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Re: [EMHL] A generalization...

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  • Barry Wolk
    ... It s a cubic, 0=cyclic sum of (u(u+v)bb+u(u+w)cc-(u+v)(u+w)aa)aa(yw-zv)yz ... It s a quartic, 0=cyclic sum of a^4 SA (bbzz-ccyy)yz ... -- Barry Wolk
    Message 1 of 13 , Mar 1, 2011
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      > Let ABC be a triangle P = (x:y:z) a point, A'B'C'
      > the cevian triangle of a point Q = (u:v:w).
      >
      > Which is the locus of P such that the circles
      > (APA'),(BPB'),(CPC') are coaxial?

      It's a cubic, 0=cyclic sum of
      (u(u+v)bb+u(u+w)cc-(u+v)(u+w)aa)aa(yw-zv)yz
      >
      > Specialcase: P,Q are isogonal conjugate points.

      It's a quartic, 0=cyclic sum of a^4 SA (bbzz-ccyy)yz

      >
      > Antreas

      --
      Barry Wolk
    • Bernard Gibert
      Dear Antreas and Barry, ... this is the circular pivotal cubic with pivot Q, isopivot igQ (inverse of the isogonal conjugate of Q) Q must be different of H.
      Message 2 of 13 , Mar 1, 2011
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        Dear Antreas and Barry,

        > > Let ABC be a triangle P = (x:y:z) a point, A'B'C'
        > > the cevian triangle of a point Q = (u:v:w).
        > >
        > > Which is the locus of P such that the circles
        > > (APA'),(BPB'),(CPC') are coaxial?
        >
        > It's a cubic, 0=cyclic sum of
        > (u(u+v)bb+u(u+w)cc-(u+v)(u+w)aa)aa(yw-zv)yz

        this is the circular pivotal cubic with pivot Q, isopivot igQ (inverse of the isogonal conjugate of Q)
        Q must be different of H.

        http://bernard.gibert.pagesperso-orange.fr/Tables/table12.html

        > >
        > > Specialcase: P,Q are isogonal conjugate points.
        >
        > It's a quartic, 0=cyclic sum of a^4 SA (bbzz-ccyy)yz

        This is Q002, one of the most extraordinary curves in the plane of ABC.

        http://bernard.gibert.pagesperso-orange.fr/curves/q002.html

        Best regards

        Bernard



        [Non-text portions of this message have been removed]
      • Antreas Hatzipolakis
        Dear Barry and Bernard A further generalization of the problem is possible if we replace the point P with three points Pa,Pb,Pc related to it. For example:
        Message 3 of 13 , Mar 2, 2011
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          Dear Barry and Bernard

          A further generalization of the problem is possible if we replace the
          point P with three points Pa,Pb,Pc related to it.

          For example:

          Let ABC be a triangle, P a point [not in Darboux cubic], P1P2P3 the
          pedal triangle of P, Q a point and A'B'C' the cevian triangle of Q.

          Denote

          Pa = BP2 /\ CP3

          Pb = CP3 /\ AP1

          Pc = AP1 /\ BP2

          Which is the locus of P such that the circumcircles of
          (APaA'),(BPbB'),(CPcC') are concurrent / have collinear
          circumcenters?

          APH



          On Wed, Mar 2, 2011 at 8:21 AM, Bernard Gibert <bg42@...> wrote:

          >
          >
          > Dear Antreas and Barry,
          >
          >
          > > > Let ABC be a triangle P = (x:y:z) a point, A'B'C'
          > > > the cevian triangle of a point Q = (u:v:w).
          > > >
          > > > Which is the locus of P such that the circles
          > > > (APA'),(BPB'),(CPC') are coaxial?
          > >
          > > It's a cubic, 0=cyclic sum of
          > > (u(u+v)bb+u(u+w)cc-(u+v)(u+w)aa)aa(yw-zv)yz
          >
          > this is the circular pivotal cubic with pivot Q, isopivot igQ (inverse of
          > the isogonal conjugate of Q)
          > Q must be different of H.
          >
          > http://bernard.gibert.pagesperso-orange.fr/Tables/table12.html
          >
          >
          > > >
          > > > Specialcase: P,Q are isogonal conjugate points.
          > >
          > > It's a quartic, 0=cyclic sum of a^4 SA (bbzz-ccyy)yz
          >
          > This is Q002, one of the most extraordinary curves in the plane of ABC.
          >
          > http://bernard.gibert.pagesperso-orange.fr/curves/q002.html
          >
          > Best regards
          >
          > Bernard
          >


