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A point $(a,b)$ is called a rational point when $a$ and $b$ are both rationals.

Find the maximum possible rational points on a circle centred at $(\pi,2)$.

Answer: $2$

My attempt:

Suppose there exist $3$ (or more) such rational points. Consider a triangle formed by three points.

It is fairly easy to prove that such a "rational triangle" has a rational point as the circumcentre.

But this contradicts the fact that the actual circumcentre is, in fact, not a rational point. Hence, there can exist at most two rational points.

The existence of two rational points:

Let the radius of the circle be $r$. Then, the equation of the circle happens to be $$(x-\pi)^2+(y-2)^2=r^2=r_r+r_i$$ Where $r_r$ denotes the rational part of $r^2$ and the $r_i$ denotes the irrational part of $r^2$.

Let $(a,b)$ be a rational point on this circle.

It is fairly easy to arrive at $$r_r=a^2+(b-2)^2$$ And $$r_i=\pi^2-2\pi a$$

Now, for certain values of $r$, each of them generates a single value each of $r_r$ and $r_i$. And, each value of $r_i$ generates a single value of $a$, which on substituting in the second equation, gives two values of $b$.

Hence, there may exist two such points.

I somehow feel this proof of mine isn't exactly correct. Please correct me if I'm wrong. Other solutions are welcome as well.

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  • $\begingroup$ You should expand on the "fairly easy" part, since that is essentially where all the proof lies... Also, to prove the existence you need only give a concrete example. For instance, you can set the radius to $\sqrt{\pi^2 + 4}$, and point out two rational points. $\endgroup$
    – Trebor
    Jun 16, 2020 at 9:51
  • $\begingroup$ BTW, what's the "rational part" of say $\sqrt{2}$? Is it $0$, or $1$ or maybe it is $1.4142$? That notion of "xyz part" of something makes sense when that part is uniquely determined, or maybe there are only very few possible values. But here, $r_r$ can be any rational number and an appropriate irrational $r_i$ does exist. So that notion doesn't make sense here, IMO. That part of the proof is indeed strange (let's say I don't understand it at all, but I don't understand everything). As Trebor said, all you need to do is find one circle that gives you 2 rational points. $\endgroup$
    – Ingix
    Jun 16, 2020 at 10:52
  • $\begingroup$ @Trebor Well, this actually came on an OMR based test, so the above is a rough sketch of the proof in my mind. $\endgroup$
    – AryanSonwatikar
    Jun 16, 2020 at 11:28
  • $\begingroup$ @Ingix Uhh, now that you mention it, I am not exactly sure as to what I mean by that irrational part. It's getting stranger by the minute. Oof. $\endgroup$
    – AryanSonwatikar
    Jun 16, 2020 at 11:32
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    $\begingroup$ You can solve that "rational part" thing with Hamel basis, but that's way past the supposed difficulty of the problem.. $\endgroup$
    – Trebor
    Jun 17, 2020 at 3:55

1 Answer 1

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$(x-\pi)^2+(y-2)^2=r^2$ and $(w-\pi)^2+(z-2)^2=r^2$ means by subtraction that $(x-\pi)^2-(w-\pi)^2$ is rational which means $x=w$ since clearly $x+w-2 \pi \ne 0$ irrational, when $x,w \in \mathbb Q$

This clearly gives at most two possible values for $y$ (since there is at most one $x$ for which the rational point is on the circle) and one can clearly get two for appropriate radiuses

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    $\begingroup$ (+1) Thanks! It actually should be "radii" instead if "radiuses". $\endgroup$
    – AryanSonwatikar
    Jun 16, 2020 at 12:43

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