Solution:

a) Let $H$ be the smallest convex set of points in the plane which contains $S$. Take 3 points $a, b, c \in S$ which lie on the boundary of $H$. (There must always be at least 3 (but not necessarily 4) such points.)

Since $a$ lies on the boundary of the convex region $H$, we can construct a chord $L$ such that no two points of $H$ lie on opposite sides of $L$. Of the two points where the perpendicular to $L$ at $a$ meets the circle, choose one which is on a side of $L$ not containing any points of $H$ and call this point $A$. Certainly $A$ is closer to $a$ than to any other point on $L$ or on the other side of $L$. Hence $A$ is closer to $a$ than to any other point of $S$. We can find the required points $B$ and $C$ in an analogous way and the proof is complete.

(a)

(b)

b) Let $P Q R$ be an equilateral triangle inscribed in the circle and let $a, b, c$ be midpoints of the three sides of $\triangle P Q R$. If $r$ is the radius of the circle, then every point on the circle is within $(\sqrt{3} / 2) r$ of one of $a, b$ or $c$. (See figure (b) above.) Now $\sqrt{3} / 2<9 / 10$, so if $S$ consists of $a, b, c$ and a cluster of points within $r / 10$ of the centre of the circle, then we cannot select 4 points from $S$ (and corresponding points on the circle) having the desired property.

Solution:

a) If all the points of $S$ lie on a line $L$, then choose any 3 of them to be $a, b, c$. Let $A$ be a point on the circle which meets the perpendicular to $L$ at $a$. Clearly $A$ is closer to $a$ than to any other point on $L$, and hence closer than any other point in $S$. We find $B$ and $C$ in an analogous way.

Otherwise, choose $a, b, c$ from $S$ so that the triangle formed by these points has maximal area. Construct the altitude from the side $b c$ to the point $a$ and extend this line until it meets the circle at $A$. We claim that $A$ is closer to $a$ than to any other point in $S$.

Suppose not. Let $x$ be a point in $S$ for which the distance from $A$ to $x$ is less than the distance from $A$ to $a$. Then the perpendicular distance from $x$ to the line $b c$ must be greater than the perpendicular distance from $a$ to the line $b c$. But then the triangle formed by the points $x, b, c$ has greater area than the triangle formed by $a, b, c$, contradicting the original choice of these 3 points. Therefore $A$ is closer to $a$ than to any other point in $S$.

The points $B$ and $C$ are found by constructing similar altitudes through $b$ and $c$, respectively.

b) See Solution 1.