the cutting angle that determines which conic appears is another, and in both cases what makes it possible is that the constraint was chosen to be the one the quantity depends on.

The drill that cuts a square hole #

The Reuleaux triangle’s practical fame comes from a drill bit. Rotate a Reuleaux triangle inside a square whose side equals the triangle’s width, keeping it in contact with all four sides, and it sweeps out almost the entire square.

Almost. It is worth being precise, because the usual telling is not.

The swept region is a square with slightly rounded corners — each corner is an arc, not a point, and the arc is a real amount of missing material: about $1.2\%$ of the square’s area fails to be reached. A hole cut this way is a rounded square and a joiner who needs a true square corner still needs a chisel.

There is a second complication that the geometry conceals. The centre of the rotating triangle does not stay still — it traces a small closed path — so the bit cannot be held in an ordinary chuck. The Watts drill, patented in 1914, uses a floating chuck that allows the bit to wander, and the mechanism is more of the invention than the shape is.

That the shape sweeps a near-square is a genuine consequence of constant width. That it drills a square hole is a claim about a hole, and holes have corners.

The same construction generalises in a way that is more surprising than the square case and much less advertised. A Reuleaux pentagon rotated inside a regular hexagon sweeps the hexagon in the same near-complete way; more generally an odd Reuleaux polygon on $n$ vertices fits a regular $(n+1)$-gon. So there is a family of these bits, and the square is simply its first member — which is the usual situation once a construction is understood rather than remembered, and the same demotion of a famous case to an instance that one cone performs on four curves.

There is also a limit on the whole idea worth stating: no rotating shape sweeps a region with a reflex corner, and none sweeps a triangle, because a rotating convex body cannot reach into an angle sharper than the one its own corners present. The bit’s corner angle for the Reuleaux triangle is $120°$, and $120°$ is what it can get into.

Where the shape is used without ceremony #

The property is quietly load-bearing in one everyday place: coins.

A vending machine measures a coin by rolling it through a slot and reading the gap. A coin of constant width passes correctly whatever its orientation, which is why non-circular coins are constant-width shapes rather than arbitrary polygons — the British twenty-pence and fifty-pence pieces are seven-sided curves of constant width, with the flats slightly bowed for exactly this reason. They look like heptagons and are not; a true heptagon would jam.

The design buys two things at once: a shape distinguishable by touch from the round coins, and a shape a machine can gauge without caring which way up it went in. That is a case where the mathematics is doing real work and nobody is asked to admire it.

The smallest one #

Among all curves of a given constant width, the circle has the largest area — which is the isoperimetric result, since they all have the same perimeter. The natural next question is which has the least, and the answer is the one this essay opened with.

The Blaschke–Lebesgue theorem says the Reuleaux triangle is the unique minimiser: of every shape that measures $w$ in every direction, it encloses the least. Its area is $\tfrac{1}{2}(\pi - \sqrt3)w^2 \approx 0.7048\,w^2$, against the circle’s $\tfrac{\pi}{4}w^2 \approx 0.7854\,w^2$ — about $10\%$ less.

So the family is bracketed at both ends by shapes on this page, and the corners are what does it. Sharpening a curve of constant width removes area and cannot remove width, and the Reuleaux triangle is as sharp as a constant-width curve is allowed to get: three corners, which is the fewest an odd polygon has.

What the picture cannot show #

The calipers in the opening figure are drawn at six angles. The claim is about all of them, and six is not all of them — the same gap as in every drawing on this site that stands in for a general statement.

Worse, a near-miss would be invisible. A shape whose width varied by half a percent would look identical at any drawn size, would roll under a plank with a wobble no eye could catch, and would satisfy every visible feature of these figures. This is why the generator measures the width in seven hundred and twenty directions and requires it to be the arc radius each time, rather than trusting the construction: the difference between constant width and nearly constant width is a real difference with no visual signature at all.

The pictures are also silent on the three-dimensional question, and it is a good one. A body of constant width — one that measures the same between every pair of parallel planes — exists, and the obvious guess is wrong: rotating a Reuleaux triangle about an axis of symmetry gives a solid of constant width, but the tetrahedral analogue built the same way as the triangle, the Reuleaux tetrahedron, is not of constant width. It misses by about $2.5\%$, and fixing it needs a small correction along three of its edges. Nothing about the flat case predicts that, and nothing in a flat drawing could.

The ladder from here #

Rungs above: Barbier’s theorem proved by the support function rather than by needles, with the width written as $h(\theta) + h(\theta + \pi)$. The support function itself, which is the right language for all of this. Smooth curves of constant width, built by choosing a function and constraining it. Blaschke–Lebesgue in full. Bodies of constant width in three dimensions, and the Meissner solids that repair the tetrahedron. The Wankel engine, whose rotor is a Reuleaux triangle doing a different job. Curves of constant brightness, a different constraint with a different answer. And the isoperimetric inequality, where the circle wins at the question this shape loses.

What the caliper was actually measuring #

The moral is about definitions rather than about shapes.

Constant width sounds like a complete description and is a very partial one. It pins down one function of the shape — how far it reaches in each direction, summed with how far it reaches in the opposite direction — and leaves everything else free. Two shapes can agree on that sum at every angle and disagree about almost everything a person would notice.

Measurements that fail to distinguish are the useful ones to collect, because each says exactly which information it throws away. A caliper throws away everything except the summed reach. A determinant throws away everything except the area factor. Euler’s count of odd vertices throws away everything except degree parity. In each case the discarded information is enormous and the retained information answers one question completely — and knowing which question is the whole skill.