Is it possible to find 8 non collinear points with their mutual distances being rational?












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Is it possible to find 8 non collinear points with their mutual distances being rational?



This came to my mind after reading this article (http://mathworld.wolfram.com/RationalDistances.html)



Note that in my question 4 points can lie on a circle.



But no three points are collinear.



Answers referencing to a article/proof are appreciated.






number-theory coordinate-systems







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Is it possible to find 8 non collinear points with their mutual distances being rational?



    This came to my mind after reading this article (http://mathworld.wolfram.com/RationalDistances.html)



    Note that in my question 4 points can lie on a circle.



    But no three points are collinear.



    Answers referencing to a article/proof are appreciated.






    number-theory coordinate-systems







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Is it possible to find 8 non collinear points with their mutual distances being rational?



      This came to my mind after reading this article (http://mathworld.wolfram.com/RationalDistances.html)



      Note that in my question 4 points can lie on a circle.



      But no three points are collinear.



      Answers referencing to a article/proof are appreciated.






      number-theory coordinate-systems







      share|cite|improve this question









      $endgroup$




      Is it possible to find 8 non collinear points with their mutual distances being rational?



      This came to my mind after reading this article (http://mathworld.wolfram.com/RationalDistances.html)



      Note that in my question 4 points can lie on a circle.



      But no three points are collinear.



      Answers referencing to a article/proof are appreciated.






      number-theory coordinate-systems




      number-theory coordinate-systems






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Oct 23 '17 at 3:48









      Agile_EagleAgile_Eagle

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          $begingroup$

          On a unit circle, you can produce countably many points so that the mutual distances are rational. That is why they exclude the circle points in their problem.



          To explain what happens here: say $z_l= e^{ialpha_k}$, $k = 1,2$. Then
          $$|z_1 - z_2| = 2 |sin(frac{alpha_1 - alpha_2}{2})|$$



          Let now $alpha_k = 2 beta_k$. Then $|z_1 - z_2| = 2 |sin(beta_1 - beta_2)| = 2 | sin beta_1 cos beta_2 - sin beta_2 cos beta_1|$.



          Take now $beta_k$ so that $tan frac{beta_k}{2}$ is rational. Then both
          $sin beta_k$, and $cos beta_k$ will be rational. Moreover, the distances between all these points will be rational.



          Therefore, for every $sin mathbb{Q}$ consider an angle $beta_s$ with $tan frac{beta_s}{2}= s$. Then the corresponding point on the circle will be
          $$P_s= e^{i 2 beta_s} =
          left(frac{1 - 6 s^2 + s^4}{(1 + s^2)^2}, frac{4 s (1 -s^2)}{(1 + s^2)^2}right)$$



          One can check that
          $$d(P_s ,P_t) = frac{4 |(s-t)(1+s t)|}{(1+s^2)(1+t^2)}$$



          We are taking squares of rational points on the circle. Their pairwise distances are also rational.



          $bf{Added:}$ Check a similar problem on this site at the link.






          share|cite|improve this answer











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            3












            $begingroup$

            On a unit circle, you can produce countably many points so that the mutual distances are rational. That is why they exclude the circle points in their problem.



            To explain what happens here: say $z_l= e^{ialpha_k}$, $k = 1,2$. Then
            $$|z_1 - z_2| = 2 |sin(frac{alpha_1 - alpha_2}{2})|$$



            Let now $alpha_k = 2 beta_k$. Then $|z_1 - z_2| = 2 |sin(beta_1 - beta_2)| = 2 | sin beta_1 cos beta_2 - sin beta_2 cos beta_1|$.



            Take now $beta_k$ so that $tan frac{beta_k}{2}$ is rational. Then both
            $sin beta_k$, and $cos beta_k$ will be rational. Moreover, the distances between all these points will be rational.



            Therefore, for every $sin mathbb{Q}$ consider an angle $beta_s$ with $tan frac{beta_s}{2}= s$. Then the corresponding point on the circle will be
            $$P_s= e^{i 2 beta_s} =
            left(frac{1 - 6 s^2 + s^4}{(1 + s^2)^2}, frac{4 s (1 -s^2)}{(1 + s^2)^2}right)$$



            One can check that
            $$d(P_s ,P_t) = frac{4 |(s-t)(1+s t)|}{(1+s^2)(1+t^2)}$$



            We are taking squares of rational points on the circle. Their pairwise distances are also rational.



            $bf{Added:}$ Check a similar problem on this site at the link.






