4. Place 2005 points on the circumference of a circle. Two points $P, Q$ are said to form a pair of neigbours if the chord $P Q$ subtends an angle of at most $10^{\circ}$ at the centre. Find the smallest number of pairs of neighbours.
4. Let $P, Q$ be a pair of neighbours Suppose $P$ has $k$ neighbours who are not neighbours of $Q$ and $Q$ has $\ell$ neighbours who are not neightbours of $P$, with $k \geq \ell$. Then by moving $P$ to a point very close to $Q$ the number of neighbouring pairs does not increase and the set of neighbours of $P$ and $Q$ coincide. Thus by repeating this procedure, we see that the minimum is attained when the points are divided into $s$ clusters, $s \leq 35$, such that two points are neighbours iff they are in the same cluster. Thus, if $n_{i}$ is the number of points in cluster $i$, then
$$
\text { minimum } \geq \sum\binom{n_{i}}{2} \geq 25\binom{57}{2}+10\binom{58}{2}=56430 .
$$

The last inequality follows from the following: If the largest of the $n_{i}$ 's, say $a$, is $>58$, then the smallest, say $b$, is $\leq 57$ and $\binom{\bullet}{2}+\binom{b}{2} \geq\binom{ e-1}{2}+\binom{b+1}{2}$ which can be proved by direct computation. This minimum is attained when there are 35 clusters, 25 with 57 points and 10 with 58 points.