# Can a shape rolling inside itself reproduce that shape?

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.

A $1 \times \frac{1}{2}$ ellipse rolling inside a $2 \times 1$ ellipse; $p$ slightly off-center.