TITLE: n-simplex in an intersection of n balls
QUESTION [15 upvotes]: Consider a $n$-simplex, $n \geq 2$ with vertices $x_i,i=1,...,n+1$. 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. It is impossible to make the question stronger by claiming 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 $n=3$).
Also, $x_0$ 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 (Thanks to Alex Ravsky for solving it).
Any help is much appreciated. Thank you.

REPLY [15 votes]: I will use the notation $\mathbf{v}_i$ for the vector from $x_0$ to $x_i$.
Assume that $x_0$ is in the interior of the simplex. (By continuity, the statement will extend to boundary points.) Then
there are some positive weights $w_1,\ldots,w_{n+1}>0$ such that $\sum\limits_{i=1}^{n+1} w_i \mathbf{v}_i=\textbf0$.
Let us build a graph on the vertices $1,\ldots,n+1$. Let the vertices $i$ and $j$ be connected if $x_0$ is inside the ball with diameter $x_ix_j$; in other words, if $\mathbf{v}_i\cdot \mathbf{v}_j\le0$.
We show that this graph is connected. Assume the contrary; then the vertices can be split in two disjoint nonempty sets, say $A$ and $B$ such that $A\cup B=\{1,2,\ldots,n+1\}$. Since there is no edge between the two vertex sets, we have $\mathbf{v}_i\cdot \mathbf{v}_j>0$ for all $i\in A$ and $j\in B$.
Consider
$$
0 = 
\left(\sum_{i\in A\cup B} w_i \mathbf{v}_i \right)^2 =
\left(\sum_{i\in A} w_i \mathbf{v}_i \right)^2 +
\left(\sum_{i\in B} w_i \mathbf{v}_i \right)^2 +
2\sum_{i\in A}\sum_{i\in B} w_iw_j(\mathbf{v}_i\cdot \mathbf{v}_j).
$$
In the last sum all terms are positive, contradiction. Now we have proved that the graph is connected.
A connected graph on $n+1$ vertices has at least $n$ edges, so at least $n$ balls contain $x_0$. Done.