The question is written like this:

Is it possible to find an infinite set of points in the plane, not all on the same straight line, such that the distance between EVERY pair of points is rational?

This would be so easy if these points could be on the same straight line, but I couldn't get any idea to solve the question above(not all points on the same straight line). I believe there must be a kind of concatenation between the points but I couldn't figure it out.

What I tried is totally mess. I tried to draw some triangles and to connect some points from one triangle to another, but in vain.

Note: I want to see a real example of such an infinite set of points in the plane that can be an answer for the question. A graph for these points would be helpful.

Ahmed Amir
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    The "rationality vs integrality" part of this article gives an example where no three of the points lie on the same line: https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Anning_theorem – Marc Oct 21 '16 at 11:33
  • Possibly related: https://math.stackexchange.com/questions/2097901/ – Watson Nov 28 '18 at 12:42
  • @Ahmed, Can you tell me which standard do you study now? What inspired you to ask this nice question? – MAS Aug 07 '19 at 16:57

3 Answers3


You can even find infinitely many such points on the unit circle: Let $\mathscr S$ be the set of all points on the unit circle such that $\tan \left(\frac {\theta}4\right)\in \mathbb Q$. If $(\cos(\alpha),\sin(\alpha))$ and $(\cos(\beta),\sin(\beta))$ are two points on the circle then a little geometry tells us that the distance between them is (the absolute value of) $$2 \sin \left(\frac {\alpha}2\right)\cos \left(\frac {\beta}2\right)-2 \sin \left(\frac {\beta}2\right)\cos \left(\frac {\alpha}2\right)$$ and, if the points are both in $\mathscr S$ then this is rational.

Details: The distance formula is an immediate consequence of the fact that, if two points on the circle have an angle $\phi$ between them, then the distance between them is (the absolute value of) $2\sin \frac {\phi}2$. For the rationality note that $$z=\tan \frac {\phi}2 \implies \cos \phi= \frac {1-z^2}{1+z^2} \quad \& \quad \sin \phi= \frac {2z}{1+z^2}$$

Note: Of course $\mathscr S$ is dense on the circle. So far as I am aware, it is unknown whether you can find such a set which is dense on the entire plane.

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  • Can you please explain this in more details, especially the tan part. – Ahmed Amir Oct 21 '16 at 12:04
  • Which part in particular? Can you get the distance formula? – lulu Oct 21 '16 at 12:17
  • No, I tried to use Pythagoras, but couldn't get the same formula. – Ahmed Amir Oct 21 '16 at 12:26
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    Suppose two points on the circle have angle $\phi$ between them. Bisect the chord connecting them to get two right triangles, each with central angle $\frac {\phi}2$. The length of the side opposite that central angle is $\sin \frac {\phi}2$. I then use the addition formula for $\sin$ applied to the angle $\frac {\alpha - \beta}2$. – lulu Oct 21 '16 at 12:33
  • It makes sense now.. Will try to make it again by myself.. Thanks – Ahmed Amir Oct 21 '16 at 12:54
  • No problem. Write again if something is still unclear. – lulu Oct 21 '16 at 12:56
  • Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/47186/discussion-between-ahmed-amir-and-lulu). – Ahmed Amir Oct 21 '16 at 17:07
  • @lulu I'm interested in how could your solution be built from rational Pythagorean triples, as $(\frac {1-z^2}{1+z^2}, \frac {2z}{1+z^2},1)$ is a rational Pythagorean triples, too. –  Oct 22 '16 at 15:23
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    @J.Pak well, the $\tan x2$ transformation is a rationalization of the circle so my set is just a rational transformation of the Pythagorean set. To be precise: take the point $(-1,0)$ on the circle and extend the line connecting it to a point $(0,z)$ for $0 – lulu Oct 22 '16 at 15:31
  • Not only are there no known examples, it is known that there are no examples, on algebraic curves other than lines and circles. – zyx Oct 23 '16 at 05:14
  • @xyz Didn't realize that. I am aware of a number of non-existence results for particular families, but not such a sweeping result. I'll go read. thanks! – lulu Oct 23 '16 at 11:45
  • @xyz For those interested, the reference result is discussed [here](https://arxiv.org/PS_cache/arxiv/pdf/0806/0806.3095v1.pdf). Proof appears to be an adaptation of Faltings' proof of the Mordell conjecture. Again, thanks for the comment. – lulu Oct 23 '16 at 11:49
  • The discussion in and below my answer might interest you. Given how restricted the solutions must be, I find it surprising that the proof does not go through to show that the known solutions are the only ones. @lulu – zyx Oct 24 '16 at 19:34

