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ORTHOLOGIC TRIANGLES, EULER LINE

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  • Antreas
    Let ABC be a triangle, A B C , A B C the medial, orthic triangles, resp. and A*,B*,C* points on AA ,BB ,CC , resp. such that: A*A/A*A = B*B/B*B = C*C/C*C =
    Message 1 of 3 , Apr 6, 2013
      Let ABC be a triangle, A'B'C', A"B"C" the medial,
      orthic triangles, resp. and A*,B*,C* points on AA",BB",CC", resp.
      such that: A*A/A*A" = B*B/B*B" = C*C/C*C" = t.

      Denote:

      Ab = (Parallel to BC through A*) /\ AC

      Ac = (Parallel to BC through A*) /\ AB

      Bc = (Parallel to CA through B*) /\ BA

      Ba = (Parallel to CA through B*) /\ BC

      Ca = (Parallel to AB through C*) /\ CB

      Cb = (Parallel to AB through C*) /\ CA

      1. Oa, Ob, Oc = the circumcenters of A'AbAc, B'BcBa, C'CaCb, resp.

      The triangles ABC, OaObOc are orthologic

      The locus of the orthologic center (OaObOc, ABC), as t varies,
      is the Euler line of ABC.

      Which is the locus of the other orthologic center (ABC, OaObOc)?

      2. Na, Nb, Nc = the NPCs centers of A'AbAc, B'BcBa, C'CaCb, resp.

      The triangles ABC, NaNbNc are orthologic

      The locus of the orthologic center (NaNbNc, ABC), as t varies,
      is the Euler line of ABC.

      Which is the locus of the other orthologic center (ABC, NaNbNc)?

      Figures:
      http://anthrakitis.blogspot.gr/2013/04/orthologic-euler-line.html

      APH
    • Francisco Javier
      Here is the circumcenters part with a generalization for any point P instead of the centroid: Let ABC a triangle and P a point.Call A B C the cevian
      Message 2 of 3 , Apr 7, 2013
        Here is the "circumcenters" part with a generalization for any point P instead of the centroid:

        Let ABC a triangle and P a point.Call A'B'C' the cevian triangle of P with respect to BC.
        The points Ba, Ca; Cb, Ab; Ac, Bc lies on sides BC, CA, AB in such a way that BBa : BaC = CCb : CbA = AAc : AcB = t : 1
        and Ca, Ab, Bc are the reflections of Ba, Cb, Ac on the midpoints of the sides.
        Let Oa, Ob, Oc be the circumcenters of A'AbAc, B'BcBa, C'CaCb respectively.
        (1) The triangles OaObOc and ABC are orthologic.
        (2) The center of orthology of OaObOc with respect to ABC lies on Euler line and divides the segment GO in the ratio (t - 2)/3.
        (3) For P=G, the isotomic conjugate of the second orthology center (of ABC with respect to OaObOc) divides the segment GK in the ratio -(t+5)/6, therefore the other center of orthology lies on Kiepert hyperbola.
        (4) For an arbitrary P, the second orthology center is a quartic through the orthocenter whose isotomic conjugate is a conic. The equation of the quartic, that of course depends on P, is quite long.


        --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
        >
        > Let ABC be a triangle, A'B'C', A"B"C" the medial,
        > orthic triangles, resp. and A*,B*,C* points on AA",BB",CC", resp.
        > such that: A*A/A*A" = B*B/B*B" = C*C/C*C" = t.
        >
        > Denote:
        >
        > Ab = (Parallel to BC through A*) /\ AC
        >
        > Ac = (Parallel to BC through A*) /\ AB
        >
        > Bc = (Parallel to CA through B*) /\ BA
        >
        > Ba = (Parallel to CA through B*) /\ BC
        >
        > Ca = (Parallel to AB through C*) /\ CB
        >
        > Cb = (Parallel to AB through C*) /\ CA
        >
        > 1. Oa, Ob, Oc = the circumcenters of A'AbAc, B'BcBa, C'CaCb, resp.
        >
        > The triangles ABC, OaObOc are orthologic
        >
        > The locus of the orthologic center (OaObOc, ABC), as t varies,
        > is the Euler line of ABC.
        >
        > Which is the locus of the other orthologic center (ABC, OaObOc)?
        >
        > 2. Na, Nb, Nc = the NPCs centers of A'AbAc, B'BcBa, C'CaCb, resp.
        >
        > The triangles ABC, NaNbNc are orthologic
        >
        > The locus of the orthologic center (NaNbNc, ABC), as t varies,
        > is the Euler line of ABC.
        >
        > Which is the locus of the other orthologic center (ABC, NaNbNc)?
        >
        > Figures:
        > http://anthrakitis.blogspot.gr/2013/04/orthologic-euler-line.html
        >
        > APH
        >
      • Francisco Javier
        The nine point centers part for P=G is (1) The triangles NaNbNc and ABC are orthologic. (2) The center of orthology of NaNbNc with respect to ABC lies on
        Message 3 of 3 , Apr 7, 2013
          The "nine point centers" part for P=G is

