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Re: A Midpoint on the Euler Line

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  • Antreas
    Another midpoint: Let ABC be a triangle, P a point and A B C the cevian triangle of P. Denote: Ma, Mb, Mc = the midpoints of AA , BB , CC resp. X = the
    Message 1 of 6 , Mar 31 6:00 AM
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      Another midpoint:

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

      Denote:

      Ma, Mb, Mc = the midpoints of AA', BB', CC' resp.

      X = the radical center of (Ma, MaB), (Mb, MbC), (Mc, McA)

      Y = the radical center of (Ma, MaC), (Mb, MbA), (Mc, McB).

      The midpoint of XY is the circumcenter O.

      Now, denote:

      Ka, Kb, Kc points on AA', BB', CC', resp. such that:

      1. KaA / KaP = KbB / KbP = KcB / KcP = k

      La, Lb, Lc on the same cevians such that:

      2. LaA' / LaP = LbB' / LbP = LcC' / LcP = l

      and Ma, Mb, Mc similarly such that:

      3. MaA / MaA' = MbB / MbB' = McC / McC' = m

      X1 = the radical center of (Ka, KaB), (Kb, MbC), (Kc, McA)

      Y1 = the radical center of (Ka, KaC), (Kb, KbA), (Kc, KcB).

      X2 = the radical center of (La, LaB), (Lb, LbC), (Lc, LcA)

      Y2 = the radical center of (La, LaC), (Lb, LbA), (Lc, LcB).

      X3 = the radical center of (Ma, MaB), (Mb, MbC), (Mc, McA)

      Y3 = the radical center of (Ma, MaC), (Mb, MbA), (Mc, McB).

      M1, M2, M3 = the midpoints of X1Y1, X2Y2, X3C3, resp.

      Which are the loci of M1,M2,M3 as k,l,m vary?

      Special Case:

      P = G. Since GA / GA' = GB / GB' = GC / GC', the three
      loci are identical: the Euler line of ABC.

      By the way, G is the ONLY one point P with the property

      PA / PA' = PB / PB' = PC / PC' ?


      Antreas


      [APH]:
      > Let ABC be a triangle and A'B'C' the medial triangle.
      >
      > Denote:
      >
      > (Oab) = the circle with diameter A'B
      > (ie the circle centered on the midpoint Oab of A'B,
      > with radius OabB=OabA')
      > (Oac) = the circle with diameter A'C
      >
      > (Obc) = the circle with diameter B'C
      > (Oba) = the circle with diameter B'A
      >
      > (Oca) = the circle with diameter C'A
      > (Ocb) = the circle with diameter C'B
      >
      > X = the radical center of the circles (Oab), (Obc), (Oca)
      >
      > Y = the radical center of the circles (Oac), (Oba), (Ocb)
      >
      > Which line is the line (XY) (ie wwhose point is trilinear polar)
      > and which point is the midpoint of the line segment XY ?
      >
      >
      > Antreas
      >
    • rhutson2
      There is, in fact, only one point inside ABC such that P is the midpoint of XY, and that point is P=X=Y=I. Randy
      Message 2 of 6 , Mar 31 7:23 PM
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        There is, in fact, only one point inside ABC such that P is the 'midpoint' of XY, and that point is P=X=Y=I.

        Randy

        --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
        >
        > The line is
        >
        > (4 a^4 - 11 a^2 b^2 - 3 b^4 - 11 a^2 c^2 + 24 b^2 c^2 -
        > 3 c^4) x + (-3 a^4 - 11 a^2 b^2 + 4 b^4 + 24 a^2 c^2 -
        > 11 b^2 c^2 - 3 c^4) y + (-3 a^4 + 24 a^2 b^2 - 3 b^4 -
        > 11 a^2 c^2 - 11 b^2 c^2 + 4 c^4) z = 0
        >
        > Its trilinear pole is not in ETC. The midpoint of XY is X3526.
        >
        > If we take A'B'C' as the pedal triangle an arbitrary point P instead of O, it seems that there is an only point inside the triangle such that P is exactly the midpoint of XY.
        >
        >
        > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
        > >
        > > Let ABC be a triangle and A'B'C' the medial triangle.
        > >
        > > Denote:
        > >
        > > (Oab) = the circle with diameter A'B
        > > (ie the circle centered on the midpoint Oab of A'B,
        > > with radius OabB=OabA')
        > > (Oac) = the circle with diameter A'C
        > >
        > > (Obc) = the circle with diameter B'C
        > > (Oba) = the circle with diameter B'A
        > >
        > > (Oca) = the circle with diameter C'A
        > > (Ocb) = the circle with diameter C'B
        > >
        > > X = the radical center of the circles (Oab), (Obc), (Oca)
        > >
        > > Y = the radical center of the circles (Oac), (Oba), (Ocb)
        > >
        > > Which line is the line (XY) (ie wwhose point is trilinear polar)
        > > and which point is the midpoint of the line segment XY ?
        > >
        > >
        > > Antreas
        > >
        >
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