8. Let $\mathcal{R}$ be the finite region in the plane bounded by the $x$-axis and the graph of the curve given by the equation $2 x^{2}+5 y=10$. Given three points in $\mathcal{R}$, which of the following statements is always true?

(A) at least two of the three points have a distance $\geq \frac{\sqrt{5}}{2}$

(B) at least two of the three points have a distance $\leq 3$

(C) the sum of the distances between the points is $\leq \frac{9}{2} \sqrt{5}$

(D) the sum of the squares of the distances between the points is $\leq 38$

(E) the product of the distances between the points is $\leq 16 \sqrt{5}$
8. The answer is (B). It is a segment of a parabola with the vertex $V=(0,2)$ and intersections with the axes $A=(-\sqrt{5}, 0)$ and $B=(\sqrt{5}, 0)$. Two of the three points must be in the same quadrant, and therefore their distance cannot exceed the length of the segment $A V$, which is 3 by the Pythagorean theorem (indeed, the part of the region in the first quadrant is entirely contained within the circle with diameter $B V$, and the part in the second quadrant is entirely contained within the circle with diameter $A V$). Note that all other answers are false:

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

(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 is equal to 40 (if one seeks 3 distinct points with a sum of the squares of the distances greater than 38, just take $A, B, C$ with $C$ very close to $B$).

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