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Berkeley Math Circle Monthly Contest 1

Geometry Difficulty 4.9 AIME Prove it United States

Problem:

A finite number of points are drawn in the plane. Prove that one can select two of them, $A$ and $B$ , such that:

a. $A$ and $B$ are not the same point.

b. No drawn point, other than $A$ itself, is closer to $A$ than $B$ is.

c. No drawn point, other than $B$ itself, is closer to $B$ than $A$ is.

Solution

Solution:

Consider all the distances $X Y$ between two different drawn points. Since a finite number of points are drawn, there are only finitely many distances, and one of them, say $d$ , is minimal. Let $A$ and $B$ be two of the drawn points such that $A B = d$ . Now, by definition, $A$ and $B$ are not the same point.

To verify part (b), let $C$ be any drawn point other than $A$ and $B$ . Then $A C \geq d = A B$ , so $C$ is not closer to $A$ than $B$ is.

The verification of part (c) is similar.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.