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Loci

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  • Antreas P. Hatzipolakis
    Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P. Ab := AB / PPa, Ac := AC / PPa A := (perpendicular to AB at Ab) / (perpendicular to AC
    Message 1 of 26 , May 2, 2005
      Let ABC be a triangle, P a point and PaPbPc the pedal
      triangle of P.

      Ab := AB /\ PPa, Ac := AC /\ PPa

      A' := (perpendicular to AB at Ab) /\ (perpendicular to AC at Ac)

      A" := the orthogonal projection of A' on BC

      A# := BC /\ AA'

      Similarly B',B",B# and C',C",C#

      Which is the locus of P such that

      1. ABC, A'B'C' are perspective

      2. ABC, A"B"C" are perspective

      3. A#,B#, C# are collinear ?


      Antreas
      --
    • xpolakis
      ... I think that the more interesting is #3. I found that the locus is Darboux cubic + (Three Altitudes) Antreas
      Message 2 of 26 , May 3, 2005
        --- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis" <xpolakis@o...> wrote:
        > Let ABC be a triangle, P a point and PaPbPc the pedal
        > triangle of P.
        >
        > Ab := AB /\ PPa, Ac := AC /\ PPa
        >
        > A' := (perpendicular to AB at Ab) /\ (perpendicular to AC at Ac)
        >
        > A" := the orthogonal projection of A' on BC
        >
        > A# := BC /\ AA'
        >
        > Similarly B',B",B# and C',C",C#
        >
        > Which is the locus of P such that
        >
        > 1. ABC, A'B'C' are perspective
        >
        > 2. ABC, A"B"C" are perspective
        >
        > 3. A#,B#, C# are collinear ?

        I think that the more interesting is #3.

        I found that the locus is

        Darboux cubic + (Three Altitudes)


        Antreas
      • Antreas Hatzipolakis
        An application to triangle geometry of a problem of Kostas Vittas (*) Let ABC be a triangle, P a point and A B C the cevian or pedal triangle of P. Denote:
        Message 3 of 26 , May 28, 2014

        An application to triangle geometry of a problem of Kostas Vittas (*)

        Let ABC be a triangle, P a point and A'B'C' the cevian or pedal triangle of P.

        Denote:

        Mab, Mac = the midpoints of BA', CA', resp.
        M1 = the midpoint of MabMac

        The circle (M1, M1Mab = M1, M1Mac) ie the circle with diameter MabMac,
        intersects the circles with diameters BA', CA' at M12, M13, resp. on the positive
        side of BC (ie the side where A is located on).

        Denote:

        A* = MabM13 /\ MacM12

        The line M12M13 intersects again the circles with diameters BA', CA' at M'12, M'13 resp.

        The lines A'A*, BM'12, CM'13 are concurrent at point A**.

        [see figure]

        Which is the locus of P such that:

        1. ABC, A*B*C*
        2, A'B'C', A*B*C*,
        3. ABC, A**B**C**
        are perspective ?

        Note: The line A**A' is the symmedian from A** of the triangle A**BC.
        If Ka is the symmedian point of A**BC, and Kb,Kc the symmedian points
        of B**CA, C**AB, resp., which is the locus of P such that
        ABC, KaKbKc are perspective?
        APH
      • Antreas Hatzipolakis
        Probably we have seen some of the following loci. Anyways here are :-) Let ABC be a triangle P a point, Na,Nb,Nc the NPC centers of PBC, PCA, PAB , resp and
        Message 4 of 26 , Jan 17, 2015
          Probably we have seen some of the following loci.
          Anyways here are :-)

          Let ABC be a triangle P a point, Na,Nb,Nc the NPC centers
          of PBC, PCA, PAB , resp and Op,Np the circumcenter, NPC
          center of NaNbNc, resp.

          Which is the locus of P such that:

          1.1. Op lies on the OP line
          1.2. Op les on the Euler line of ABC

          2.1. Np lies on the OP line
          2.2. Np les on the Euler line of ABC
          ?

          APH
        • Antreas Hatzipolakis
          [APH]: Probably we have seen some of the following loci. Anyways here are :-) Let ABC be a triangle P a point, Na,Nb,Nc the NPC centers of PBC, PCA, PAB , resp
          Message 5 of 26 , Jan 18, 2015

             

            [APH]:

            Probably we have seen some of the following loci.
            Anyways here are :-)

            Let ABC be a triangle P a point, Na,Nb,Nc the NPC centers

            of PBC, PCA, PAB , resp and Op,Np the circumcenter, NPC

            center of NaNbNc, resp.

            Which is the locus of P such that:

            1.1. Op lies on the OP line

            1.2. Op les on the Euler line of ABC

            2.1. Np lies on the OP line
            2.2. Np les on the Euler line of ABC
            ?

            APH


            [César Lozada]:


            1.1)  Circum-quintic:

            Q11 = CyclicSum[(v*w*(-v*cos(2*B+C)+w*cos(B+2*C))*sin(A)+

                                               u*((2*cos(A)*cos(B-C)+1)*v*w+u^2*cos(3*A))*sin(B-C))*v*w ] = 0

            Through vertices of cevian of N and  ETC´s: 3, 4, 5, 80, 1113, 1114

             

            1.2)  Excentral-circum-quintic:

            Q12 = CyclicSum[(v*w*(cos(C)*(1+2*cos(2*B))*v-w*cos(B)*(1+2*cos(2*C)))*sin(A)+

                                               u*(-3*u^2*cos(A)+v*w*(1+2*cos(2*A)))*sin(B-C))*v*w] = 0

            Through vertices of cevian of X(265) and  ETC´s: 1, 4, 5, 30, 1113, 1114, 1138

             

            2.1) Circum-quintic:

                    Q21 = CyclicSum[(v*w*(-w*cos(B)+v*cos(C))*sin(A)+

                                                      u*((2*cos(A)*cos(B-C)+cos(2*A)+2)*v*w+(cos(3*A)-2*cos(A))*u^2)*sin(B-C))*v*w] = 0

            Through vertices of cevian of N and  ETC´s: 3, 4, 1113, 1114

             

            2.2) Circum-quintic:

                     Q22=CyclicSum[(-v*w*((2*cos(C)+cos(2*B+C))*v+(-2*cos(B)-cos(B+2*C))*w)*sin(A)+

                                                     u*((4*cos(A)*cos(B-C)+cos(2*A))*v*w+(cos(A)+2*cos(3*A))*u^2)*sin(B-C))*v*w]

            Through vertices of cevian of X(30) and  ETC´s: 3, 4, 30, 1113, 1114, 5627

             

            César Lozada

             


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