          (1) The triangles NaNbNc and ABC are orthologic.
          (2) The center of orthology of NaNbNc with respect to ABC lies on Euler line and divides the segment GO in the ratio (2t - 1)/3.
          (3) For P=G, the isotomic conjugate of the second orthology center (of ABC with respect to NaNbNc) divides the segment GK in the ratio (t-7)/12, therefore the other center of orthology lies on Kiepert hyperbola.

          The locus for P such that NaNbNc and ABC are orthologic is the Lucas cubic.



          --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
          >
          > Here is the "circumcenters" part with a generalization for any point P instead of the centroid:
          >
          > Let ABC a triangle and P a point.Call A'B'C' the cevian triangle of P with respect to BC.
          > The points Ba, Ca; Cb, Ab; Ac, Bc lies on sides BC, CA, AB in such a way that BBa : BaC = CCb : CbA = AAc : AcB = t : 1
          > and Ca, Ab, Bc are the reflections of Ba, Cb, Ac on the midpoints of the sides.
          > Let Oa, Ob, Oc be the circumcenters of A'AbAc, B'BcBa, C'CaCb respectively.
          > (1) The triangles OaObOc and ABC are orthologic.
          > (2) The center of orthology of OaObOc with respect to ABC lies on Euler line and divides the segment GO in the ratio (t - 2)/3.
          > (3) For P=G, the isotomic conjugate of the second orthology center (of ABC with respect to OaObOc) divides the segment GK in the ratio -(t+5)/6, therefore the other center of orthology lies on Kiepert hyperbola.
          > (4) For an arbitrary P, the second orthology center is a quartic through the orthocenter whose isotomic conjugate is a conic. The equation of the quartic, that of course depends on P, is quite long.
          >
          >
          > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
          > >
          > > Let ABC be a triangle, A'B'C', A"B"C" the medial,
          > > orthic triangles, resp. and A*,B*,C* points on AA",BB",CC", resp.
          > > such that: A*A/A*A" = B*B/B*B" = C*C/C*C" = t.
          > >
          > > Denote:
          > >
          > > Ab = (Parallel to BC through A*) /\ AC
          > >
          > > Ac = (Parallel to BC through A*) /\ AB
          > >
          > > Bc = (Parallel to CA through B*) /\ BA
          > >
          > > Ba = (Parallel to CA through B*) /\ BC
          > >
          > > Ca = (Parallel to AB through C*) /\ CB
          > >
          > > Cb = (Parallel to AB through C*) /\ CA
          > >
          > > 1. Oa, Ob, Oc = the circumcenters of A'AbAc, B'BcBa, C'CaCb, resp.
          > >
          > > The triangles ABC, OaObOc are orthologic
          > >
          > > The locus of the orthologic center (OaObOc, ABC), as t varies,
          > > is the Euler line of ABC.
          > >
          > > Which is the locus of the other orthologic center (ABC, OaObOc)?
          > >
          > > 2. Na, Nb, Nc = the NPCs centers of A'AbAc, B'BcBa, C'CaCb, resp.
          > >
          > > The triangles ABC, NaNbNc are orthologic
          > >
          > > The locus of the orthologic center (NaNbNc, ABC), as t varies,
          > > is the Euler line of ABC.
          > >
          > > Which is the locus of the other orthologic center (ABC, NaNbNc)?
          > >
          > > Figures:
          > > http://anthrakitis.blogspot.gr/2013/04/orthologic-euler-line.html
          > >
          > > APH
          > >
          >
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