Mean vector minimized angle $\DeclareMathOperator*{\argmin}{arg\,min}$Suppose I have $n$ vectors $\vec{v}_1,\cdots,\vec{v}_n \in \mathbb{R}^N$. If I define the mean vector as:
$$\hat{\vec{v}}=\frac{1}{n}\sum_{i=1}^{n}\vec{v}_i$$
And the unit vector, which minimizes the largest possible angle between any of the $\vec{v}_i$ and itself, as:
$$\vec{v}^\theta=\argmin_{|\vec{v}|=1}\bigg(\max_{i\in\{1,\cdots,n\}}\bigg(\angle\vec{v}\vec{v}_i\bigg)\bigg)$$
where $\angle \vec{v} \vec{v}_i$ is the (unsigned) angle between $\vec{v}$ and $\vec{v}_i$; are the following two statements equivalent? And if so why?

*

*$\hat{\vec{v}}$ is non-zero and for all $i \in \{1,\cdots,n\}$ it is true that $\angle \hat{\vec{v}} \vec{v}_i<\frac{\pi}{2}$.

*$\vec{v}^\theta$ is uniquely defined and for all $i \in \{1,\cdots,n\}$ it is true that $\angle \vec{v}^\theta\vec{v}_i<\frac{\pi}{2}$.

When I draw it always seems true but how to rigoursly prove it?
 A: $\def\u{\vec{u}}\def\v{\vec{v}}\def\paren#1{\left(#1\right)}$These two propositions are not equivalent. Here is an explicit counter-example for $N = 2$ and $n = 2$: Take $\v_1 = (x_1, y_1) \in (-∞, 0) × (0, +∞)$ and $\v_2 = (x_2, 0)$ where $x_2 > -\dfrac{x_1^2 + y_1^2}{x_1}$. It is easy to see that $\v^θ$ is the bisector of $\angle \v_1\v_2$, so$$
\angle \v^θ\v_1 = \angle \v^θ\v_2 = \frac{1}{2} \angle \v_1\v_2 < \frac{π}{2},
$$
but$$
\hat{\v} · \v_1 = \frac{1}{2} (x_1 + x_2, y_1) · (x_1, y_1) = \frac{1}{2} (x_1^2 + y_1^2 + x_1 x_2) < 0,
$$
i.e. $\angle \hat{\v}\v_1 > \dfrac{π}{2}$.
For general $N$ and $n$, the key observation is that the lengths of $\v_k$'s does not affect $\v^θ$ but does affect $\hat{\v}$. Therefore, if $\v_1, \cdots, \v_n$ satisfy the following conditions:

*
*All $\v_k$'s are in one half-space, i.e. there exists $\u$ such that $\u · \v_k > 0$ for all $k$;

*The length of one of $v_k$'s is far larger than others' (w.l.o.g. suppose it is $\v_1$), i.e.\begin{gather*}
\|\v_1\| \gg \max_{k ≠ 1} \|\v_k\|; \tag{1}
\end{gather*}

*One of $\v_k$'s ($k ≠ 1$) satisfies $\v_1 · \v_k < 0$ (w.l.o.g. suppose it is $\v_2$);

then by the definition of $\v^θ$,$$
\max_k \angle \v^θ\v_k \leqslant \max_k \angle \u\v_k < \frac{π}{2}, 
$$
but\begin{gather*}
\hat{\v} ≈ \frac{1}{n} \v_1 \Longrightarrow \hat{\v} · \v_2 ≈ \frac{1}{n} \v_1 · \v_2 < 0. \tag{2}
\end{gather*}
To be more rigorous, defining $\u_k = \dfrac{\v_k}{\|\v_k\|}$ for all $k$, (1) can be expressed as\begin{gather*}
\|\v_1\| > -\frac{1}{\u_1 · \u_2} \sum_{k ≠ 1} \|\v_k\|, \tag{1$'$}
\end{gather*}
and thus\begin{gather*}
\hat{\v} · \v_2 = \frac{1}{n} \paren{ \sum_{k = 1}^n \v_k } · \v_2 \leqslant \frac{1}{n} \paren{ \v_1 · \v_2 + \sum_{k ≠ 1} \|\v_k\| \|\v_2\| }\\
= \frac{1}{n} \paren{ \|\v_1\| \|\v_2\| · (\u_1 · \u_2) + \sum_{k ≠ 1} \|\v_k\| \|\v_2\| } < 0. \tag{2$'$}
\end{gather*}
