## Problem B2

If $\mathrm{P}$ and $\mathrm{Q}$ are two points in the plane, let $\mathrm{m}(\mathrm{PQ})$ be the perpendicular bisector of $\mathrm{PQ} . \mathrm{S}$ is a finite set of $n>1$ points such that: (1) if $P$ and $Q$ belong to $S$, then some point of $m(P Q)$ belongs to S, (2) if PQ, $\mathrm{P}^{\prime} \mathrm{Q}^{\prime}, \mathrm{P}^{\prime \prime} \mathrm{Q}^{\prime \prime}$ are three distinct segments, whose endpoints are all in S, then if there is a point in all of $\mathrm{m}(\mathrm{PQ}), \mathrm{m}\left(\mathrm{P}^{\prime} \mathrm{Q}^{\prime}\right), \mathrm{m}\left(\mathrm{P}^{\prime \prime} \mathrm{Q}^{\prime \prime}\right)$ it does not belong to $\mathrm{S}$. What are the possible values of $n$ ?

## Answer

$\mathrm{n}=3$ (equilateral triangle), 5 (regular pentagon).


## Solution

There are $n(n-1) / 2$ pairs of points. Each has a point of $S$ on its bisector. But each point of $S$ is on at most two bisectors, so $2 n \geq n(n-1) / 2$. Hence $n \leq 5$.

The equilateral triangle and regular pentagon show that $\mathrm{n}=3,5$ are possible.

Consider $\mathrm{n}=4$. There are 6 pairs of points, so at least one point of $\mathrm{S}$ must be on two bisectors. wlog $\mathrm{A}$ is on the bisectors of $\mathrm{BC}$ and $\mathrm{BD}$. But then it is also on the bisector of CD.

Contradiction.

