/sci/ - Science & Math

Previous Thread >>10589427

Well, a circle is uniquely determined by three points, so we can use the three rational givens, and show that the resulting circle has a small enough radius.

>>10592377
The minimum greatest of the three possible separations is when the points are evenly spaced around the circle (because if one pair was closer together, another pair would necessarily be further apart). Inscribing an equilateral triange ABC in the circle and extending lines from any two of these points through the origin O shows that half this length is equal to r cos(pi/3), so the minimum greatest distance is 2r cos(pi/3) completely disregarding the fact that the points have integer coefficients. 2r cos(pi/3) > r^(1/3) for r > 1, so the case remains to be prove for r < 1.
Within the circle of radius r = 1, there exist only 5 points with integer coefficients, and one (the origin) cannot lie on a circle with radius greater than zero. All the others (1,0), (-1,0), (0,1), (0,-1) lie on the circle of radius one, so on circles with radius less than 1 but greater than 0 no points with integer coefficients exist.

>>10592810
not sure why my brain was liking the word "coefficients" so much those should all be "coordinates"

>>10592810
nice demonstration friend

the problem is very weirdly formulated. Why would you pick r^1/3 as a lower bound when r cos pi(/3) is more accurate ?

>>10592839
I'm worrying I might have misunderstood the wording of the question. It's posed with a lot of English and not much math so I'm guessing there might be some implicit quantifiers I missed

>>10592810
>The minimum greatest of the three possible separations is when the points are evenly spaced around the circle.
No, that is the maximum of the least separation. For real numbers the three points can be chosen very close together.

>>10592377
Some progress:
Estimate the radius of a triangle of sizes lenght d,d and angle between them [math]180-\alpha[/math].
For [math]\alpha[/math] small you get something of the order [math]d/sin(\alpha)[/math] (times constant).
Now estimate how small the angle can get for integer valued points (in terms of the size of the sides). Say the vertice is in the origin. An the two points are vectors a,b.

From the formula [math]-cos(\alpha)=<a,b>/|a||b|[/math] you get
[math]\alpha^2~1-cos(\alpha)=(|a||b|-<a,b>)/|a||b|
[/math]
I think but have not proved yet that the nonzero values coming from integers of
|a||b|-<a,b> are bounded below by a constant
(for example this is the case for (0,d) (1,-d), d large).

Assuming this is the case one can make the radius of given triangle bigger by increasing the angle to near 180, and increasing the sides if necessary so that previous estimates give an upper bound of size O(d^2). For d sufficiently large we are done, the other cases is a finite check for the computer.

>>10593137
I no longer think
[math]|a||b|-<a,b>[/math] is bounded below by a constant. This is because
[math] (|a|^2|b|^2-<a,b>^2)/(|a||b|+<a,b>)=
(a_1b_2-a_2b_1)^2/(|a||b|+<a,b>)[/math]
So if I was looking for "near misses" I would look for a_1b_2-a_2b_1=some small number (1 might work). However this also shows that
the radius is of the order O(d^3). In fact I think the limsup radius/d^3 is [math]1/2\sqrt{2}[/math] by following the above estimations. However the calculations are fairly annoying and I dont want to double check them. As long as this constant is <1 we are done.

>>10593233
I no longer think
|a||b|−<a,b> is bounded below by a constant. This is because
[math](|a|^2|b|^2−<a,b>2)/(|a||b|+<a,b>)=(a_1b_2−a_2b_1)2/(|a||b|+<a,b>)[/math]So if I was looking for "near misses" I would look for a_1b_2−a_2b_1=some small number (1 might work).However this also shows that the radius is of the order O(d^3).In fact I think the limsup of radius/d^3 is [math]1/2\sqrt{2}[/math] by following the above estimations. However the calculations are fairly annoying and I dont want to double check them. As long as this constant is <1 we are done.
(Forgot an opening math for the latex code)