          [Non-text portions of this message have been removed]
        • Nikolaos Dergiades
          Dear friends, does anybody can give a simple proof to the following inequality? If p, q, r are arbitrary positive real numbers a,b,c the sides, F the area of
          Message 4 of 13 , Mar 5, 2011
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            Dear friends,
            does anybody can give a simple proof to the
            following inequality?
            If p, q, r are arbitrary positive real numbers
            a,b,c the sides, F the area of triangle ABC
            and U = p/(q+r), V = q/(r+p), W = r/(p+q)
            then prove that
            U(bc)^2 + V(ca)^2 + W(ab)^2 >= 8F^2.

            Best regards
            Nikos Dergiades
          • Francisco Javier
            Dear Nikos, By Nesbitt inequality, we get that U+V+W = 3/2, thus by Chebyshev inequality we have U(bc)^2 + V(ca)^2 + W(ab)^2 ... hence is enough to prove that
            Message 5 of 13 , Mar 5, 2011
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              Dear Nikos,

              By Nesbitt inequality, we get that U+V+W >= 3/2, thus by Chebyshev inequality we have

              U(bc)^2 + V(ca)^2 + W(ab)^2
              >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2)
              >= (1/2)((bc)^2 + (ca)^2 + (ab)^2),

              hence is enough to prove that

              (bc)^2 + (ca)^2 + (ab)^2 >= 16 F^2,

              but (bc)^2 + (ca)^2 + (ab)^2 - 16 F^2 equals to

              (1/2) ((b^2-c^2)^2 + (c^2-a^2)^2 + (a^2-b^2)^2),

              which is always positive.



              --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@...> wrote:
              >
              > Dear friends,
              > does anybody can give a simple proof to the
              > following inequality?
              > If p, q, r are arbitrary positive real numbers
              > a,b,c the sides, F the area of triangle ABC
              > and U = p/(q+r), V = q/(r+p), W = r/(p+q)
              > then prove that
              > U(bc)^2 + V(ca)^2 + W(ab)^2 >= 8F^2.
              >
              > Best regards
              > Nikos Dergiades
              >
            • RIJUL
              Dear Franscisco, How can you say that by Chebyshev Inequality, U(bc)^2 + V(ca)^2 + W(ab)^2 = (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2) ?? U,V,W and bc,ca,ab are
              Message 6 of 13 , Mar 5, 2011
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                Dear Franscisco,

                How can you say that by Chebyshev Inequality, U(bc)^2 + V(ca)^2 + W(ab)^2 >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2) ??
                U,V,W and bc,ca,ab are not necessarily similarly sorted, aren't they?


                --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
                >
                > Dear Nikos,
                >
                > By Nesbitt inequality, we get that U+V+W >= 3/2, thus by Chebyshev inequality we have
                >
                > U(bc)^2 + V(ca)^2 + W(ab)^2
                > >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2)
                > >= (1/2)((bc)^2 + (ca)^2 + (ab)^2),
                >
                > hence is enough to prove that
                >
                > (bc)^2 + (ca)^2 + (ab)^2 >= 16 F^2,
                >
                > but (bc)^2 + (ca)^2 + (ab)^2 - 16 F^2 equals to
                >
                > (1/2) ((b^2-c^2)^2 + (c^2-a^2)^2 + (a^2-b^2)^2),
                >
                > which is always positive.
                >
                >
                >
                > --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@> wrote:
                > >
                > > Dear friends,
                > > does anybody can give a simple proof to the
                > > following inequality?
                > > If p, q, r are arbitrary positive real numbers
                > > a,b,c the sides, F the area of triangle ABC
                > > and U = p/(q+r), V = q/(r+p), W = r/(p+q)
                > > then prove that
                > > U(bc)^2 + V(ca)^2 + W(ab)^2 >= 8F^2.
                > >
                > > Best regards
                > > Nikos Dergiades
                > >
                >
              • Francisco Javier
                Rijul, you are right, I didn t take this into account. My apologies.
                Message 7 of 13 , Mar 5, 2011
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                  Rijul, you are right, I didn't take this into account.

                  My apologies.

