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[EMHL] Areas and curves

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  • Milorad Stevanovic
    Dear friends, Let A1,B1,C1 be the feets of perpendiculars from point P(x:y:z) to the sides BC,CA,AB and F1=area(PBA1),F2=area(PA1C),
    Message 1 of 3 , May 12, 2003
      Dear friends,

      Let A1,B1,C1 be the feets of perpendiculars
      from point P(x:y:z) to the sides BC,CA,AB
      and F1=area(PBA1),F2=area(PA1C),
      F3=area(PCB1),F4=area(PB1A),
      F5=area(PAC1),F6=area(PC1B).
      Then we have
      1.F1+F3+F5=F2+F4+F6
      iff P is on Stammler hyperbola.
      2.F1*F3*F5=F2*F4*F6
      iff P is on Darboux cubic.
      3.F1*F3+F3*F5+F5*F1=F2*F4+F4*F6+F6*F2
      iff P is on conic
      Cyclic sum(bc(b^2-c^2)(bcxx-2aayzcosA))=0.

      As I remember,I certainly saw one of first two
      propositions in Hyacinthos message.
      Maybe,all of them are known.

      It could be of interest to find some other
      relations for these six areas which could
      give some interesting properties of curves.

      Best regards
      Sincerely
      Milorad R.Stevanovic





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    • Bernard Gibert
      Dear Milorad, [MS] ... Are you sure about this one ? I found the McCay cubic... ... 4. F1^2+F3^2+F5^2= F2^2+F4^2+F6^2 gives a beautiful cubic with asymptotes
      Message 2 of 3 , May 16, 2003
        Dear Milorad,

        [MS]
        > Let A1,B1,C1 be the feets of perpendiculars
        > from point P(x:y:z) to the sides BC,CA,AB
        > and F1=area(PBA1),F2=area(PA1C),
        > F3=area(PCB1),F4=area(PB1A),
        > F5=area(PAC1),F6=area(PC1B).
        > Then we have
        > 1.F1+F3+F5=F2+F4+F6
        > iff P is on Stammler hyperbola.
        > 2.F1*F3*F5=F2*F4*F6
        > iff P is on Darboux cubic.
        > 3.F1*F3+F3*F5+F5*F1=F2*F4+F4*F6+F6*F2
        > iff P is on conic
        > Cyclic sum(bc(b^2-c^2)(bcxx-2aayzcosA))=0.

        Are you sure about this one ?
        I found the McCay cubic...

        > It could be of interest to find some other
        > relations for these six areas which could
        > give some interesting properties of curves.

        4. F1^2+F3^2+F5^2= F2^2+F4^2+F6^2
        gives a beautiful cubic with asymptotes parallel to those of McCay and
        concuring at O which is the center of symmetry of the curve.
        I usually call this type of cubic a $\mathcal{K}_{60}^{++}$.

        O is an inflexion point with tangent the Euler line.

        it passes through the in/excenters with tangents concuring at G (the
        polar conic of G is the Stammler hyperbola) and their reflections in O.

        it meets the circumcircle at the second intersections of the bisectors.

        barycentric equation :

        b^2c^2[(b^2-c^2) x^3 + a^2 y z(y-z)] + cyclic = 0

        Best regards

        Bernard

        [Non-text portions of this message have been removed]
      • Milorad Stevanovic
        Dear Bernard, [MS] ... [BG] ... Yes,in third case the solution is the McCay cubic.The solution is not so easy to obtain. When I passed from case 2 to case
        Message 3 of 3 , May 16, 2003
          Dear Bernard,

          [MS]
          > Let A1,B1,C1 be the feets of perpendiculars
          > from point P(x:y:z) to the sides BC,CA,AB
          > and F1=area(PBA1),F2=area(PA1C),
          > F3=area(PCB1),F4=area(PB1A),
          > F5=area(PAC1),F6=area(PC1B).
          > Then we have
          > 1.F1+F3+F5=F2+F4+F6
          > iff P is on Stammler hyperbola.
          > 2.F1*F3*F5=F2*F4*F6
          > iff P is on Darboux cubic.
          > 3.F1*F3+F3*F5+F5*F1=F2*F4+F4*F6+F6*F2
          > iff P is on conic
          > Cyclic sum(bc(b^2-c^2)(bcxx-2aayzcosA))=0.
          [BG]
          >Are you sure about this one ?
          >I found the McCay cubic...

          Yes,in third case the solution is
          the McCay cubic.The solution is not so easy to obtain.
          When I passed from case 2 to case three I lost some
          factors.Sorry about mistake.

          > It could be of interest to find some other
          > relations for these six areas which could
          > give some interesting properties of curves.

          4. F1^2+F3^2+F5^2= F2^2+F4^2+F6^2
          >gives a beautiful cubic with asymptotes parallel to those of McCay and
          >.concuring at O which is the center of symmetry of the curve.
          >.I usually call this type of cubic a $\mathcal{K}_{60}^{++}$.

          >O is an inflexion point with tangent the Euler line.

          .it passes through the in/excenters with tangents concuring at G (the
          >polar conic of G is the Stammler hyperbola) and their reflections in O.

          >it meets the circumcircle at the second intersections of the bisectors.

          .barycentric equation :

          >b^2c^2[(b^2-c^2) x^3 + a^2 y z(y-z)] + cyclic = 0

          It seems that this cubic could be very important.
          I am happy about this discovery.
          Congratulations.

          Best regards
          Sincerely
          Milorad R.Stevanovic





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