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For $X=(x_1,\dots,x_n) \in (\mathbb{R}^2)^n$, the generated zonotope (zonogon in 2D) is defined by $ Z(X) := \{\sum_{k=1}^n \sigma_k x_k : \sigma_1,\dots,\sigma_n \in [0,1] \}. $

Which $X$ generate the regular $2n$-gon inscribed in the unit disk centered at the origin?

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    $\begingroup$ You take vectors at angles $\frac{0}{n} \pi, \frac{1}{n} \pi, \ldots, \frac{n-1}{n} \pi$ of appropriate equal length (which should be easy to work out based on the unit disk requirement, but I am too lazy to do it...) $\endgroup$
    – Sam Hopkins
    Commented Jan 6, 2023 at 19:00
  • $\begingroup$ Thanks. See edit. $\endgroup$
    – BerndM
    Commented Jan 6, 2023 at 20:05
  • $\begingroup$ The vertices of this $2n$-gon are $(\cos(\theta),\sin(\theta))$ for $\theta=\frac{0}{2n}\pi, \frac{1}{2n}\pi,\ldots, \frac{2n-1}{2n}\pi$. Choose $n+1$ consecutive of these points and make your vectors be the consecutive differences. Anyways, I don't think this is a research-level math question, so probably belongs on math.stackexchange.com instead... $\endgroup$
    – Sam Hopkins
    Commented Jan 6, 2023 at 20:09
  • $\begingroup$ By the way, your definition of zonotope implicitly assumes that the generating vectors are based at the origin, but with this restriction you will never be able to get something centered at the origin. Better to define a zonotope to be a Minkowski sum of line segments. $\endgroup$
    – Sam Hopkins
    Commented Jan 6, 2023 at 21:29
  • $\begingroup$ For $n=3$, take $X=( (1,0), (\cos\frac{2\pi}3,\sin\frac{2\pi}3), (\cos\frac{4\pi}3,\sin\frac{4\pi}3) )$. $\endgroup$
    – BerndM
    Commented Jan 6, 2023 at 21:42

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