Loading ...
Sorry, an error occurred while loading the content.

THREE CIRCLES

Expand Messages
  • Antreas Hatzipolakis
    1. Let A be a fixed point on the sideline BC of triangle ABC. To construct point A on BC such that the NPC centers of ABA , ACA , ABA , ACA are concyclic.
    Message 1 of 4 , Sep 18, 2013
    • 0 Attachment
      1. Let A' be a fixed point on the sideline BC of triangle ABC.
      To construct point A" on BC such that the NPC centers of
      ABA', ACA', ABA", ACA" are concyclic.

      The point A" is the trace on BC of the isogonal cevian of AA'
      (ie A" = BC /\ (reflection of AA' in the int. angle bisector of  A)

      2. Let P,Q be two isogonal conjugate points and A'B'C', A"B"C"
      the cevian triangles of P,Q, resp.

      Denote: 
      Oa = the center of the circle passing through the
      NPC centers of ABA', ACA', ABA", ACA"
      Similarly Ob, Oc.

      For P = O, Q = H, the triangles ABC, OaObOc are orthologic.

      [Cesar Lozada]:
          OrthologicCenters( P=O ) = { Po, X(546)=midpoint of (H,N) }
             Po = a*((b^2+c^2)*a^2-(b^2-c^2)^2)/((b^2+c^2)*a^6-(b^4+c^4-4*b^2*c^2)*a^4-(b^2+c^2)*   (b^2-c^2)^2*a^2+(b^2-c^2)^4) :: (trilinears)
             ETC(Po) =  4.088637332117895, -1.05148019490158, 2.481548925092599
             Po not in any line through  ETC centers

      3. Let P be a point. 

      In the triangle PBC, let A',A" be the traces on BC of the (isogonal) cevians
      through the circumcenter, orthocenter, resp.  of PBC.
      [ie If O1,H1 the circumcenter, orthocenter, resp. of PBC, then
      A' = BC /\ PO1, A" = BC /\ PH1].
      Let Oa be the center of the circle passing through the NPC centers
      of the triangles PBA', PCA', PBA", PCA" 
      Similarly Ob, Oc.

      The triangles OaObOc, ABC are orthologic.
      The orthologic center P1 = (OaObPc, ABC) is lying on the line OP.
      The orthologic center P2 = (ABC, OaObOc) is lying on ?????

      4. Let P, Q be two isogonal conjugate points.
      Let Oa,Ob,Oc be the centers respective to P (above #3)
      and O'a,O'b,O'c  the centers respective to Q (above #3)
      and let P1, P2 be the orthologic centers (OaObOc, ABC), (ABC, OaObOc), resp
      and Q1,Q2 the orthologic centers (O'aO'bO'c, ABC), (ABC, O'aO'bO'c)

      Questions:
      i) Are the six centers Oa,Ob,Oc, O'a,O'b, O'c lying on a conic?
      ii) Are the four orthologic centers P1,P2,Q1,Q2 concyclic?

      Antreas






    • Antreas Hatzipolakis
      Let ABC be a triangle. Denote: Ab = AC / (perpendicular bisector of BC) Ac = AB / (perpendicular bisector of BC) Similarly Bc, Ba and Ca, Cb. Ba, Ca, Cb, Bc
      Message 2 of 4 , Oct 26, 2014
      • 0 Attachment



        Let ABC be a triangle.

        Denote:
        Ab = AC /\ (perpendicular bisector of BC)
        Ac = AB /\ (perpendicular bisector of BC)

        Similarly Bc, Ba and Ca, Cb.

        Ba, Ca, Cb, Bc are concyclic.

        Reference (wording mine):

        http://www.artofproblemsolving.com/Forum/viewtopic.php?t=608711&p3635815

        Let Oa be the center of the circle (Ba, Ca, Cb, Bc)
        Simiarly Ob, Oc.

        Are the triangles ABC, OaObOc perspective or orthologic ?


        APH
      • Antreas Hatzipolakis
        ... [Peter Moses]: perspective at X(184) orthologic at X(4) and {{3,64},{4,54},{5,182},{6,1598},{20,110},{22,5562},{23,5889},{24,185},{25,389},{30,156},...}
        Message 3 of 4 , Oct 26, 2014
        • 0 Attachment


          [APH]:



          Let ABC be a triangle.

          Denote:
          Ab = AC /\ (perpendicular bisector of BC)
          Ac = AB /\ (perpendicular bisector of BC)

          Similarly Bc, Ba and Ca, Cb.

          Ba, Ca, Cb, Bc are concyclic.

          Reference (wording mine):

          http://www.artofproblemsolving.com/Forum/viewtopic.php?t=608711&p3635815

          Let Oa be the center of the circle (Ba, Ca, Cb, Bc)
          Simiarly Ob, Oc.

          Are the triangles ABC, OaObOc perspective or orthologic ?


          APH

          [Peter Moses]:

          perspective at X(184)
          orthologic at X(4) and {{3,64},{4,54},{5,182},{6,1598},{20,110},{22,5562},{23,5889},{24,185},{25,389},{30,156},...}
           
          Best regards,
          Peter.


        • Antreas Hatzipolakis
          ... Synthetic proofs of the fact that the perspector is X(184) by Luis Gonzalez and Telv Cohl
          Message 4 of 4 , Oct 26, 2014
          • 0 Attachment




            [APH]:



            Let ABC be a triangle.

            Denote:
            Ab = AC /\ (perpendicular bisector of BC)
            Ac = AB /\ (perpendicular bisector of BC)

            Similarly Bc, Ba and Ca, Cb.

            Ba, Ca, Cb, Bc are concyclic.

            Reference (wording mine):

            http://www.artofproblemsolving.com/Forum/viewtopic.php?t=608711&p3635815

            Let Oa be the center of the circle (Ba, Ca, Cb, Bc)
            Simiarly Ob, Oc.

            Are the triangles ABC, OaObOc perspective or orthologic ?


            APH

            [Peter Moses]:

            perspective at X(184)
            orthologic at X(4) and {{3,64},{4,54},{5,182},{6,1598},{20,110},{22,5562},{23,5889},{24,185},{25,389},{30,156},...}
             
            Best regards,
            Peter.




            Synthetic proofs of the fact that the perspector is X(184) by Luis Gonzalez and Telv Cohl

            www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=608711&p=3636331


          Your message has been successfully submitted and would be delivered to recipients shortly.