IMC 2015 · Day 1 · P5

Let $v_i = \overrightarrow{B A_i}$. The condition $\angle A_i B A_j > 90^\circ$ is equivalent with $v_i \cdot v_j < 0$. Since $B$ is an interior point of the simplex, there are some weights $w_1, \dots, w_{n+1} > 0$ with $\sum\limits_{i=1}^{n+1} w_i v_i = 0$.

Let us build a graph on the vertices $1, \dots, n+1$. Let the vertices $i$ and $j$ be connected by an edge if $v_i \cdot v_j < 0$. We show that this graph is connected. Since every connected graph on $n + 1$ vertices has at least $n$ edges, this will prove the problem statement.

Suppose the contrary that the graph is not connected; then the vertices can be split in two disjoint nonempty sets, say $V$ and $W$ such that $V \cup W = \{1, 2, \dots, n+1\}$. Since there is no edge between the two vertex sets, we have $v_i \cdot v_j \ge 0$ for all $i \in V$ and $j \in W$. Consider $$0 = \left( \sum_{i \in V \cup W} w_i v_i \right)^2 = \left( \sum_{i \in V} w_i v_i \right)^2 + \left( \sum_{i \in W} w_i v_i \right)^2 + 2 \sum_{i \in V} \sum_{j \in W} w_i w_j (v_i \cdot v_j).$$ Notice that all terms are nonnegative on the right-hand side. Moreover, $\sum\limits_{i \in V} w_i v_i \ne 0$ and $\sum\limits_{i \in W} w_i v_i \ne 0$, so there are at least two strictly nonzero terms, contradiction.

Remark 1. The number $n$ in the statement is sharp; if $v_{n+1} = (1, 1, \dots, 1)$ and $v_i = (\underbrace{0, \dots, 0}_{i-1}, -1, \underbrace{0, \dots, 0}_{n-i})$ for $i = 1, \dots, n$ then $v_i \cdot v_j < 0$ holds only when $i = n+1$ or $j = n+1$.

Remark 2. The origin of the problem is here: http://math.stackexchange.com/questions/476640/ n-simplex-in-an-intersection-of-n-balls/789390