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I am trying to prove the following statement. Any suggestions or references are highly appreciated.

Consider $n$ points in $R^2$, i.e., $x_i\in R^2, i=1,\ldots, n.$ Suppose the centroid (or center of mass with unit mass) denoted as $\bar{x}=\frac{1}{n}\sum_{k=1}^n x_k$ coincides with the circumcenter (namely we assume there exists a circumcircle for the points $x_i, i=1,\ldots,n$), then there exists $\alpha\in R$ such that \begin{align} x_{i+1}-x_i + x_{i-1} - x_i & = \alpha (\bar{x}-x_i), \forall i=2,\ldots,n-1, \\ x_{2}-x_1 + x_{n} - x_1 & = \alpha (\bar{x}-x_1), \\ x_{1}-x_n + x_{n-1} - x_n & = \alpha (\bar{x}-x_n). \end{align}

Chandler
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  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. – José Carlos Santos Aug 21 '18 at 09:51

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This does not look true as stated

Consider the six points $(5,0),(4,3),(-4,3),(-5,0),(-4,-3),(4,-3)$ which clearly have a centroid and circumcentre at $\bar{x}=(0,0)$. Consider $x_i=(-5,0)$ and then $x_i=(-4,-3)$:

  • $(-4,-3)-(-5,0)+(-4,3)-(-5,0) = (2,0) \qquad$ $= \frac25\big((0,0)-(-5,0)\big)$

  • $(4,-3)-(-4,-3)+(-5,0)-(-4,-3) = (7,3) \qquad$ $ \not= \frac25\big((0,0)-(-4,-3)\big)$

Henry
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  • Thank you Henry for the comments. Indeed, the previous statement is not precise. What I was thinking is the vector $x_{i+1}-x_i+x_{i-1}-x_i$ is on the same line as $\bar{x}-x_i$. So the parameter $\alpha$ should be changed to $\alpha_i$ which can be different for different $x_i$. – Chandler Aug 21 '18 at 10:14
  • @Chandler - my second example has $x_{i+1}-x_i+x_{i-1}-x_i$ having a different direction from $\bar{x}-x_i$ – Henry Aug 21 '18 at 10:15
  • I see it now. Thank you so much about this counterexample. – Chandler Aug 21 '18 at 10:25