I'm looking for a proof for this problem on simplex which I think it is true



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> **Question.** $A\_0A\_1...A\_n$ is a simplex in the Euclidean space $\Bbb E^n$. $G$ is its centroid and its center circumscribed sphere is $O.$ Let $O\_i$ be the centers circumscribed sphere of the simplex $GA\_{0}A\_{1}...A\_{i-1}A\_{i+1}...A\_n$. Prove that $O$ is centroid of the simplex $O\_0O\_1...O\_n.$
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Centroid and center circumscribed spheres in simplex
Let $O$ be the origin, $\vec{OA\_i}=a\_i$, $\vec{OO\_i}=p\_i$, $\vec{OG}=a$.


Then $\|a\_i\|=1$ and $\sum a\_i/(n+1)=a$. We want to prove that $\sum p\_i=0$.


We know that for $i \neq j$:
\begin{align}
\|p\_i-a\|^2 &= \|p\_i-a\_j\|^2\\
(p\_i,a\_j) &= (p\_i,a)+\frac{\|a\_j\|^2-\|a\|^2}{2}.
\end{align}
Summing up over all $j\ne i$:
\begin{align}
(p\_i,(n+1)a-a\_i) &= n(p\_i,a)+\frac{n-n\|a\|^2}{2} \\
(p\_i,a\_i) &= (p\_i,a)+\frac{-n+n\|a\|^2}{2} \\
\sum\_i(p\_i,a\_j) &= (p\_j,a\_j)+\sum\_{k\ne i}(p\_k,a\_j)=\sum\_i (p\_i,a).
\end{align}
Now $\sum\_i p\_i$ is orthogonal to $a-a\_i$, but these vectors generate the whole space and we conclude that $\sum\_i p\_i=0$.


