# Can every simplex be a regular simplex by pointwise scaling?

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 $n>3.$

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.
• +1. I wonder: What if we get to scale relative to a point (not necessarily $O$) of our choosing?
• 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.