Jordan curve : Farthest point iteration "convergence"

Question

$J$ is a plane simple closed curve.

We build a sequence $(A_n)_n$ as follow:

$A_0$ is a point on $J$.

For $n$ natural number $n>0$ :

$A_n$ is any point in the set of points $X$ on $J$ with the distance $A_{n-1}X$ maximal.

• Can we prove ou disprove the following?

There exists a natural number $N$ and a point $A \in J$, for which:

$(A_{2n})_n=A $ , $\forall n >N$

OR

$(A_{2n+1})_n=A$ , $\forall n >N$

Answer 1

Such a sequence need not have either an even or odd indexed subsequence that is eventually constant.

For example, given an equilateral triangle as our plane closed curve, the sequence that takes each vertex in steady rotation meets the maximal distant point requirement . However the even and odd indexed subsequences are not eventually constant. Indeed neither of them converges, as they are essentially steady rotations through the same three vertices (but in reverse order).