From what I have seen (and to some extend read), a curve of constant width generated from a polygon with an even number of sides is not possible. Wikipedia cites an Oxford University paper when it says

Curves of constant width can be generated by joining circular arcs centered on the vertices of a regular or irregular convex polygon with an odd number of sides (triangle, pentagon, heptagon, etc.)

It says curves of width can be generated with polygons with an odd number of sides, it does not explicitly rule out even numbers of sides. I suppose this can be asked in two questions:

Question 1 Are all curves of constant width generateable with the "Reaulaux Method"?

It seems that every document I read mentions curves of constant width generated with circular arcs, which does make sense. Is this the only possible method? Examples can be seen on the Wikipedia Article. I'm going to say that a circle can be generated with this method as well. What is the justification for the correct answer?

Question 2 Must a curve of constant width be generated with an odd number of sides?

This question is harder to answer if the answer to question 1 is no, so if that is the case this can be left out (unless it is still easy to answer, that is). Mrf's answer provided insight that could be lead to a proof that regular polygons used to generate with the "Reulaux Method" must have an odd number of sides, but is this also true for irregular polygons? I'm guessing yes but what I need is a proof

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    $\begingroup$ Sorry but maybe you can explain a little more exact. I find this hard to understand $\endgroup$ – Bernd Jan 9 '14 at 7:54
  • $\begingroup$ @Bernd thanks for the heads up, I reworded the question completely. $\endgroup$ – Pinpickle Jan 9 '14 at 8:03
  • $\begingroup$ can you add a link to the wikipedia article, maybe it is just because the number of "inside sides " is equal to the number of "outside sides" so the total number is always even. $\endgroup$ – Willemien Jan 9 '14 at 13:17
  • $\begingroup$ @Willemien I have added the link in the question, you can also have it here $\endgroup$ – Pinpickle Jan 9 '14 at 22:56
  • $\begingroup$ The question is not succinct enough, given an curve of constant width how do you define the polygon that generated it? , for example at en.wikipedia.org/wiki/Reuleaux_polygon#Reuleaux_polygons in en.wikipedia.org/wiki/File:Reuleaux_kite.svg why is the blue kite not the polygon that generated the curve? $\endgroup$ – Willemien Jan 10 '14 at 11:20

As far as I understand the questions, you mean by the "Reuleaux-method" that you take a polygon and add circular arcs to obtain a convex body of constant width. Assuming this, the answer to question 1 should be negative:

In the book "How round is your circle" there is another method to construct convex bodies of constant width: See on the How round is your circle webpage, the example titled "Half a convex shape". In this example, half an ellipse is completed to a convex body. So for curvature reasons this body can not be generated by the "Reuleaux-method".


Just for fun, I made a little animation of the half-convex example linked to by @Alexander Schmeding. The upper half is an ellipse with minor-to-major axis ratio ranging from $b = 1/2$ to $b = \sqrt{2}$. The curve can be parametrized as $$f_b(\theta) = \begin{cases} (\cos \theta, b \sin \theta), & 0 \le \theta \le \pi, \\ {\displaystyle \frac{2 (b \cos \theta, \sin \theta)}{\sqrt{b^2 \cos^2 \theta + \sin^2 \theta}}} - (\cos \theta, b \sin \theta), & \pi < \theta \le 2\pi. \end{cases}$$ In Mathematica, we can generate an interactive plot with

F[t_, b_] := Piecewise[{{{Cos[t], b Sin[t]}, 0 <= t < Pi},
             {2 {b Cos[t], Sin[t]}/Sqrt[(b Cos[t])^2 + Sin[t]^2]
               - {Cos[t], b Sin[t]}, Pi <= t <= 2 Pi}}]

Manipulate[Show[ParametricPlot[F[t, b], {t, 0, 2 Pi}, 
                PlotRange -> {{-1, 1}, {-1.5, 1.5}}, PlotStyle -> Black], 
           Graphics[Flatten[{Opacity[0.5], {Hue[#/Pi], 
           Line[{F[#, b], F[# + Pi, b]}]} & /@ (Range[n] Pi/n)}]]]
       {{b, 1}, 1/2, Sqrt[2]}, {n, 1, 75, 1}]

And we get this:

enter image description here

What I found really interesting is how the curve looks like it's the same near the extremes of the animation, but it obviously isn't so from the definition of $f$ itself. Bonus points if you can parametrize the envelope of normals for $b \in [1/2, \sqrt{2}]$. And more bonus points if you can compute the enclosed area.


It depends on what you mean by the number of sides of a curve.

