, 2003

Solution:

The answer is (B). This is a parabolic segment with vertex $V=(0,2)$ and intersections with the axes $A=(-\sqrt{5}, 0)$ and $B=(\sqrt{5}, 0)$. Two of the three points must lie in the same quadrant, and hence their distance cannot exceed the length of the segment $A V$, which is 3 by the Pythagorean theorem (indeed, the part of the region contained in the first quadrant is entirely contained in the circle with diameter $B V$, and the part contained in the second quadrant is entirely contained in the circle with diameter $A V$). Note that all the other answers are false:

(A) at least two of the three points have distance $\geq \frac{\sqrt{5}}{2}$: obviously the points can be taken as close together as one likes!

(C) the sum of the distances between the points is $\leq \frac{9}{2} \sqrt{5}$: just take $A, B$ and $V$ to get $2 \sqrt{5}+6>\frac{9}{2} \sqrt{5}$.

(D) the sum of the squares of the distances between the points is $\leq 38$: taking $A, B, B$ the sum of the squares of the distances equals 40 (if one is looking for 3 distinct points with sum of the squares of the distances greater than 38, it suffices to take $A, B, C$ with $C$ very close to $B$).

(E) the product of the distances between the points is $\leq 16 \sqrt{5}$: again taking $A, B$ and $V$ one gets $18 \sqrt{5}>16 \sqrt{5}$.