Library / All-Soviet-Union Mathematical Olympiad /134 of 185

All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 5.7 AIME, harder Prove it Soviet Union

Problem:

Given a finite set $X$ of points in the plane. $S$ is a set of vectors $AB$ where $(A, B)$ are some pairs of points in $X$ . For every point $A$ the number of vectors $AB$ (starting at $A$ ) in $S$ equals the number of vectors $CA$ (ending at $A$ ) in $S$ . Show that the sum of the vectors in $S$ is zero.

Solution

Solution:

Let $X = \{P_1, P_2, \ldots, P_n\}$ . For each vector $AB$ in $S$ , let $\vec{v}_{AB}$ denote the vector from $A$ to $B$ .

The sum of all vectors in $S$ is:
$\sum_{AB \in S} \vec{v}_{AB} = \sum_{AB \in S} (\vec{OB} - \vec{OA})$
where $O$ is the origin and $\vec{OA}$ is the position vector of $A$ .

This can be rewritten as:
$\sum_{AB \in S} \vec{OB} - \sum_{AB \in S} \vec{OA}$

For each point $A$ , let $d^+_A$ be the number of vectors in $S$ starting at $A$ (i.e., the number of $AB$ in $S$ for fixed $A$ ), and $d^-_A$ be the number of vectors in $S$ ending at $A$ (i.e., the number of $CA$ in $S$ for fixed $A$ ).

Then:
$\sum_{AB \in S} \vec{OA} = \sum_{A \in X} d^+_A \vec{OA}$
$\sum_{AB \in S} \vec{OB} = \sum_{B \in X} d^-_B \vec{OB}$

So the total sum is:
$\sum_{B \in X} d^-_B \vec{OB} - \sum_{A \in X} d^+_A \vec{OA} = \sum_{A \in X} (d^-_A - d^+_A) \vec{OA}$

But by the given, for every $A$ , $d^+_A = d^-_A$ , so $d^-_A - d^+_A = 0$ for all $A$ .

Therefore, the sum is zero.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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