TITLE: Finding the middle of n points
QUESTION [7 upvotes]: Sorry for my possible bad English, I have a problem that I spent a bit of time on and I have been blocked on it for a couple of hours, I'll try to translate it as best as I can:
Given an integer $n\ge3$, and given $A_1,\ldots,A_n$ points on a plane, under what condition(s) can we find points $P_1,\ldots,P_n$ such that $A_1$ is the midpoint of $[P_1,P_2]$, $A_2$ is the midpoint of $[P_2,P_3]$, ... and $A_n$ is the midpoint of $[P_n,P_1]$?
Please tell me if I wasn't clear enough in my translation, and thank you very much in advance!

REPLY [2 votes]: You can always do this when $n$ is odd, but if $n$ is even you can only do this if the respective centroids (or sums) of the even $A_i$ points and the odd $A_i$ points are coincident.
Suppose we have $P_1$ defined somehow. Then
\begin{align}
P_2 &= 2A_1-P_1 \\
P_3 &= 2A_2-P_2 = 2A_2 - 2A_1 + P_1 \\
P_4 &= 2A_3-P_3 = 2A_3 - 2A_2 + 2A_1 - P_1 \\
\end{align}
and consider that we need $P_{n+1}$ as defined by continuation to be equal to $P_1$.
When $n$ is odd, we have $P_1 = 2A_n-2A_{n-1}+ 2A_{n-2}-\ldots-2A_2+2A_1 - P_1$
and so we see that we need $P_1 = A_n - A_{n-1} + A_{n-2}-\ldots- A_2+ A_1$
When $n$ is even , we have $P_1 = 2A_n-2A_{n-1}+ 2A_{n-2}-\ldots+2A_2-2A_1 + P_1$,
which is only possible if $ 2A_n-2A_{n-1}+ 2A_{n-2}-\ldots+2A_2-2A_1 = 0$
or as stated  $ A_n+ A_{n-2}+\ldots+A_2 = A_{n-1}+ A_{n-3}+\ldots+ A_1$
(and in this case , the location of $P_1$ is arbitrary).
Notice this result holds regardless of the dimension of space the $A_i$ points are embedded in.