Answer: any $\mathrm{C}$ such that $\angle \mathrm{ACB}=180^{\circ} \mathrm{r} / \mathrm{s}$, with $\mathrm{r}$ and $\mathrm{s}$ relatively prime integers and $\mathrm{s}$ not a power of 2.

Let $\angle \mathrm{AC}_{\mathrm{n}} \mathrm{B}=\mathrm{x}_{\mathrm{n}}$, where the angle is measured clockwise, so that $\mathrm{x}_{\mathrm{n}}$ is positive on one side of $A B$ and negative on the other side. Then $x_{n}$ uniquely identifies $C_{n}$ on the perpendicular bisector.

We have $\mathrm{x}_{\mathrm{n}+1}=2 \mathrm{x}_{\mathrm{n}}$. To make this work in all cases we have to take it mod $180^{\circ}$ (so that if $A C_{n} B$ is obtuse, then $C_{n+1}$ lies on the other side of $A B$). If $x_{n}$ is eventually periodic then $x_{m+1}$ $=\mathrm{x}_{\mathrm{n}+1}$, for some $\mathrm{n}>\mathrm{m}$, so $\left(2^{\mathrm{n}}-2^{\mathrm{m}}\right) \mathrm{x}_{1}=0 \bmod 180$. Hence $\mathrm{x}_{1}=180 \mathrm{r} / \mathrm{s}$ for some relatively prime integers $\mathrm{r}, \mathrm{s}$. Also $\mathrm{s}$ cannot be a power of 2 for then we would have $\mathrm{x}_{\mathrm{k}}=180 \mathrm{r}$ for some $\mathrm{k}$, in which case the sequence would terminate rather than be periodic.

Conversely, suppose $\mathrm{x}_{1}=180 \mathrm{r} / \mathrm{s}$, with $\mathrm{r}$ and $\mathrm{s}$ relatively prime and $\mathrm{s}$ not a power of 2. Then $\mathrm{x}_{\mathrm{n}+1}=1802^{\mathrm{n}} \mathrm{r} / \mathrm{s}$ cannot be $0 \bmod 180$, so the sequence does not terminate. Put $\mathrm{s}=2^{\mathrm{b}} \mathrm{c}$, with $\mathrm{c}$ odd. Let $\mathrm{d}=\varphi(\mathrm{c})$, where $\varphi(\mathrm{m})$ is Euler's phi function, so that $2^{\mathrm{d}}=1 \mathrm{mod} \mathrm{c}$. Then $\mathrm{x}_{\mathrm{b}+1}=180 \mathrm{r} / \mathrm{c}$ $\bmod 180$ and $2^{\mathrm{b}+\mathrm{d}}=2^{\mathrm{b}} \bmod \mathrm{c}$, so $\mathrm{x}_{\mathrm{b}+\mathrm{d}+1}=180 \mathrm{r} / \mathrm{c}$ mod 180. Hence the sequence is periodic.