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Find the maximum number of rational points on the circle with center $(0,\sqrt3)$


Let the equation of the circle be $x^2+(y-\sqrt3)^2=r^2$

Let $(a,b)$ be any rational point on the circle $x^2+(y-\sqrt3)^2=r^2$.
Then $a^2+(b-\sqrt3)^2=r^2$.
$a^2+b^2+3-2b\sqrt3=r^2$

How can i find the maximum number of rational points from this equation.I have no idea.Can someone please elaborate?

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  • $\begingroup$ what is the value of $r$ ? $\endgroup$
    – DeepSea
    Jan 10, 2016 at 13:05
  • $\begingroup$ Only the center of the circle is mentioned in the question. $\endgroup$
    – diya
    Jan 10, 2016 at 13:06
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    $\begingroup$ This may be useful $\endgroup$
    – Varun Iyer
    Jan 10, 2016 at 13:09
  • $\begingroup$ Make your y term rational $\endgroup$ Jan 10, 2016 at 13:09
  • $\begingroup$ What if both the co-ordinates of the center is irrational? What would be the maximum number of rational points ? $\endgroup$
    – Shashaank
    Sep 26, 2022 at 12:51

2 Answers 2

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Let $\sigma$ be the nontrivial automorphism of $\Bbb Q[\sqrt 3]$.

If $P$ is a point with rational coordinates, the circle $C_1$ with center $(0,\sqrt 3)$ passing through $P$ is given by an equation with coefficients in $\Bbb Q[\sqrt 3]$.

But if $Q$ has rational coordinates and is on $C_1$, then $\sigma(Q)$ is on $\sigma(C_1)$, so $Q$ is also on the circle of center $(0,-\sqrt 3)$ passing through $P$.

Since those two circles intersect at at most two points, (well, $P$ and its symmetric opposite the $y$ axis), there can be at most two rational points on $C_1$.


Alternatively, if you have $3$ rational points on a circle then its center (the circumcenter of the triangle) has to have rational coordinates too, and then it can't be $(0,\sqrt 3)$.

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  • $\begingroup$ i am not sure how that's necessary since this answer already shows that for any rational point $P(x,y)$, its symmetric $(-x,y)$ will be on the same circle. $\endgroup$
    – mercio
    Jan 10, 2016 at 23:35
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For any pair of rational points, the straight line passing through the two may be expressed by an equation with rational coefficients.

The midpoint of the segment joining two rational points is itself a rational point. I guess you know that a line perpendicular to a line with slope $m$ always has a slope $\frac {1}{m}$ implying that the slopes of the two lines are either both rational or both irrational. Therefore, the perpendicular bisector of a segment joining two rational points always has an equation with rational coefficients.

The solution to a system of two linear equations with rational coefficients, if exists, is necessarily rational. Which is to say, that if two straight lines that are expressed by linear equations with rational coefficients intersect, their point of intersection is necessarily rational.

Assume that there is a circle with irrational center on which there are three distinct rational points, say, $A, B, C$. You might be knowing that a three non collinear points always determine a unique circle. Hence the perpendicular bisectors of $AB$ and $BC$ would meet at the circumcenter of our triangle which would be a rational point in contradiction with the conditions of the problem. Hence the maximum number of rational points can only be 2.

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  • $\begingroup$ Just because 3 isn't the answer? Does the maximum number of points become 2? What is your argument for telling that the maximum number of rational points is 2? $\endgroup$
    – Shashaank
    Sep 26, 2022 at 12:49

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