Vectors with pairwise bounded angular distance Take a subset $V$ of $\mathbb{R}^n$ for which no two vectors can have an angle greater than $\pi/2$  between them. Does it follow that $V$ is contained in a hypercone? To be specific, is there a vector $u\in \mathbb{R}^n$ such that for all $v\in V$ the angle between $u$ and $v$ is at most $\pi/4$? 
It seems fairly intuitive that this should be so, but I couldn't think of an argument. If not, a counter example would be nice. 
 A: This is actually not the case. Consider $\{e_{1}=(1,0,0),e_{2}=(0,1,0),e_{3}=(0,0,1)\}\subset\mathbb{R}^{3}$. Clearly 
$$\frac{\langle e_{i},e_{j}\rangle}{\|e_{i}\|\|e_{j}\|}=\cos(\theta)\geq0$$
for all $i,j\in\{1,2,3\}$ where $\theta$ is the angle between $e_{i}$ and $e_{j}$. Now suppose there is some $u=(u_{1},u_{2},u_{3})$ such that the angle between $u$ and $e_{i}$ is at most $\pi/4$, i.e. 
$$\frac{\langle e_{i},u\rangle}{\|e_{i}\|\|u\|}\geq\frac{1}{2}\sqrt{2}$$
Clearly $x,y,z>0$. We may assume wlog that $u_{1}^{2}+u_{2}^{2}+u_{3}^{2}=3$. Then $u_{i}\leq 1$ for some $i$. We find 
$$\frac{1}{2}\sqrt{2}\leq\frac{\langle e_{i},u\rangle}{\|e_{i}\|\|u\|}=\frac{u_{i}}{\sqrt{3}}\leq\frac{1}{3}\sqrt{3}$$
which is a contradiction.
A: The answer is negative in all its generality because. Here is a counter-example : take as subset the three vectors of the canonical basis $e_1,e_2,e_3$ (see figure), 

Fig. 1 : The intersection of the three cones should have contained a non zero vector $V$.
We are looking for a vector $V \neq 0$ with coordinates $(x,y,z)$  belonging to the intersection of the three cones ("solid cones" = their surface and their interior) with resp. equations (see explanation below) : 
$$\begin{cases}x^2+y^2&\leq&z^2\\x^2+z^2&\leq&y^2\\y^2+z^2&\leq&x^2\\\end{cases}\tag{1}$$
But no such vector exists : the intersection is reduced to the null vector ; this is convincing on Fig. 1 but an algebraic proof is at our fingertips : indeed, if the coordinates $(x,y,z)$ of $V$ satisfy these three inequations, they satisfy their sum, i.e., are such that $2(x^2+y^2+z^2) \leq (x^2+y^2+z^2)$ which is possible only for the null vector.
Explanation of inequalities (1) :
Let us take the example of inequality 
$$x^2+y^2\leq z^2.\tag{2}$$
We start from the constraint : 
$$0 \leq \alpha \leq \pi/4 \ \iff \ \cos(\alpha) \geq \cos(\pi/4),$$
itself equivalent to the following condition involving a dot product :
$$\cos(\alpha) = \dfrac{V \cdot e_3}{\|V\|\|e_3\|}=\dfrac{0x+0y+1z}{\sqrt{x^2+y^2+z^2} . 1} \geq \dfrac{1}{\sqrt{2}}$$
squaring, we get : $\dfrac{z^2}{x^2+y^2+z^2} \geq \dfrac12$
which itself is equivalent to inequality (2), which was our objective.
Remark 1 : more generaly, this brings a counter-example even for a set of $n$ vectors ; as long as this set contains these 3 vectors, there is no hope...
Remark 2 : this solution has much  in common with that of @Floris Claassens, but thinking in terms of cones brings a global view. 
