Track / Stage 6 / 85 of 400 #1085 of 2000

Problem 1085

National olympiad, first round

Geometry Difficulty 6.1 Prove it

$10 \cdot 5$ Given $n$ points in space, any three of which form a triangle with one interior angle greater than $120^{\circ}$ . Prove that the points can be labeled as $A_{1}, A_{2}, \cdots, A_{n}$ , such that each $\angle A_{i} A_{j} A_{k}$ is greater than $120^{\circ}$ , where $1 \leqslant i<j<k \leqslant n$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

[Proof] Let the longest line among the connections between $n$ points be denoted as $A_{1} A_{n}$ , and label the remaining points according to their distance from $A_{1}$ in ascending order as $A_{2}, A_{3}, \cdots, A_{n-1}$ , i.e., $A_{1} A_{2} < A_{1} A_{3} < \cdots < A_{1} A_{n-1}$ . Since $A_{1} A_{n}$ is the longest, we have $\angle A_{1} A_{2} A_{n} > 120^{\circ}$ . On the other hand, because $A_{1} A_{n}$ is the longest, we have $\angle A_{1} A_{i} A_{n} > 120^{\circ}$ for $i < n$ and $\angle A_{1} A_{j} A_{n} > 120^{\circ}$ for $j < n$ . Therefore, $\angle A_{i} A_{1} A_{n} > 120^{\circ}$ . Thus, when $1 < i < j < k < n$ , we have $\angle A_{1} A_{i} A_{k} > 120^{\circ}$ and $\angle A_{1} A_{j} A_{k} > 120^{\circ}$ . Since the sum of the three dihedral angles (including the degenerate case of three rays emanating from a point on a plane) is no more than $360^{\circ}$ , we have $\angle A_{j} A_{i} A_{k} < 120^{\circ}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.