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Re: [EMHL] Re: Euler Lines, Circumcircles (corrected)

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  • Antreas Hatzipolakis
    Dear Nikos, We can replace the incenter I with any other fixed point and ask for the loci, but I guess they are complicated in that general case. Probably we
    Message 1 of 9 , Feb 9, 2013
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      Dear Nikos,

      We can replace the incenter I with any other fixed point and ask for the
      loci,
      but I guess they are complicated in that general case.

      Probably we get interesting loci for points on the Neuberg cubic (like I)

      How about H instead of I ?

      I will ask about H in an other post.

      APH


      On Fri, Feb 8, 2013 at 10:32 PM, Nikolaos Dergiades <ndergiades@...>wrote:

      > **
      >
      >
      > Dear Antreas,
      >
      > the locus is the conic
      >
      > b^3 c x^2 - b c^3 x^2 - 2 a^3 c x y - a^2 b c x y + a b^2 c x y +
      > 2 b^3 c x y + 2 a c^3 x y - 2 b c^3 x y - a^3 c y^2 + a c^3 y^2 +
      > 2 a^3 b x z - 2 a b^3 x z + a^2 b c x z + 2 b^3 c x z -
      > a b c^2 x z - 2 b c^3 x z + 2 a^3 b y z - 2 a b^3 y z - 2 a^3 c y z -
      > a b^2 c y z + a b c^2 y z + 2 a c^3 y z + a^3 b z^2 - a b^3 z^2=0
      >
      > Yes . The circumcenter of O1O2O3 is on the line L for all S's.
      >
      > The three centers Oa, Ob, Oc are the mid points of the arcs
      > BC, CA, AB of the circumcircle of ABC.
      > The points O1, O2, O3 are the mid points of SOa, SOb, SOc.
      > If K is the mid point of SO then KO1 = OOa/2 = R/2.
      > Hence K is the center of the circle O1O2O3 with radious R/2.
      >
      > Best regards
      > Nikos
      >
      > > GENERALIZATION
      > >
      > > Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of
      > > ABC, IBC, ICA, IAB, resp. and S a point.
      > >
      > > Denote:
      > >
      > > L, La, Lb ,Lc = the lines SO, SOa, SOb, SOc.
      > >
      > > Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.
      > >
      > > Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.
      > >
      > > Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.
      > >
      > > O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb,
      > > resp.
      > >
      > > 1. For which S's the triangles ABC, O1O2O3 are perspective?
      > > (Locus)
      > >
      > > 2. The circumcenter of O1O2O3 is on the line L for all S's
      > > (??).
      > >
      >


      [Non-text portions of this message have been removed]
    • Antreas
      Case of H: Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of ABC, HBC, HCA, HAB, resp. and S a point. Denote: L, La, Lb ,Lc = the lines SO, SOa, SOb,
      Message 2 of 9 , Feb 9, 2013
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        Case of H:

        Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of
        ABC, HBC, HCA, HAB, resp. and S a point.

        Denote:

        L, La, Lb ,Lc = the lines SO, SOa, SOb, SOc.

        Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.

        Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.

        Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.

        O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb,
        resp.

        1. For which S's the triangles ABC, O1O2O3 are perspective?
        [they are homothetic, I think]
        2. For which S's the circumcenter of O1O2O3 is on the line L ?

        Interesting Cases:

        1/ S is lying on the Euler Line of ABC (so L = SO = Euler line).
        I think that for both 1,2 the answer is the Euler line!

        2/ L, La, Lb, Lc = the Brocard axes of ABC, HBC, HCA, HAB, concurrent
        at a point S.
        The triangles ABC, O1O2O3 are perspective, but the circumcenter of O1O2O3 is not on L.

        APH
      • Nikolaos Dergiades
        Dear Antreas, Let P be an arbitrary point Let ABC be a triangle, O, Oa, Ob, Oc the circumcenters of ABC, PBC, PCA, PAB, resp. and S a point. Denote: L, La, Lb
        Message 3 of 9 , Feb 10, 2013
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          Dear Antreas,
          Let P be an arbitrary point
          Let ABC be a triangle, O, Oa, Ob, Oc the circumcenters of
          ABC, PBC, PCA, PAB, resp. and S a point.
          Denote:
          L, La, Lb ,Lc = the lines SO, SOa, SOb, SOc.
          Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.
          Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.
          Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.
          O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb,
          resp. and O' the circumcenter of O1O2O3.
          Fact1 = the triangles ABC, O1,O2,O3 are perspective
          Fact2 = O' lies on L

          The locus of S for Fact1 is a conic and
          the locus of S for Fact2 is a line.

          If P = I = Ia = Ib = Ic Fact2 holds for every S

          If P = H then Fact1 holds for every S and Fact2 if S lies on
          the Euler line.

          ND


          [APH]
          > Case of H:
          >
          > Let ABC be a triangle, O ,Oa, Ob, Oc the circumcenters of
          > ABC, HBC, HCA, HAB, resp. and S a point.
          >
          > Denote:
          >
          > L, La, Lb ,Lc = the lines SO, SOa, SOb, SOc.
          >
          > Ab, Ac = the orthogonal projections of Oa on Lb,Lc, resp.
          >
          > Bc, Ba = the orthogonal projections of Ob on Lc,La, resp.
          >
          > Ca, Cb = the orthogonal projections of Oc on La,Lb, resp.
          >
          > O1, O2, O3 = the circumcenters of OaAbAc, ObBcBa, OcCaCb,
          > resp.
          >
          > 1. For which S's the triangles ABC, O1O2O3 are perspective?
          > [they are homothetic, I think]
          > 2. For which S's the circumcenter of O1O2O3 is on the line L
          > ?
          >
          > Interesting Cases:
          >
          > 1/ S is lying on the Euler Line of ABC (so L = SO = Euler
          > line).
          > I think that for both 1,2 the answer is the Euler line!
          >
          > 2/ L, La, Lb, Lc = the Brocard axes of ABC, HBC, HCA, HAB,
          > concurrent
          > at a point S.
          > The triangles ABC, O1O2O3 are perspective, but the
          > circumcenter of O1O2O3 is not on L.
          >
          > APH
          >
          >
          >
          > ------------------------------------
          >
          > Yahoo! Groups Links
          >
          >
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          >
          >
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