Q. Is the circle the only shape that, when rolling inside itself, has a point that draws out a scaled copy of itself?

Let $C$ be a simple, closed, smooth curve in the plane. (Likely "smooth" can be replaced by $C^2$.) Let $B = s C$, $s \in (0,1)$ be a scaled copy of $C$, and $p$ some fixed point inside $B$. Finally, let $A$ be the trace of $p$ as $B$ rolls (without slippage) inside $C$. An example of what it means to "roll" is provided below: tangents match at the contact point, and arc length rolled matches arc length traversed.

My question is whether it is possible for $A$ to be equal to $t C$, $t \in (0,1)$—a scaled copy of $C$—for any curve $C$ that is not a circle? Clearly the circle has this property, if $p$ is chosen to be the center of $B$. $C$ could conceivably be nonconvex, and perhaps smoothness need not be assumed.

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      A $1 \times \frac{1}{2}$ ellipse rolling inside a $2 \times 1$ ellipse; $p$ slightly off-center.
      (Reload to repeat animation.)

  • $\begingroup$ In order for the trace to be convex and possibly smooth, some assumptions need to be made, such as the curvature of C is greater than the curvature of B at all points of contact. Gerhard "Or Maybe Always Less Than" Paseman, 2017.03.08. $\endgroup$ – Gerhard Paseman Mar 9 '17 at 1:07
  • $\begingroup$ In fact, curvature may be the key here. I bet Robert Bryant could tell you a slick proof that the only possibilities in any dimension are a an n-sphere rolling inside a larger n-sphere, and also that the necessary point must be the center. Gerhard "Not Good At Differential Geometry" Paseman, 2017.03.08. $\endgroup$ – Gerhard Paseman Mar 9 '17 at 1:14
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    $\begingroup$ maybe checking whether a closed Pythagorean Hodograph Curve with that properties exists is easier. $\endgroup$ – Manfred Weis Mar 10 '17 at 10:37
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    $\begingroup$ the curves that are generated by rolling one curve along another, are called Roulettes. The Wiki article also contains a characterization of those curves; may be that provides the lever to decide the question. $\endgroup$ – Manfred Weis Mar 11 '17 at 14:58
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    $\begingroup$ Probably not hard to show that the circle is the only curve that works at any scale s. $\endgroup$ – Ian Agol Apr 11 '17 at 14:40

Refering only to the question as formulated in the title and, ignoring the clarification that aims at giving a precise definition, the answer is yes, namely the non-smooth Reuleaux Triangle if one is willing to accept the Envelope Curve as the curve that is generated by the rolling curve. Further restrictions are of course that in the initial position a corner of the rolling curve must coincide with a corner of the containing curve and, that the rolling curve must be a copy of the containing curve that is shrunk by an integer multiple of 3 in the case of the Reuleaux triangle.

Smooth Reuleaux triangles will fail to generate a scaled copy of itself because the envelope is an offset-curve, which in turn means that a constant value is subtracted from all curvature radii, whereas are a scaled copy would require that all radii are divided by a constant value.


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