Finding the middle of n points

Question

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!

Answer 1

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.

Answer 2

A remark; not an answer.

If you had posed the question for the line rather than the plane, then $A$ needs to be the Voronoi Diagram for $P$. For example, for $$ A = \left( \frac{1}{2}, 3, 6 \right) \;, $$ we need $$ P = ( 0, 1, 5, 7 ) \;. $$ So $A$ must represent a possible $1$-dimensional Voronoi Diagram.

I have ignored the wrap-around requirement that $A_n=(P_n+P_1)/2$, and instead used the more natural (to me) $A_n=(P_n+P_{n+1})/2$. So $|P|=|A|+1$.

Answer 3

So you'd like to see $A_1 = P_1/2 + P_2/2, A_2 = P_2/2 + P_3/2, \dots, A_n = P_n/2 + P_1/2$. What if we try this in $p$ dimensions (you asked about $p = 2$) and represented the components of $A_1$ as $A_{11}, \dots, A_{1p}$, and so forth for all $A$s, then the same thing for the $P$s. You'd have

\begin{equation} A := \begin{bmatrix} A_{11} & A_{12} & \cdots & A_{1p} \\ \vdots & & & \vdots \\ A_{n1} & A_{n2} & \cdots & A_{np} \end{bmatrix}. \end{equation} and \begin{equation} P := \begin{bmatrix} P_{11} & P_{12} & \cdots & P_{1p} \\ \vdots & & & \vdots \\ P_{n1} & P_{n2} & \cdots & P_{np} \end{bmatrix}. \end{equation}

The coordinates of $A_i$ are represented by row $i$ of $A$, etc.

Looking at it this way, the question you're asking is whether it's possible to find $P$ satisfying $A = WP$ where $W$ is the $n \times n$ matrix

\begin{equation} W := \begin{bmatrix} 1/2 & 1/2 & 0 & \dots & 0 \\ 0 & 1/2 & 1/2 & \dots & 0 \\ \vdots & & & & \vdots \\ 1/2 & 0 & 0 & \dots & 1/2 \end{bmatrix}. \end{equation}

You'd be home-free if you could show the determinant of $W$ is nonzero.