Some people are playing paintball against each other. In a given situation, the distances between them are all different. In this case, everyone shoots at the person closest to them. What is the maximum number of people who can shoot at the same person?
Solution. We will prove that at most five can shoot at the same person.

Suppose that (at least) six shoot at the same point, $A$. Based on the diagram, let $\alpha_{i}$ denote the angle $A_{i} A A_{i+1}$. If $\alpha_{i}>60^{\circ}(1 \leq i, i \in \mathbb{N})$, then the sum of these angles would be greater than $360^{\circ}$, which is impossible.

![](https://cdn.mathpix.com/cropped/2024_05_02_345dbb1fd6f4cbba5585g-1.jpg?height=295&width=526&top_left_y=308&top_left_x=797)

If one of the $\alpha_{i}$ is not greater than $60^{\circ}$: $\alpha_{j} \leq 60^{\circ}$. Since $A$ is the target, $A_{j} A_{j+1}$ is the longest side of the triangle $A_{j} A A_{j+1}$, so $\alpha_{j}$ is the largest angle of the triangle, which is impossible.

Therefore, at most five can shoot at the same person, and this is possible: stand 5, 6, 7, 8, and 9 units away from $A$ in a line with the other players. If $A$ looks at the closest player, then in sequence, they can look at each with turns of $70^{\circ}, 70^{\circ}, 70^{\circ}, 70^{\circ}$. Using the cosine rule, it can be verified that the distances between the players are different for each pair, and that the adjacent players are farther from each other than from $A$.

![](https://cdn.mathpix.com/cropped/2024_05_02_345dbb1fd6f4cbba5585g-1.jpg?height=434&width=534&top_left_y=954&top_left_x=790)