Reflection of a rational point with respect to line having its equation with rational coefficient gives a rational point

What is the number of rational points on a circle with centre $(\sqrt 2, \sqrt 3)$ and with any radius

In this the centre is not rational then also how they decided the answer as 'at most 1'

Original source : https://i.stack.imgur.com/3aMvD.gif

  • 4
    $\begingroup$ If you could, please properly translate the question and post it here, instead of an image. It may be clearer and people can search for it later $\endgroup$ – AspiringMathematician Apr 14 '17 at 5:02
  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Apr 14 '17 at 5:15
  • $\begingroup$ I aprove what @Guilherme K. Nakassima says, you should know, as you are not at all a newcomer, that it is important, for searches, that all is written, and not under the form of a picture, whenever possible? $\endgroup$ – Jean Marie Apr 14 '17 at 7:16
  • $\begingroup$ Which institute's practice paper is this question from? $\endgroup$ – Navin Apr 17 '17 at 19:12
  • $\begingroup$ @navinstudent Allen carrer institute $\endgroup$ – Koolman Apr 18 '17 at 2:39

Suppose for the sake of contradiction that the circle contains two rational points.

If we have two rational points $A(x_1, y_1)$ and $B(x_2, y_2)$ on the circle, then the perpendicular bisector of $AB$ has equation $$y - \frac{y_1 + y_2}{2} = -\frac{x_2 - x_1}{y_2 - y_1}\left(x - \frac{x_1 + x_2}{2}\right)$$ which can be written as $ax + by = c$ for rational $a, b, c$.

But this perpendicular bisector must pass through $(\sqrt2, \sqrt3)$, the center of the circle, so we have $a \sqrt 2 + b \sqrt3 = c$. This is a contradiction: there is no rational relation between $\sqrt 2$ and $\sqrt 3$.

Therefore at most one rational point can lie on the circle.

  • $\begingroup$ I don't think it is necessary to pass through $(\sqrt 2 , \sqrt 3)$ $\endgroup$ – Koolman Apr 14 '17 at 5:11
  • $\begingroup$ The perpendicular bisector of any chord of a circle is a diameter of the circle, so it must pass through the center. (The perpendicular bisector of $AB$ consists of all the points equidistant from $A$ and $B$, and the center of the circle is certainly equidistant from $A$ and $B$.) $\endgroup$ – Misha Lavrov Apr 14 '17 at 5:11
  • $\begingroup$ But the answer is given as 'at most 1' $\endgroup$ – Koolman Apr 14 '17 at 5:13
  • $\begingroup$ we assume that there's two points, and get a contradiction, so the assumption must be false, and there can be at most one point. (I've edited my answer to clarify.) $\endgroup$ – Misha Lavrov Apr 14 '17 at 5:14
  • $\begingroup$ @Koolman the answer shows two or more has contradiction (directly shows for two, but indicates for more, because you could draw more of those bisectors if you have more than two points). Thus, this proves at most 1 $\endgroup$ – Yujie Zha Apr 14 '17 at 5:15

Let $(a,b)$ and $(c,d)$ be distinct rational points on the circle.

$$\begin{array}{rcl} (a-\sqrt2)^2 + (b-\sqrt3)^2 &=& (c-\sqrt2)^2 + (d-\sqrt3)^2 \\ a^2 + b^2 - 2a\sqrt2 - 2b\sqrt3 &=& c^2 + d^2 - 2c\sqrt2 - 2d\sqrt3\\ a^2 + b^2 - c^2 - d^2 &=& 2(a-c)\sqrt2 + 2(b-d)\sqrt3\\ (a^2 + b^2 - c^2 - d^2)^2 &=& 8(a-c)^2 + 12(b-d)^2 + 8(a-c)(b-d)\sqrt6 \\ \sqrt6 &=& \dfrac{(a^2 + b^2 - c^2 - d^2)^2 - 8(a-c)^2 - 12(b-d)^2}{8(a-c)(b-d)}\\ \end{array}$$

Contradicting the fact that $\sqrt6$ is irrational, unless $a=c$ or $b=d$.

I believe you can deal with the cases:

  • Case 1: $a=c$ and $b \ne d$
  • Case 2: $a \ne c$ and $b=d$

To complete the proof.

  • $\begingroup$ From case 1 $b+d =2\sqrt 3$ . And from case 2$a+c=2\sqrt 3$ $\endgroup$ – Koolman Apr 14 '17 at 5:07

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