Yes, it's possible. For instance, you could start with $(0,1)$ and $(0,0)$, and then put points along the $x$-axis, noting that there are infinitely many different right triangles with rational sides and one leg equal to $1$. For instance, $(3/4,0)$ will have distance $5/4$ to $(0,1)$.

This means that most if the points are on a single line (the $x$-axis), but one point, $(0,1)$, is not on that line.

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  • Thanks for your answer.. But how could you find the point (3/4, 0) and how can I find the other similar infinite points? – Ahmed Amir Oct 21 '16 at 11:48
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    Let $(a,b,c)$ be a Pythagorean triple. Then $\left(\frac ab\right)^2+1^2=\left(\frac cb\right)^2$, so the distance from $(a/b,0)$ to $(0,1)$ is rational (since it equals $c/b$). There are infinitely many Pythagorean triples, and each one gives you a point $(a/b,0)$. – Arthur Oct 21 '16 at 13:32
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    The Pythagorean triple makes it so simple.. It's totally clear now. – Ahmed Amir Oct 21 '16 at 17:02
  • @Arthur Apparently, the above solution of lulu could be also built from rational Pythagorean triples, as $(\frac {1-z^2}{1+z^2}, \frac {2z}{1+z^2},1)$ is a rational Pythagorean triples, too. –  Oct 22 '16 at 14:02

There is essentially only one known infinite rational-distance set, built from rational Pythagorean triples, and all other examples are derived from this by inversions (with rational radius and center one of the points in the set), isometries, dilation, and taking subsets.

There are no other examples on algebraic curves (Solymosi and de Zeeuw 2008, http://arxiv.org/abs/0806.3095).

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    Yeah,, but what makes the unit circle example special is that there are no more than two points on the same straight line.. While in the essential example(built from Pythagorean triples) all the points are on the same line except one on top of them.. Anyway, thanks for your addition. – Ahmed Amir Oct 22 '16 at 07:53
  • @zyx Apparently, the above solution of lulu could be also built from rational Pythagorean triples, as $(\frac {1-z^2}{1+z^2}, \frac {2z}{1+z^2},1)$ is a rational Pythagorean triples, too. –  Oct 22 '16 at 14:03
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    Yes, the words "built from Pythagorean triples" were written for the purpose of including both constructions and not treating either one as more basic than the other. The solution with a unit circle is to take 2S where S is the subset (and subgroup) of angles with rational sine and cosine, also known as rational Pythagorean triples. @J.Pak – zyx Oct 22 '16 at 16:56
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    @Ahmed: In inversive geometry, lines and circles are the "same thing" -- in particular, I believe for any specific line and circle, there is an inversion that gives a bijection between them. (if you include the point at infinity on the line) –  Oct 23 '16 at 09:08
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    Another indication of the the inversive nature of the problem is that it is natural to allow $\infty$ as a rational distance, and add one point at infinity to the plane, that can be included (or not) in any rational distance set. In addition to making inversions bijective, this restores symmetry in some constructions. @Hurkyl – zyx Oct 24 '16 at 19:31
  • @Hurkyl I don't know what is inversive geometry.. But I understand that you want to say that the two examples (Unit circle & the Pythagorean triples) are the same cause we can "convert" in somehow a circle to a line or as you have said "Lines and Circles are the same thing". – Ahmed Amir Oct 26 '16 at 17:31