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Re: [EMHL] Rational distances from the vertices of aquilateral triangle

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  • Nikolaos Dergiades
    Dear Antreas, I wrote ... Not always. Consider the trivial case when the point P with distances a, b, c from the vertices of the equilateral triangle, lies on
    Message 1 of 13 , Aug 17 8:22 AM
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      Dear Antreas,
      I wrote
      > in the equation
      > 3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2
      > the four variables are equivalent
      > and we can interchange them.

      Not always.
      Consider the trivial case when the point P
      with distances a, b, c from the vertices of the
      equilateral triangle, lies on one side of the
      equilateral triangle that has side length d.
      A general solution of this Diophantine equation
      is given by the formulas
      a = m^2 + 3n^2
      b = Abs((m + n)(m - 3n))
      c = Abs(4mn)
      d = Abs((m + 3n)(m - n))
      For m = 2, n = 1 we get
      a = 7, b = 3, c = 8, d = 5
      Which means that the side of the equilateral
      triangle can be 5, 3, 8 but it can't be 7.
      It is interesting the figure.
      Construct the equilateral triangle A1P2P3 with side 8
      On P2P3 take P1 such that P2P1 = 5 and P1P3 = 3
      and on the equilateral triangle A1P2P3 construct
      two equilateral triangles A3P2P1 and A2P1P3.
      We have A1P1 = A2P2 = A3P3 = 7 and the points
      P1, P2, P3 are the required points for the 3 equilateral
      triangles. The conclusion is that there is not fourth
      equilateral triangle with side 7. The number 7 in this
      tetrad is always the distance of P from the opposite vertex.
      Best regards
      Nikos
    • Ignacio Larrosa Cañestro
      Dear Nikos, ... there are equilateral tringle with side 7 and points that distfrom its vertices 3, 4 and 5. See:
      Message 2 of 13 , Aug 17 10:26 AM
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        Dear Nikos,

        El 17/08/2012 17:22, Nikolaos Dergiades escribió:
        > Dear Antreas,
        > I wrote
        >> in the equation
        >> 3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2
        >> the four variables are equivalent
        >> and we can interchange them.
        > Not always.
        > Consider the trivial case when the point P
        > with distances a, b, c from the vertices of the
        > equilateral triangle, lies on one side of the
        > equilateral triangle that has side length d.
        > A general solution of this Diophantine equation
        > is given by the formulas
        > a = m^2 + 3n^2
        > b = Abs((m + n)(m - 3n))
        > c = Abs(4mn)
        > d = Abs((m + 3n)(m - n))
        > For m = 2, n = 1 we get
        > a = 7, b = 3, c = 8, d = 5
        > Which means that the side of the equilateral
        > triangle can be 5, 3, 8 but it can't be 7.
        > It is interesting the figure.
        > Construct the equilateral triangle A1P2P3 with side 8
        > On P2P3 take P1 such that P2P1 = 5 and P1P3 = 3
        > and on the equilateral triangle A1P2P3 construct
        > two equilateral triangles A3P2P1 and A2P1P3.
        > We have A1P1 = A2P2 = A3P3 = 7 and the points
        > P1, P2, P3 are the required points for the 3 equilateral
        > triangles. The conclusion is that there is not fourth
        > equilateral triangle with side 7. The number 7 in this
        > tetrad is always the distance of P from the opposite vertex.
        > Best regards
        > Nikos
        >
        >

        there are equilateral tringle with side 7 and points that distfrom its
        vertices 3, 4 and 5. See:

        http://www.xente.mundo-r.com/ilarrosa/GeoGebra/DistEnterasTriangEquil.html

        From a equilateral triangle with side 8, you can construc 12 of thats
        triangles of side 7.



        --
        Best regards,

        Ignacio Larrosa Cañestro
        A Coruña (España)
        ilarrosa@...
        http://www.xente.mundo-r.com/ilarrosa/GeoGebra/
      • Nikolaos Dergiades
        Dear Ignacio, thank you very much. You are right. What I said is nonsense. I was deceived thinking that the fourth case of equilateral triangle with side 7
        Message 3 of 13 , Aug 17 12:51 PM
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          Dear Ignacio,
          thank you very much. You are right.
          What I said is nonsense.
          I was deceived thinking that the fourth
          case of equilateral triangle with side 7
          would be a trivial case as the other three.
          Best regards
          Nikos Dergiades

          > there are equilateral tringle with side 7 and points that
          > distfrom its
          > vertices 3, 4 and 5. See:
          >
          > http://www.xente.mundo-r.com/ilarrosa/GeoGebra/DistEnterasTriangEquil.html
          >
          > From a equilateral triangle with side 8, you can construc
          > 12 of thats
          > triangles of side 7.

          > Best regards,
          >
          > Ignacio Larrosa Cañestro
          > A Coruña (España)
          > ilarrosa@...
          > http://www.xente.mundo-r.com/ilarrosa/GeoGebra/
          >
          >
          >
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