MHB Can three points on a circle have a distance between them of less than r^(1/3)?

[Ackbach]

Here is this week's POTW:

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Three distinct points with integer coordinates lie in the plane on a circle of radius $r>0$. Show that two of these points are separated by a distance of at least $r^{1/3}$.

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[Ackbach]

Re: Problem Of The Week # 263 - May 15, 2017

This was Problem A-5 in the 2000 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg for his correct answer, which follows:

Spoiler

By a translation, we can assume that one of the three points is at the origin. The other two points are then $(a,b)$ and $(c,d)$, where $a,b,c,d$ are all integers. Let $l$ be the maximum distance between the points. Then $a^2+b^2 \leqslant l^2$ and $c^2+d^2 \leqslant l^2$. Also, the three points are not collinear, which implies that $\dfrac ba \ne \dfrac dc$. (That assumes that $a$ and $c$ are nonzero. A minor modification would be needed to deal with the case where one of them is zero.) So $bc-ad\ne0$. Since $a,b,c,d$ are all integers, it follows that $|bc-ad| \geqslant1.$The perpendicular bisector of the line from $(0,0)$ to $(a,b)$ has equation $y - \frac12b = -\frac ab\bigl(x - \frac12a\bigr)$, or $-2ax + 2by = a^2 + b^2$. In the same way, the perpendicular bisector of the line from $(0,0)$ to $(c,d)$ has equation $-2cx + 2dy = c^2+d^2$. Those two lines intersect at the centre of the circle. Solving the two equations, you see that the centre is at the point $(x,y)$ given by $$2(bc-ad)x = d(a^2+b^2) - b(c^2+d^2), \qquad 2(bc-ad)y = c(a^2+b^2) - a(c^2+d^2).$$ Since $1\leqslant |bc-ad|$, it follows that $$4x^2 \leqslant \bigl(d(a^2+b^2) - b(c^2+d^2)\bigr)^2 \leqslant (d^2 + b^2)\bigl((a^2+b^2)^2 + (c^2+d^2)^2 \bigr) \leqslant 2(d^2 + b^2)l^4$$ (the middle of those three inequalities coming from Cauchy–Schwarz). Therefore $x^2 \leqslant \frac12(d^2 + b^2)l^4$, and in the same way $y^2 \leqslant \frac12(c^2+a^2)l^2$.

The radius of the circle is given by $r^2 = x^2+y^2 \leqslant \frac12\bigl((a^2 +b^2 + c^2 + d^2\bigr)l^4 \leqslant \frac12(2l^2)l^4 = l^6$, from which $l \geqslant r^{1/3}.$