= Solution
For $y\ne0$, the squared <Euclidean norm> of the <orthogonal projection> residual is nonnegative:
$$
0\le\left|x-\frac{x\cdot y}{|y|^2}y\right|^2
=|x|^2-\frac{(x\cdot y)^2}{|y|^2}.
$$
This proves the <Cauchy-Schwarz inequality> $|x\cdot y|\le|x||y|$. Equality holds exactly when $x$ is a scalar multiple of $y$. If $y=0$, equality is automatic. Equivalently, equality means the two vectors are <linearly dependent>, including the cases with a zero vector.
For a point $C$ equidistant from the coordinate vertices,
$$
|C-e_i|^2=|C|^2-2C_i+1
$$
must be independent of $i$. Thus all $C_i$ are equal, and their sum is one. The <centroid> is therefore
$$
\boxed{C=\frac1n(1,\ldots,1),\qquad |OC|=\frac1{\sqrt n}.}
$$
For every $x$ in the base <probability simplex>, the <Cauchy-Schwarz inequality> gives $1=(1,\ldots,1)\cdot x\le\sqrt n\,|x|$. Equality forces the same coordinates as $C$. Thus $C$ is the unique <closest point of a probability simplex to the origin>.
The angle $\alpha$ between $OC$ and any edge from the origin to $e_i$ satisfies
$$
\cos\alpha=\frac{C\cdot e_i}{|C|\,|e_i|}=\frac1{\sqrt n}.
$$
Consequently
$$
\boxed{\alpha=\arccos(n^{-1/2})\longrightarrow\frac\pi2,\qquad |OC|\longrightarrow0.}
$$
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