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?

Would you take an (acrobatic) exercise of riding the bike backwards with the front wheel turned back? If you succeed, and managed to keep the front wheel going along a straight line, then the rear wheel would go along a tractrix – despite which direction you chose to ride.
