, 2003

Solution:
We will construct the set $T$ in the following way: Assume the points of $S$ are in the $xy$-plane and let $P$ be a point in $S$ with maximum $y$-coordinate. This point $P$ will be a member of the set $T$ and now, from $S$, we will remove $P$ and all points in $S$ which are less than $\sqrt{3}$ units from $P$. From the remaining points we choose one with maximum $y$-coordinate to be a member of $T$ and remove from $S$ all points at distance less than $\sqrt{3}$ units from this new point. We continue in this way, until all the points of $S$ are exhausted. Clearly any two points in $T$ are at least $\sqrt{3}$ units apart. To show that $T$ has at least $n / 7$ points, we must prove that at each stage no more than $6$ other points are removed along with $P$.

At a typical stage in this process, we've selected a point $P$ with maximum $y$-coordinate, so any points at distance less than $\sqrt{3}$ from $P$ must lie inside the semicircular region of radius $\sqrt{3}$ centred at $P$ shown in the first diagram below. Since points of $S$ are at least $1$ unit apart, these points must lie outside (or on) the semicircle of radius $1$. (So they lie in the shaded region of the first diagram.) Now divide this shaded region into $6$ congruent regions $R_{1}, R_{2}, \ldots, R_{6}$ as shown in this diagram.

We will show that each of these regions contains at most one point of $S$. Since all $6$ regions are congruent, consider one of them as depicted in the second diagram below. The distance between any two points in this shaded region must be less than the length of the line segment $AB$. The lengths of $PA$ and $PB$ are $\sqrt{3}$ and $1$, respectively, and angle $APB=30^{\circ}$. If we construct a perpendicular from $B$ to $PA$ at $C$, then the length of $PC$ is $\cos 30^{\circ}=\sqrt{3} / 2$. Thus $BC$ is a perpendicular bisector of $PA$ and therefore $AB=PB=1$. So the distance between any two points in this region is less than $1$. Therefore each of $R_{1}, \ldots, R_{6}$ can contain at most one point of $S$, which completes the proof.