Closed orbits of a geometrically defined dynamical system

Question

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I am not sure of the answer but after a few examples, I think that the process repeats $\frac{360}{\angle(\text{line,X-axis})}$. And perhaps it's possible for all angles like perhaps the irrational values of $\theta$. I guess all the Rationals will lead to a repetition. Not sure though.

For example,in the picture, we have begun with a point P on the X-axis and then continued till we reached P again. Here, the angle between line l and the x-axis is 60∘ and we see 6 steps required to reach back to P.

Answer 1

Assume $0<\theta<{\pi\over2}$.

Claim. The resulting orbit is closed iff $\theta$ is a rational multiple of $\pi$.

Proof. We draw our figure in the complex plane. Let $\ell_1$ be the real axis, and choose $\ell_2$ through the origin. We denote the points on $\ell_1$ by $p_k$ and the points on $\ell_2$ by $q_k$ and then obtain an orbit of the form $$(p_0,q_0,p_1,q_1,p_2, \ldots)\ .$$ Define $$z_k:=q_k-p_k\ ,\qquad{\rm resp.,}\qquad\vec z_k:=\vec{p_k q_k}\ .$$ Note that for all $k\geq0$ one has $|z_k|=L$ with a given $L>0$. Furthermore $p_k$ and $q_k$ are uniquely defined by $z_k$: Place the tail of $\vec z_k$ anywhere on $\ell_1$ and then translate $\vec z_k$ horizontally until its head is lying on $\ell_2$. The final position of the tail then is $p_k$, and the final position of the head is $q_k$. It follows that it is sufficient to analyze the sequence $(z_k)_{k\geq0}$.

Inspecting the orbit construction shows that $$p_1-q_0=\overline{q_0-p_0}\ .$$ In the same way as $p_1-q_0$ results from conjugating $q_0-p_0$ with respect the real axis $\ell_1$ the next difference $q_1-p_1$ results from "conjugating" $p_1-q_0$ with respect to $\ell_2$. It follows that $$z_1=q_1-p_1=e^{i\theta}\>\overline{e^{-i\theta}(p_1-q_0)}=e^{2i\theta}(q_0-p_0)=e^{2i\theta} z_0\ .$$ But this implies $$z_{k+1}=e^{2i\theta} z_k\qquad(k\geq0)\ ,$$ and then, by induction, $$z_k=e^{2ik\theta}z_0\qquad(k\geq0)\ .\qquad\square$$ If $\theta={m\over n}\pi$ in lowest terms then the period of the orbit is $n$, independently of the choice of $p_0\in \ell_1$ and $q_0\in\ell_2$, thereby fixing an $L=|q_0-p_0|>0$.

Answer 2

Assuming $P=(x,0)\ne (0,0)$ and $\theta \not \in \{0,\pi /2\} $, whether you return to $P$ does not depend on $x$ but on the angle $\psi$ between the $X$-axis and the segment joining $P$ to the first point chosen on $L.$ Because we can multiply all distances by a positive constant without affecting the outcome.

Using co-ordinates we can show by computing the successive points in the sequence that if you return to $P$ then there is a $2$-variable polynomial $f$ such that $f( \cos \psi,\sin \psi)=0$ where the co-efficients of $f$ belong to the field $F$ generated by $\{\cos \theta,\sin \theta\}.$

I haven't checked but I think that $\deg (f)\ne 0.$ This implies that $\sin^2 \psi$ is algebraic over $F$, and $F$ is countable, so there are only countably many possible $\psi$ for which you return to $P$ (for a given $\theta $) but uncountably many $\psi$ where you don't.

Perhaps there is an ingenious geometric solution ?