The barycenter of a regular tetrahedron coincides with the center of its circumsphere This is a self-answered question, after some playing around. I would be happy to see alternative solutions.
Let $x_1,x_2,x_3,x_4 \in \mathbb{R}^3$ be
the vertices of a regular tetrahedron, i.e. $|x_i-x_j|$ is constant for $i \neq j$.
The $x_i$ lie on a sphere, so after translation and scaling we maw assume $|x_i|=1$; in particular the center of the sphere is at the origin of coordinates.
How to prove that $\sum_i x_i=0$?
I would like to find slick elegant proofs.
 A: Since the segment from a vertex of a regular tetrahedron to the center of its circumsphere is orthogonal to the opposing face, we have $\{x_4\}^\perp=\text{span}(x_1-x_2, x_1-x_3)$.
Thus, we can define an orthogonal map $Q:\mathbb{R}^3 \to \mathbb{R}^3$ in the following way:
$Qx_4=x_4$. Since we want $Q$ to be an isometry, we have $Q(\{x_4\}^\perp)=\{Qx_4\}^\perp=\{x_4\}^\perp$, or $$Q\big(\text{span}(x_1-x_2, x_1-x_3)\big)=\text{span}(x_1-x_2, x_1-x_3).$$
Now, on $\text{span}(x_1-x_2, x_1-x_3)$, we define $Q$ to be a rotation of $\pi/3$ (around $x_4$), i.e. a cyclic rotation of the vertices $x_1,x_2,x_3$.
Write $L=\sum_{i=1}^4x_i$. Then we have
$$
QL=\sum_{i=1}^4 Qx_i=x_4+\sum_{i=1}^3 Qx_i=x_4+\sum_{i=1}^3 x_i=\sum_{i=1}^4 x_i=L.
$$
So, $L$ is an eigenvector of $Q$. Since $x_4$ is an eigenvector and the eigenspace of every orthogonal map which is not the identity is one dimensional, we must have $L \in \text{span}(x_4)$.
But $x_4$ was an arbitrary vertex of the tetrahderon, so $L \in \cap_{i=1}^4\text{span}(x_i)=\{0\}$ as required.