                  --- In Hyacinthos@yahoogroups.com, "RIJUL" <rijul.champ@...> wrote:
                  >
                  > Dear Franscisco,
                  >
                  > How can you say that by Chebyshev Inequality, U(bc)^2 + V(ca)^2 + W(ab)^2 >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2) ??
                  > U,V,W and bc,ca,ab are not necessarily similarly sorted, aren't they?
                  >
                  >
                  > --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@> wrote:
                  > >
                  > > Dear Nikos,
                  > >
                  > > By Nesbitt inequality, we get that U+V+W >= 3/2, thus by Chebyshev inequality we have
                  > >
                  > > U(bc)^2 + V(ca)^2 + W(ab)^2
                  > > >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2)
                  > > >= (1/2)((bc)^2 + (ca)^2 + (ab)^2),
                  > >
                  > > hence is enough to prove that
                  > >
                  > > (bc)^2 + (ca)^2 + (ab)^2 >= 16 F^2,
                  > >
                  > > but (bc)^2 + (ca)^2 + (ab)^2 - 16 F^2 equals to
                  > >
                  > > (1/2) ((b^2-c^2)^2 + (c^2-a^2)^2 + (a^2-b^2)^2),
                  > >
                  > > which is always positive.
                  > >
                  > >
                  > >
                  > > --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@> wrote:
                  > > >
                  > > > Dear friends,
                  > > > does anybody can give a simple proof to the
                  > > > following inequality?
                  > > > If p, q, r are arbitrary positive real numbers
                  > > > a,b,c the sides, F the area of triangle ABC
                  > > > and U = p/(q+r), V = q/(r+p), W = r/(p+q)
                  > > > then prove that
                  > > > U(bc)^2 + V(ca)^2 + W(ab)^2 >= 8F^2.
                  > > >
                  > > > Best regards
                  > > > Nikos Dergiades
                  > > >
                  > >
                  >
                • Nikolaos Dergiades
                  Dear Franscisco, I knew this approach but it is not always correct because Chebyshev s Inequality 3(a1b1 + a2b2 + a3b3) = (a1 + a2 + a3)(b1 + b2 + b3) holds
                  Message 8 of 13 , Mar 5, 2011
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                    Dear Franscisco,
                    I knew this approach but it is not
                    always correct because
                    Chebyshev's Inequality
                    3(a1b1 + a2b2 + a3b3) >= (a1 + a2 + a3)(b1 + b2 + b3)
                    holds as said RIJUL only when
                    a1 > a2 > a3 we have b1 > b2 > b3.

                    Best regards
                    Nikos Dergiades

                    > Dear Franscisco,
                    >
                    > How can you say that by Chebyshev Inequality, U(bc)^2 +
                    > V(ca)^2 + W(ab)^2  >= (1/3)(U+V+W)((bc)^2 + (ca)^2 +
                    > (ab)^2) ??
                    > U,V,W and bc,ca,ab are not necessarily similarly sorted,
                    > aren't they?
                    >
                    >
                    > --- In Hyacinthos@yahoogroups.com,
                    > "Francisco Javier" <garciacapitan@...> wrote:
                    > >
                    > > Dear Nikos,
                    > >
                    > > By Nesbitt inequality, we get that U+V+W >= 3/2,
                    > thus by Chebyshev inequality we have
                    > >
                    > >    U(bc)^2 + V(ca)^2 + W(ab)^2
                    > > >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2)
                    > > >= (1/2)((bc)^2 + (ca)^2 + (ab)^2),
                    > >
                    > > hence is enough to prove that   
                    > >
                    > > (bc)^2 + (ca)^2 + (ab)^2 >= 16 F^2,
                    > >
                    > > but (bc)^2 + (ca)^2 + (ab)^2 - 16 F^2 equals to
                    > >
                    > > (1/2) ((b^2-c^2)^2 + (c^2-a^2)^2 + (a^2-b^2)^2),
                    > >
                    > > which is always positive.
                    > >
                    > >
                    > >
                    > > --- In Hyacinthos@yahoogroups.com,
                    > Nikolaos Dergiades <ndergiades@> wrote:
                    > > >
                    > > > Dear friends,
                    > > > does anybody can give a simple proof to the
                    > > > following inequality?
                    > > > If p, q, r are arbitrary positive real numbers
                    > > > a,b,c the sides, F the area of triangle ABC
                    > > > and U = p/(q+r), V = q/(r+p), W = r/(p+q)
                    > > > then prove that
                    > > > U(bc)^2 + V(ca)^2 + W(ab)^2 >= 8F^2.
                    > > >
                    > > > Best regards
                    > > > Nikos Dergiades
                    > > >
                    > >
                    >
                    >
                    >
                    >
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