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Let $X$ be a non convex subset of $\mathbb{R}^n$. Is it possible that the sets $$S_2=\left\{\dfrac{1}{2}x_1+\dfrac{1}{2}x_2,x_i\in X\right\}$$ and $$S_3=\left\{\dfrac{1}{3}x_3+\dfrac{1}{3}x_4+\dfrac{1}{3}x_5,x_i\in X\right\}$$ are equal? If yes is there any characterization of these sets $X$?

Thomas Andrews
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Michael
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    If $X$ has more than one point, say that $x\neq y\in X$, we have that $\frac12x+\frac12x=\frac13y+\frac13y+\frac13y$, a contradiction. – ajotatxe Oct 25 '16 at 17:23
  • I will edit my question, since I had something else in my head. – Michael Oct 25 '16 at 17:29

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If two different points $x,y\in X$, then $$\frac12x+\frac12x=\frac13y+\frac13y+\frac13y$$ that is $$x=y$$ a contradiction.

Then $X$ is empty or has only one point, so it is hardly non convex.

ajotatxe
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