Circle's slightly harder challenge

[davee123]

-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.

I don't quite follow this assumption-- Let's suppose you're dropping down points A, B, and C, and putting B directly on the Y axis as you state. Hence the coordinates for A will have a negative X value, and the coordinates for C will have a positive X value. I'm with you so far. But putting down A and C as you mention above seems like it's not covering all the cases.

Let's suppose that B = (0,10). Now you decide to drop point A at (-1,13). By your assumption, we should be putting C = (1,13). But then we've ruined the fact that the longest side is AC, and it's instead a tie between AB and BC. Instead, we could place point C = (1,6).

Now, in theory, you get the same circle if you place A at (-3,9) and C at (4,11), and yes, technically, the *distance* between A and C gets shorter if you place C at (3,9) instead of (4,11). But you've also decreased the radius of the circumcircle, which (in theory) may have suddenly crossed some threshhold.

I suppose if you can prove via calculus that the rate of decrease in the distance of AC is less than the rate of decrease of the cube root of the radius, and show what the minimal case is, you've got a solution.

DaveE

 

[daniel_i_l]

I think that the problem is that we're both looking at it from opposite perspectives.
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. if they're below then you can always "reflect" them up without changing the solution.
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.

 

[davee123]

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.

Yeah, I think if you assume that the origin of the circle is on an integer coordinate, and that the middle point of the three points shares one coordinate with the origin of the circle (making the radius an exact integer), then I expect your proof holds.

I was tempted to try a similar strategy by assuming that two of the points were necessarily separated by a distance of 1, trying to reduce the overall distance to the 3rd point. But that didn't seem to be necessarily true, so I abandoned that one...

DaveE