Consider any $n$-simplex, $n \geq 2$. For each edge $(i,j)$, consider $n$-ball $B\_{ij}$ such that vertices $x\_i$ and $x\_j$ are antipodal on this ball. Fix a point $x\_0$ in the simplex. The question: is $x\_0$ in at least $n$ balls?


**Some notes.** The question cannot be strengthened by claiming, for example, that there is vertex $i=i(x\_0)$ such that $x\_0$ is in all balls $B\_{ij},j \neq i$. Also, it is not true for $n+1$ balls (there are counterexamples for both cases if $n=3$).


Just in case, $x\_0$ is in ball $B\_{ij}$ if and only if $\angle x\_ix\_0x\_j \geq \pi/2$, or equivalently, vertices $x\_i$ and $x\_j$ are on opposite sides of the hyperplane $H\_i$ that contains $x\_0$ and is orthogonal to the line through $x\_i$ and $x\_0$ (and similarly for $H\_j$).


Also, this question is a stronger version of [that question](https://mathoverflow.net/questions/140081/cover-of-a-n-simplex-with-balls) (thanks to [zeb](https://mathoverflow.net/users/2363/zeb) for solving it).


Any help is much appreciated. Thank you.



n-simplex in an intersection of n balls
Yes, because a connected graph ($i\sim j$ iff $\angle x\_ix\_0x\_j$ is more than $\frac\pi 2$) with $n+1$ vertices has at least $n$ edges.
The graph is connected because (assuming WLOG that $x\_0=0$ is strictly inside the simplex) there exists a linear combination $\sum\_j y\_j=0$ where $y\_j=c\_jx\_j$, $c\_j>0$. If there are two not connected components, just compute the scalar product of the corresponding sums in two ways: one using the edge definition and one using the fact that the sums are just the opposites of each other.


I do like AoPS style questions, but, honestly, they belong there, not here! 


