Ambiguous Curve: can you follow the bicycle? Let $\alpha:[0,1]\to \mathbb R^2$ be a smooth closed curve parameterized by the arc length. We will think of $\alpha$ like a back track of the wheel of a bicycle. If we suppose that the distance between the two wheels is $1$ then we can describe the front track by 
$$\tau(t)=\alpha(t)+\alpha'(t)\;.$$
Suppose we know the two (back and front) trace of a bicycle. Can you determine the orientation of the curves? For example if $\alpha$ was a circle the answer is no. 
More precisely the question is:
Is there a smooth closed curve parameterized by the arc length $\alpha$ such that   
$$\tau([0,1])=\gamma([0,1])$$  
where $\gamma(t)=\alpha(1-t)-\alpha'(1-t)$?
If trace of $\alpha$ is a circle we have  $\tau([0,1])=\gamma([0,1])$. Is there another?

Take a figure eight curve, made of two tangent circles, not necesarily of the same size. 
Or an arbitrary chain of tangent circles. 