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              On a unit circle, you can produce countably many points so that the mutual distances are rational. That is why they exclude the circle points in their problem.



              To explain what happens here: say $z_l= e^{ialpha_k}$, $k = 1,2$. Then
              $$|z_1 - z_2| = 2 |sin(frac{alpha_1 - alpha_2}{2})|$$



              Let now $alpha_k = 2 beta_k$. Then $|z_1 - z_2| = 2 |sin(beta_1 - beta_2)| = 2 | sin beta_1 cos beta_2 - sin beta_2 cos beta_1|$.



              Take now $beta_k$ so that $tan frac{beta_k}{2}$ is rational. Then both
              $sin beta_k$, and $cos beta_k$ will be rational. Moreover, the distances between all these points will be rational.



              Therefore, for every $sin mathbb{Q}$ consider an angle $beta_s$ with $tan frac{beta_s}{2}= s$. Then the corresponding point on the circle will be
              $$P_s= e^{i 2 beta_s} =
              left(frac{1 - 6 s^2 + s^4}{(1 + s^2)^2}, frac{4 s (1 -s^2)}{(1 + s^2)^2}right)$$



              One can check that
              $$d(P_s ,P_t) = frac{4 |(s-t)(1+s t)|}{(1+s^2)(1+t^2)}$$



              We are taking squares of rational points on the circle. Their pairwise distances are also rational.



              $bf{Added:}$ Check a similar problem on this site at the link.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                On a unit circle, you can produce countably many points so that the mutual distances are rational. That is why they exclude the circle points in their problem.



                To explain what happens here: say $z_l= e^{ialpha_k}$, $k = 1,2$. Then
                $$|z_1 - z_2| = 2 |sin(frac{alpha_1 - alpha_2}{2})|$$



                Let now $alpha_k = 2 beta_k$. Then $|z_1 - z_2| = 2 |sin(beta_1 - beta_2)| = 2 | sin beta_1 cos beta_2 - sin beta_2 cos beta_1|$.



                Take now $beta_k$ so that $tan frac{beta_k}{2}$ is rational. Then both
                $sin beta_k$, and $cos beta_k$ will be rational. Moreover, the distances between all these points will be rational.



                Therefore, for every $sin mathbb{Q}$ consider an angle $beta_s$ with $tan frac{beta_s}{2}= s$. Then the corresponding point on the circle will be
                $$P_s= e^{i 2 beta_s} =
                left(frac{1 - 6 s^2 + s^4}{(1 + s^2)^2}, frac{4 s (1 -s^2)}{(1 + s^2)^2}right)$$



                One can check that
                $$d(P_s ,P_t) = frac{4 |(s-t)(1+s t)|}{(1+s^2)(1+t^2)}$$



                We are taking squares of rational points on the circle. Their pairwise distances are also rational.



                $bf{Added:}$ Check a similar problem on this site at the link.






                share|cite|improve this answer











                $endgroup$



                On a unit circle, you can produce countably many points so that the mutual distances are rational. That is why they exclude the circle points in their problem.



                To explain what happens here: say $z_l= e^{ialpha_k}$, $k = 1,2$. Then
                $$|z_1 - z_2| = 2 |sin(frac{alpha_1 - alpha_2}{2})|$$



                Let now $alpha_k = 2 beta_k$. Then $|z_1 - z_2| = 2 |sin(beta_1 - beta_2)| = 2 | sin beta_1 cos beta_2 - sin beta_2 cos beta_1|$.



                Take now $beta_k$ so that $tan frac{beta_k}{2}$ is rational. Then both
                $sin beta_k$, and $cos beta_k$ will be rational. Moreover, the distances between all these points will be rational.



                Therefore, for every $sin mathbb{Q}$ consider an angle $beta_s$ with $tan frac{beta_s}{2}= s$. Then the corresponding point on the circle will be
                $$P_s= e^{i 2 beta_s} =
                left(frac{1 - 6 s^2 + s^4}{(1 + s^2)^2}, frac{4 s (1 -s^2)}{(1 + s^2)^2}right)$$



                One can check that
                $$d(P_s ,P_t) = frac{4 |(s-t)(1+s t)|}{(1+s^2)(1+t^2)}$$



                We are taking squares of rational points on the circle. Their pairwise distances are also rational.



                $bf{Added:}$ Check a similar problem on this site at the link.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Oct 23 '17 at 4:46

























                answered Oct 23 '17 at 4:23









                Orest BucicovschiOrest Bucicovschi

                28.5k31746




                28.5k31746






























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