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”.
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 $|alpha(t) times dot alpha(t)|=R$.