If $k$ is an odd integer, and $p(t) = a\cos^2(kt/2) + b$ then $$ \begin{cases} x(t) = p(t)\cos t - p'(t)\sin t \\ y(t) = p(t)\sin t + p'(t)\cos t \end{cases} $$ where $0 \le t \le 2\pi$ is a curve of constant width. Choosing $k=3$, $a=3$, $b=1$ you end up with non-convex curve of constant width

How many sides does that curve have? More details can be found in this document

  • $\begingroup$ Don't curves of constant width have to be convex? The document you cited says the support function $p$ has to be convex and even says itself that particular values for $k$, $a$ and $b$, outlining a case where $k=3$, as it does in your curve. I don't think it's a curve of constant width. $\endgroup$ – Pinpickle Jan 10 '14 at 2:47
  • $\begingroup$ I guess it's a matter of taste $\endgroup$ – mrf Jan 10 '14 at 6:40
  • $\begingroup$ That document has provided some really useful insights into this though. For example, it seems as though having $k$ as an even number will create a non-constant width function ($p(t)+p(t+\pi)$). Coming up with a proof for that would be easy enough probably but using that assumes that this creates every possible curve of constant width - which it may very well do. $\endgroup$ – Pinpickle Jan 10 '14 at 9:03

Concerning your first question: It is easy to produce curves of constant width without corners. See here, in particular the upper half of page 107:


This paper is in German. The relevant statement on page 107 is the following:

Put $${\bf u}(\phi):=(\cos\phi,\sin\phi),\qquad {\bf u}'(\phi):=(-\sin\phi,\cos\phi)$$ and consider the parametric representation $$\gamma:\quad {\bf z}(\phi):= p(\phi){\bf u}(\phi)+p'(\phi){\bf u}'(\phi)\qquad(\phi\in{\mathbb R}/(2\pi))\ ,\tag{1}$$ where the function $p\in C^2$ fulfills the following conditions: $$p(\phi)+p''(\phi)>0,\qquad p(\phi)+p(\phi+\pi)\equiv b\ .\tag{2}$$ Then $\gamma$ is a closed convex curve of constant width $b$, and $(1)$ is a regular parametrization of $\gamma$. In fact from $(1)$ one computes $${\bf z}'(\phi)=\bigl(p(\phi)+p''(\phi)\bigr){\bf u}'(\phi)\ ,$$ so that $$|{\bf z}'(\phi)|=p(\phi)+p''(\phi)>0,\qquad\arg\bigl({\bf z}'(\phi)\bigr)=\phi+{\pi\over2}\ .$$ It follows that the radius of curvature $\rho$ along $\gamma$ is given by $$\rho(\phi)=p(\phi)+p''(\phi)\ .$$ When $p$ is sufficiently smooth, then so is $\rho$, whereas in the case of curves obtained by the "Reuleaux method" the radius of curvature is piecewise constant and has jump discontinuities at the hinges of the construction.

On the other hand the curves obtained by the "Reuleaux method" have a representation of the form $(1)$ as well. The function $p$ is then only continuous and piecewise of the form $$p(\phi)=\rho_i+a_i\cos\phi+b_i\sin(\phi)\qquad(\alpha_{i-1}<\phi<\alpha_i)\ .$$

$C^\infty$-functions $p$ fulfilling the conditions $(2)$ are, e.g., the trigonometric polynomials $$p(\phi):={b\over2}+\sum_{j=1}^N\biggl(a_j\cos\bigl(2j+1)\phi\bigr)+b_j\sin\bigl((2j+1)\phi\bigr)\biggr)\ ,$$ where the $|a_j|$ and $|b_j|$ are sufficiently small.

  • $\begingroup$ Thanks for deciphering that paper. What I meant in my first question was not whether you could have shapes of constant width without corners, but if you could have curves of constant width that cannot be generated with a set of circular arcs whose origins make up the vertices of a convex polygon (which I've dubbed the 'Reulaux Method'). These curves that you have listed, can they be generated in such a way? $\endgroup$ – Pinpickle Jan 14 '14 at 8:03

Question 1 The answer is no. Take the Minkowski sum of a Reuleaux triangle and a small ball. You'll get a Reuleaux triangle with rounded corners.

On the other hand there exist general Reuleaux polygons which are curves of constant width obtained as an intersection of a finite number of circles. First let's note that Reuleaux polygons need not be regular. If we have more than $5$ sides a Reuleaux polygon may be non-regular. It is possible to prove that Reuleaux polygons are dense in the class of curves of constant width: i.e. all curves of constant width can be approximated as well as you want with a Reuleaux polygon.

Question 2 As the answer to the first question is no, the answer to this question is also negative.

Nevertheless, a Reuleaux polygon (described above) must be generated by an even number of arcs. This is due to the fact that to every vertex we must be able to associate an opposite arc. This is possible only if we have an odd number of arcs.


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