A polygon with constant angular momentum bounds a circle

Question

Let $\alpha:[0,L] \to \mathbb{R}^2$ be a piecewise affine map satisfying $\alpha(0)= \alpha(L)$ and $|\dot \alpha|=1$. Supopse that $\alpha(t) \times \dot \alpha(t)$ is constant.

How to prove that $\operatorname{Image}(\alpha)$ is a tangential polygon, i.e. a polygon whose edges are all tangent to a fixed circle, centered at the origin?

It suffices to prove that for each subinterval $[a,b] \subseteq [0,L]$ where $\alpha|_{[a,b]}$ is affine, there exists a $t_0 \in (a,b)$ such that $\dot \alpha(t_0) \perp \alpha(t_0)$.

Indeed, if this is the case, then $|\alpha(t_0)|=|\alpha(t) \times \dot \alpha(t)|=C$ is independent of the segment $[a,b]$ chosen. Thus, every "edge" $\alpha([a,b])$, contains a point $P_{a,b}=\alpha(t_0)$ on the circle with radius $C$, and the edge is perpendicular to radius at $P_{a,b}$, i.e. it is tangent to the circle at $P_{a,b}$.

I am not sure how to prove the bold statement. I think we need to use somehow the fact that the polygon "closes".

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The converse implication is easy:

If there exists such a circle with radius $R$, then $|\alpha(t) \times \dot \alpha(t)|=R$ is constant: Indeed, suppose that $\alpha(t_0)$ lies on the circle -- so it is a tangency point.

Then $\dot \alpha(t_0) \perp \alpha(t_0)$, and $|\alpha(t_0)|=R$.

Let $t$ satisfies $\dot \alpha(t)=\dot \alpha(t_0)$, i.e. $\alpha(t)$ belongs to the same edge as $\alpha(t_0)$. Then

$\alpha(t)=\alpha(t_0)+\beta(t)$, where $\beta(t) || \dot \alpha(t_0)$, so $$ \alpha(t) \times \dot \alpha(t)=\big( \alpha(t_0)+\beta(t) \big) \times \dot \alpha(t_0)=\alpha(t_0) \times \dot \alpha(t_0), $$ which implies $|\alpha(t) \times \dot \alpha(t)|=R$.

Answer 1

Suppose $|\alpha(t)\times\dot{\alpha}(t)|=C$. Then this means that each affine piece of the path, extended to a line if necessary, is tangent to the circle of radius $C$ centred at the origin.

Suppose further $\alpha(t)\times\dot{\alpha}(t)>0$ throughout. Then $\alpha$ moves along each tangent keeping the circle to its left. So at each vertex, $\alpha$ moves from one tangent to the circle to the other one, and one knows in which direction along it to move.

I believe that's as far as you can go however. I imagine the trace of such an $\alpha$ should fall under some definition of tangential polygon. Do note that the condition you hoped to prove need not hold. For example if $\alpha$ is as in my picture, moving from $A$ to $B$, from $B$ to $C$ etc., then while the angular momentum is constant, at no point does $\alpha$ touch the circle.

Answer 2

Working in the complex plane.

Consider a sub-segment $[a,b] \subseteq [0,L]$ where $\alpha(t)$ lies on a line. By hypothesis, it exists $(\omega, k) \in \mathbb U \times \mathbb R$ such that

$$\omega \overline{\alpha} + \overline{\omega} \alpha = k.$$ By differentiation, this implies

$$\omega \overline{\dot\alpha} + \overline{\omega} \dot \alpha =0.$$

We also have $\dot \alpha\overline{\dot \alpha} = 1$ and $\overline{\alpha }\dot \alpha - \alpha\overline{\dot \alpha} = 2iK $ where $K \in \mathbb R$ which follows from the condition $ \alpha \times \dot \alpha$ constant. The distance of the line on which "the mobile" is running to the origin is equal to $\vert k \vert$.

From there we get $$\left(\omega \overline{\dot \alpha}\right)^2 + 1=0$$ and finally that $\alpha(t) = \pm i\omega t + a$ where $a \in \mathbb C$.

We will be able to conclude the desired result if we prove that $k$ only depends on $K$.

Which is the case as $k= \omega \bar a + \bar \omega a$ while $2iK=\pm i (\bar a \omega + a \bar \omega)$.

Note: we proved above in several lines the "evident fact" that a mobile running at a constant speed equal to $1$ along a line in the complex plane has an equation of the form $\alpha(t) = vt+ a$ where $(v,a) \in \mathbb U \times \mathbb C$.