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edit: double post, sorry.

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- Thread starter dontdisturbmycircles
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- #26

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edit: double post, sorry.

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- #27

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dontdisturbmycircles said:if D > (2r)^(1/3) then D must also be greater than r^(1/3). right?

Oh yeah, duh.

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.

Huh-- I had never heard of this one before... Interesting.

DaveE

- #28

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Yea, I admit, it was a hard one... :P

- #29

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- #30

daniel_i_l

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I think that you can solve it without pick's theorem, here's how i did it:

-first put the first point on the circle, to make it easier lets assume that it's on the y-axis with the height of r which is an integer, this doesn't change anything cause i can always rotate the coordinate system so that it is on the y-axis.

-now we have to put down two more integer points, one with a positive x and one with a negative one (if you want to put them on the same side then just choose a different point as the "midpoint" so that there will be one point on the positive y-axis and two more on either side of it)

-since it's obvious that the longest line will be between the two points on either side of the middle we have to prove that that distance will always be bigger that r^1/3.

-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.

-so the distance d = 2x where x is the distance of the points from the y-axis.

-so we have to prove that for every right-angle triangle with 3 integer sides, the smaller side (x) will always be bigger that 0.5r^1/3 where r is the diagonal.

-to prove this we use the following equation:

x^2 = 2ny + n^2 for every right angle integer triangle where x is the smaller side, y is the bigger side and n is r-y. it is easy to see that this is true by drawing r^2 as a square with a side the length of r and then to fill in it's bottom right corner with a square the size of y (the bigger side) all the area left is equal to x^2.

-using this it's easy to show that x > r^1/3 and then d > x > r^1/3

Q.E.D

what do you think?

-first put the first point on the circle, to make it easier lets assume that it's on the y-axis with the height of r which is an integer, this doesn't change anything cause i can always rotate the coordinate system so that it is on the y-axis.

-now we have to put down two more integer points, one with a positive x and one with a negative one (if you want to put them on the same side then just choose a different point as the "midpoint" so that there will be one point on the positive y-axis and two more on either side of it)

-since it's obvious that the longest line will be between the two points on either side of the middle we have to prove that that distance will always be bigger that r^1/3.

-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.

-so the distance d = 2x where x is the distance of the points from the y-axis.

-so we have to prove that for every right-angle triangle with 3 integer sides, the smaller side (x) will always be bigger that 0.5r^1/3 where r is the diagonal.

-to prove this we use the following equation:

x^2 = 2ny + n^2 for every right angle integer triangle where x is the smaller side, y is the bigger side and n is r-y. it is easy to see that this is true by drawing r^2 as a square with a side the length of r and then to fill in it's bottom right corner with a square the size of y (the bigger side) all the area left is equal to x^2.

-using this it's easy to show that x > r^1/3 and then d > x > r^1/3

Q.E.D

what do you think?

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- #31

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-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

- #32

daniel_i_l

Gold Member

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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.

- #33

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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 seperated 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

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