Prove that all the five sequences converge to the same point $P \in \mathbb{R}^3$. Let five sequences $A_n, B_n, C_n, D_n, E_n \in \mathbb{R}^3$ be constructed as follows: $A_0, B_0, C_0, D_0$ and $E_0$ are some given points of the space and $A_{n+1}, B_{n+1}, C_{n+1}, D_{n+1}, E_{n+1}$ are midpoints of the intervals $A_nB_n, B_nC_n, C_nD_n,D_nA_n$ respectively for $n \geq 0$. Prove that all the five sequences converge to the same point $P \in \mathbb{R}^3$.
Observation: If all five initial points $A_0,B_0,C_0,D_0,E_0$ lie on the same plane, then clearly the five sequences converge to the same point because the area of the polygon with vertices as elements from the five sequences converge to $0$. 
Hoewever, if they are not coplanar, then I don't know how to proceed from here Even my observation above is not considered as a rigorous proof. 
Can anyone give some hints?
 A: Partial answer (The estimate below are a bit hard to make precise. Feel free to edit/criticize/downvote):
Let $O = \frac{1}{5}(A_n+B_n+C_n+D_n+E_n)$ be the center of mass of the five points $A_n, B_n, C_n , D_n, E_n$. Note that $O$ is really independent of $n$ as the next five points are midpoints of the previous five. By a translation of the initial data, we assume $O$ is the origin. 
Let $I_n = \max\{|A_n|, |B_n|, |C_n|, |D_n|, |E_n|\}$. It's obiouvs that $I_n\ge I_{n+1}$. To show your claim, it suffices to show that $I_n \to 0$. (Thus $P = O$)
One is tempting to show that $I_{n+1} <(1-\epsilon) I_n$ for some $\epsilon$ small independent of $n$. But this cannot be done in general. Indeed we do have 
Claim:
$I_{n+6} < (1-\epsilon) I_n$ for some $\epsilon>0$ small, for all $n$. 
To show this, WLOG let $I_n = |A_n|$. 


*

*If $\left|\frac{1}{2}(A_n+ B_n) \right| \le (1-\epsilon)I_n$, we go on to the next step (in the next step there would be at most 4 elemnet with $|\cdot|\ge (1-\epsilon)I_n$). 

*If not, then we can choose $\epsilon$ small so that
$$\left|\frac{1}{2}(A_n+ B_n) \right| \ge (1-\epsilon)|A_n|$$
only if $\cos \theta_{A_nB_n} >1-\delta$ and $|B_n| > (1-\delta)|A_n|$, where $\theta_{A_nB_n}$ is the angle between $A_n$ and $B_n$. ($\delta$ here is small, depending on $\epsilon$). 
Now as $O$ is the center of mass of the five points, $C_n$ will not be closed to $B_n$, and $E_n$ not to $A_n$. We choose $\epsilon>0$ small so that 
$$\left| \frac{1}{2}(A_n +E_n) \right|, \left|\frac{1}{2}(B_n+ C_n) \right| \le (1-2\epsilon)|A_n|$$
Keep this in mind, we go to the next $(n+1)$ step. 
Now we have $|A_{n+1}| \ge (1-\epsilon)|I_n|$, $|B_{n+1}|, |E_{n+1}|\le (1-2\epsilon)|I_n|$. So $|A_{n+2}| , |E_{n+2}| \le (1-\epsilon) I_n$. In partucular in the $n+2$ step, at most three elements have $|\cdot | \ge (1-\epsilon) I_n$. 
Then it is obvious that in the $n+6$ step, we have 
$$I_{n+6} < (1-\epsilon)I_n$$
(Remark I suspect one can show $I_{n+2} <(1-\epsilon)I_n$)
