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Re: [EMHL] Collinear Points

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  • Antreas P. Hatzipolakis
    ... Variations: 1. Let A B C be the cevian triangle of P. Denote Ab = CC / Parallel to BB from A Ac = BB / Parallel to CC from A L1 = The line AbAc.
    Message 1 of 10 , Jan 20, 2004
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      [APH]:

      >Recent addition in Bernard's site:
      >
      >6. Let P be a point with traces A_P, B_P, C_P. Denote by L1 the line
      >joining the perpendicular feet of A_P on the cevians BB_P and CC_P.
      >Let L2, L3 be the two lines analogously defined. The triangle bounded
      >by L1, L2, L3 is perspective with ABC if and only if P lies on the
      >Darboux cubic (together with the line at infinity, the three circles
      >with diameters BC, CA, AB, and the Yiu quintic Q006 ((Paul Yiu,
      >Hyacinthos #1967).
      >
      > http://perso.wanadoo.fr/bernard.gibert/Exemples/k004.html
      >
      >Now, denote
      >
      >A* = L1 /\ BC
      >
      >B* = L2 /\ CA
      >
      >C* = L3 /\ AB
      >
      >Which is the locus of P such that A*,B*,C* are collinear?

      Variations:

      1. Let A'B'C' be the cevian triangle of P.

      Denote

      Ab = CC' /\ Parallel to BB' from A'

      Ac = BB' /\ Parallel to CC' from A'

      L1 = The line AbAc. Similarly the lines L2, L3.

      Which is the locus of P such that the triangle bounded by L1,L2,L3
      is perspective with ABC?
      [Special case: L1,L2,L3 are concurrent]

      2. Let A'B'C' be the cevian triangle of P.

      Denote

      Ab = CC' /\ Parallel to AB from A'

      Ac = BB' /\ Parallel to AC from A'

      L1 = The line AbAc. Similarly the lines L2, L3.

      Which is the locus of P such that the triangle bounded by L1,L2,L3
      is perspective with ABC?
      [Special case: L1,L2,L3 are concurrent]

      Also same questions for A*,B*,C* as defined above.


      Greetings from Athens

      Antreas


      --
    • Antreas Hatzipolakis
      Let ABC be a triangle P a fixed point and Q a variable point Denote: A1 = (Perpendicular to AQ through P) / BC Similarly B1, C1. The locus of Q such that
      Message 2 of 10 , Jan 3
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        Let ABC be a triangle P a fixed point and Q a variable point

        Denote:

        A1 = (Perpendicular to AQ through P)  /\ BC
        Similarly B1, C1.

        The locus of Q such that A1,B1, C1 are collinear is
        the rec. circumhyperbola through P. (+ ?)

        Reference

        Let P,P* be two isogonal conjugate points

        A1 = (Perpendicular to AP* through P)  /\ BC
        Similarly B1, C1.

        Which is the locus of P such that A1,B1,C1 are collinear?

        Happy New Year 2015

        APH

      • Antreas Hatzipolakis
        From: César Lozada Which is the locus of P such that A1,B1,C1 are collinear? Locus={Circumcircle} / {McCay Cubic} If P on circumcircle then A1,B1,C1 and P
        Message 3 of 10 , Jan 3
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          From: César Lozada


           

          Which is the locus of P such that A1,B1,C1 are collinear?

           

          Locus={Circumcircle} \/ {McCay Cubic}
          If P on circumcircle then A1,B1,C1 and P are collinear.

           

          César Lozada

           

          De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com]
          Enviado el: Sábado, 03 de Enero de 2015 07:46 a.m.
          Para: anopolis@yahoogroups.com; Hyacinthos
          Asunto: [EGML] Collinear Points

           

           


          Let ABC be a triangle P a fixed point and Q a variable point

          Denote:

          A1 = (Perpendicular to AQ through P)  /\ BC

          Similarly B1, C1.

          The locus of Q such that A1,B1, C1 are collinear is

          the rec. circumhyperbola through P. (+ ?)

          Reference

          Let P,P* be two isogonal conjugate points

          A1 = (Perpendicular to AP* through P)  /\ BC
          Similarly B1, C1.

          Which is the locus of P such that A1,B1,C1 are collinear?

          Happy New Year 2015

          APH

           




          --
        • Antreas Hatzipolakis
          From: César Lozada [Attachment(s) from =?UTF-8?Q?C=C3=A9sar_Lozada?= included below] A conjectured relation with circumnormal
          Message 4 of 10 , Jan 3
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            From: César Lozada


             
            [Attachment(s) from =?UTF-8?Q?C=C3=A9sar_Lozada?= included below]

             

            A conjectured relation with circumnormal triangle.

             

            See attached image.

            César Lozada

             

            De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com]
            Enviado el: Sábado, 03 de Enero de 2015 12:15 p.m.
            Para: Anopolis@yahoogroups.com
            Asunto: RE: [EGML] Collinear Points

             

             

            Which is the locus of P such that A1,B1,C1 are collinear?

             

            Locus={Circumcircle} \/ {McCay Cubic}
            If P on circumcircle then A1,B1,C1 and P are collinear.

