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

Question

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

Answer 1

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:

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.

Answer 2

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".

Answer 3

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

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

Answer 4

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:

http://www.math.ethz.ch/~blatter/Konstante_Breite.pdf

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.

Answer 5

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.

Moreover, you can observe that if you move the points in a Reuleaux polygon so that one arc vanishes, then the two arcs neighboring the opposite vertex will merge, forming a single arc.