edit: double post, sorry.
dontdisturbmycircles said:if D > (2r)^(1/3) then D must also be greater than r^(1/3). right?
Davee, you can use Pick's theorem to prove that the area of a triangle whose vertices have integer coordinates is at least 1/2.
-it's also obvious that for the shortest line both points will have the same height and opposite x's ( (x,y) , (-x,y) ) because if this isn't true and one of the points has a smaller x that the other one than we can aways make the line shorter by making the point with the bigger x have the same value (with the opposite sign) as the smaller one.
in the solution that i posted above i first assumed that there was some circle centered at (0,0) with a point at B(0,b). now there're two more points to put down A(-x,a) and C(x,c). if both a and c are above the x axis then AC is the longest line.
after looking this over a little more i saw that in my proof i never assumed that the coords were integers! the only restraint on the points needed to solve the problem is that none of the points are closer that sqrt(2) to each other which is true for integer points on a circle. because if this is true than the min distance on the y-axis (or the x-axis) is always atleast 1. with this restraint the proof is the same.