             

            César Lozada

             

            De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com]
            Enviado el: Sábado, 03 de Enero de 2015 07:46 a.m.
            Para: anopolis@yahoogroups.com; Hyacinthos
            Asunto: [EGML] Collinear Points

             

             


            Let ABC be a triangle P a fixed point and Q a variable point

            Denote:

            A1 = (Perpendicular to AQ through P)  /\ BC

            Similarly B1, C1.

            The locus of Q such that A1,B1, C1 are collinear is

            the rec. circumhyperbola through P. (+ ?)

            Reference

            Let P,P* be two isogonal conjugate points

            A1 = (Perpendicular to AP* through P)  /\ BC
            Similarly B1, C1.

            Which is the locus of P such that A1,B1,C1 are collinear?

            Happy New Year 2015

            APH

             



          • Antreas Hatzipolakis
            Let ABC be a triangle and A B C the pedal triangle of I (aka Intouch triangle). Denote: N* = the NPC center of A B C O* = the circumcenter of the pedal
            Message 5 of 10 , Feb 18
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              Let ABC be a triangle and A'B'C' the pedal triangle of I (aka Intouch triangle).

              Denote:

              N* = the NPC center of A'B'C'

              O* = the circumcenter of the pedal triangle of N* wrt A'B'C'

              The points N*, O* and Feuerbach point of ABC are collinear.

              References:
              Buratino Giggle [=Tran Quang Hung]
              https://www.facebook.com/groups/439719556180075/permalink/444075269077837/
              http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&p=3751766#p3751766

              APH





            • Antreas Hatzipolakis
              [APH]: Let ABC be a triangle and A B C the pedal triangle of I (aka Intouch ... [Randy Hutson]: Dear Antreas, N* = X(942) = INVERSE-IN-INCIRCLE OF X(36). O* =
              Message 6 of 10 , Feb 18
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                [APH]:

                Let ABC be a triangle and A'B'C' the pedal triangle of I (aka Intouch triangle).

                Denote:

                N* = the NPC center of A'B'C'

                O* = the circumcenter of the pedal triangle of N* wrt A'B'C'

                The points N*, O* and Feuerbach point of ABC are collinear.

                References:
                Buratino Giggle [=Tran Quang Hung]
                https://www.facebook.com/groups/439719556180075/permalink/444075269077837/
                http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&p=3751766#p3751766





                 
                [Randy Hutson]:

                Dear Antreas,

                N* = X(942) = INVERSE-IN-INCIRCLE OF X(36).
                O* = midpoint of X(942) and [X(3649) = KS(INTOUCH TRIANGLE)]

                The common line id X(11)X(113).

                Best regards,
                Randy



              • Antreas Hatzipolakis
                ... Given that the Feuerbach point of ABC wrt Intouch triangle is X(110) of the intouch triangle, it is equivalent to (by taking the intouch triangle as
                Message 7 of 10 , Feb 19
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                  [APH]:
                  Let ABC be a triangle and A'B'C' the pedal triangle of I (aka Intouch triangle).

                  Denote:

                  N* = the NPC center of A'B'C'

                  O* = the circumcenter of the pedal triangle of N* wrt A'B'C'

                  The points N*, O* and Feuerbach point of ABC are collinear.

                  References:
                  Buratino Giggle [=Tran Quang Hung]
                  https://www.facebook.com/groups/439719556180075/permalink/444075269077837/
                  http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&p=3751766#p3751766




                  Given that the Feuerbach point of ABC wrt Intouch triangle is X(110) of the intouch triangle,
                  it is equivalent to (by taking the intouch triangle as reference triangle):

                  Let ABC be a triangle, A'B'C' the pedal triangle of N and O' the circumcenter of A'B'C'
                  The line NO' passes through X(110)

                  Question: Is the other than X(110) intersection of NO' and circumcircle an interesting point?

                  APH
                • Antreas Hatzipolakis
                  ... *********************************************** [César Lozada] It is X(1141). [Peter Moses] Hi Antreas, X(1141). Best regards Peter. [APH]: [APH]: Let ABC
                  Message 8 of 10 , Feb 19
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                    [APH]:


                    [APH]:

                    Let ABC be a triangle and A'B'C' the pedal triangle of I (aka Intouch triangle).

                    Denote:

                    N* = the NPC center of A'B'C'

                    O* = the circumcenter of the pedal triangle of N* wrt A'B'C'

                    The points N*, O* and Feuerbach point of ABC are collinear.

                    References:
                    Buratino Giggle [=Tran Quang Hung]
                    https://www.facebook.com/groups/439719556180075/permalink/444075269077837/
                    http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&p=3751766#p3751766




                    Given that the Feuerbach point of ABC wrt Intouch triangle is X(110) of the intouch triangle,
                    it is equivalent to (by taking the intouch triangle as reference triangle):

                    Let ABC be a triangle, A'B'C' the pedal triangle of N and O' the circumcenter of A'B'C'
                    The line NO' passes through X(110)

                    Question: Is the other than X(110) intersection of NO' and circumcircle an interesting point?



                    ***********************************************


                    [César Lozada]

                    It is X(1141).


                    [Peter Moses]

                    Hi Antreas,
                     
                    X(1141).
                     
                    Best regards
                    Peter.
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