, 2017

Solution:

It is easy to verify that the smallest circle enclosing all the points will either have some 2 points in $S$ as its diameter, or will be the circumcircle of some 3 points in $S$ who form an acute triangle.

Now, clearly $\frac{D}{R} \leq 2$. Indeed consider the two farthest pair of points $S_{1}, S_{2}$. Then $D=|S_{1} S_{2}| \leq 2R$, as both points $S_{1}, S_{2}$ are inside a circle of radius $R$. We can achieve this upper bound by taking $S$ to have essentially only 2 points, and the remaining 2015 points in $S$ are at the same place as these 2 points.

For the other direction, I claim $\frac{D}{R} \geq \sqrt{3}$. Recall that the smallest circle is either the circumcircle of 3 points, or has some 2 points as the diameter. In the latter case, say the diameter is $S_{1} S_{2}$. Then $D \geq |S_{1} S_{2}| = 2R$, so $\frac{D}{R} \geq 2$ in that case. Now say the points $S_{1}, S_{2}, S_{3}$ are the circumcircle. WLOG, say that $S_{1} S_{2}$ is the longest side of the triangle. As remarked above, we can assume this triangle is acute. Therefore, $\frac{\pi}{3} \leq \angle S_{1} S_{3} S_{2} \leq \frac{\pi}{2}$. By the Law of Sines we have that
$$D \geq |S_{1} S_{2}| = 2R \sin \angle S_{1} S_{3} S_{2} \geq 2R \sin \frac{\pi}{3} = R \sqrt{3}$$
This completes the proof. To achieve equality, we can take $S$ to have 3 points in the shape of an equilateral triangle.