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There is a $n$-simplex with $n+1$ vertices $\{\mathrm P_i\} \quad (i=0, \cdots, n).$ (That is, P_i are not co-hyperplanar points.) And the origin $O$ is inside of the simplex. Is there a collection of real numbers $\{k_i\in \mathbb R: i=0, \cdots, n\}$ such that $\{k_i\mathrm P_i\}$ forms a regular $n$-simplex?

It seems to be true when $n \le 2$, but I cannot generalize it to case of $n>3.$

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No. For $n=3$, suppose $P_0$, $P_1$, and $P_2$ lie on $x$, $y$, and $z$ axes respectively. Then $k_0P_0$, $k_1P_1$, and $k_2P_2$ form an equilateral triangle only when they are all equidistant from the origin. Suppose they are $(a,0,0)$, $(0,a,0)$, and $(0,0,a)$. Then the fourth vertex of the simplex must lie along the line $x=y=z$. If $P_3$ does not lie on this line, you are out of luck.

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  • $\begingroup$ +1. I wonder: What if we get to scale relative to a point (not necessarily $O$) of our choosing? $\endgroup$
    – Blue
    Oct 6 '15 at 8:41
  • $\begingroup$ So simple counterexample... Thanks. And I wonder the answer of above question - if we can choose the origin as we like? $\endgroup$
    – Proia Lie
    Oct 6 '15 at 9:16
  • $\begingroup$ Half-answering the question in my comment: The scaling-center (call it $P$) needs to be such that unit vectors $\overrightarrow{PP_i}/|\overrightarrow{PP_i}|$ determine a regular $n$-simplex. For the triangle, $P$ is the Fermat point. I would guess that generalizations of this point (or proof that they don't exist) for higher-dimensional simplices are discussed somewhere in the literature. $\endgroup$
    – Blue
    Oct 6 '15 at 10:01
  • $\begingroup$ To be honest, I thought this problem studying the Fermat point in Euclidean space. For a triangle whose angles are all less than 120 deg, we can find a regular triangle whose center is its Fermat point. (And if k_i < 0, we can reduce the condition of angles.) And I think: How about in higher dimension? $\endgroup$
    – Proia Lie
    Oct 6 '15 at 10:17

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