3. Let $S$ be a set of 2017 distinct points in the plane. Let $R$ be the radius of the smallest circle containing all points in $S$ on either the interior or boundary. Also, let $D$ be the longest distance between two of the points in $S$. Let $a, b$ are real numbers such that $a \leq \frac{D}{R} \leq b$ for all possible sets $S$, where $a$ is as large as possible and $b$ is as small as possible. Find the pair $(a, b)$.
Answer: $(\sqrt{3}, 2)$
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 $D=\left|S_{1} S_{2}\right| \leq 2 R$, 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 $D \geq\left|S_{1} S_{2}\right|=2 R$, 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
$$
D \geq\left|S_{1} S_{2}\right|=2 R \sin \angle S_{1} S_{3} S_{2} \geq 2 R \sin \frac{\pi}{3}=R \sqrt{3} .
$$

This completes the proof. To achieve equality, we can take $S$ to have 3 points in the shape of an equilateral triangle.