A compass moving on two straight lines Let $a$ and $b$ be two straight lines intersecting at $O$ with an angle $\theta$ between them. Two sequence of points $\{A_n\} \in a$ and $\{B_n\} \in b$ are such that $A_nB_n=A_{n+1}B_n=A_{n+1}B_{n+1}=d$ for a constant $d$ and all $n \geq 1$ and that $A_n \neq A_{n+1}$, $B_n \neq B_{n+1}$ for all $n$.
Question


*

*Under what conditions (on $\theta$, $A_1$, and $d$) is $\{A_n\}$ a finite set? How to work out $card(\{A_n\})$ and $card(\{B_n\})$, the number of points in those two sets?

*Under what conditions is $\{A_n\}$ an infinite set?

*How is the arrangement of $\{A_n\}$ and $\{B_n\}$ changed when the position of $A_1$ changes?

 A: Choose a direction on $a$ and $b$ and set $a_n=\pm OA_n$, $b_n=\pm OB_n$, where the sign is positive if $A_n$ and $B_n$ are on the positive side of $O$. 
As $B_n$ is on the perpendicular bisector of $A_nA_{n+1}$, and $A_{n+1}$ is on the perpendicular bisector of $B_nB_{n+1}$,
we have the recursive relations:
$$
b_n={a_n+a_{n+1}\over2\cos\theta},\quad
a_{n+1}={b_n+b_{n+1}\over2\cos\theta}.
$$
We can then eliminate $b_n$ to find the single recursion:
$$
a_{n+2}=2\cos2\theta\,a_{n+1}-a_n.
$$
This can be solved to yield (notice that I start with $n=0$):
$$
a_n=a_0\cos2n\theta+{a_1-a_0\cos2\theta\over\sin2\theta}\sin2n\theta.
$$
Finally, we can substitute $a_1$ in the above equation with its expression in terms of $a_0$ and $d$ (of course one needs $d\ge a_0\sin\theta$):
$$
a_1=a_0\cos2\theta\pm2\cos\theta\sqrt{d^2-a_0^2\sin^2\theta},
$$
obtaining:
$$
a_n=a_0\cos2n\theta\pm{\sqrt{d^2-a_0^2\sin^2\theta}\over\sin\theta}\sin2n\theta.
$$
The choice of the sign depends on the choice made at the first step, because given $A_0$ we have two possible choices for $B_0$. Notice that this formula also works for $\theta=\pi/2$, even if the original formula didn't.
It is then apparent that $\{a_n\}$ is finite if and only if $\theta=\pi/k$, with $k\in\mathbb{N}$, $k\ge2$.

