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The Ehrmann points of a triangle

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  • Darij Grinberg
    Let ABC be a triangle and AA , BB , CC its altitudes. Let Ab, Ac, Bc, Ba, Ca, Cb be the projections of A , A , B , B , C , C on AB, CA, BC, AB, CA, BC,
    Message 1 of 14 , Aug 1, 2003
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      Let ABC be a triangle and AA', BB', CC' its altitudes. Let Ab,
      Ac, Bc, Ba, Ca, Cb be the projections of A', A', B', B', C',
      C' on AB, CA, BC, AB, CA, BC, respectively.

      It is well-known that the points Ab, Ac, Bc, Ba, Ca, Cb lie on
      the Taylor circle of triangle ABC.

      It is also important that the lines AbAc, BcBa, CaCb are
      antiparallel to BC, CA, AB, whereas the lines BaCa, CbAb,
      AcBc are parallel to BC, CA, AB.

      Triangles AB'C' and AAcAb are homothetic; triangles HB'C' and
      A'AcAb are homothetic, where H is the orthocenter of triangle
      ABC.

      We have the following theorems:

      (1a) The Euler lines of triangles HB'C', HC'A', HA'B' concur
      at one point, namely the Schiffler point of triangle
      A'B'C'. It lies on the Euler line of triangle A'B'C'.

      (1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur
      at one point on the nine-point circle of triangle ABC.

      Note that (1a) follows from the definition of the Schiffler
      point (H is the incenter of triangle A'B'C'), and (1b) is a
      result of Victor Thebault. Both Euler lines of triangles
      HB'C' and AB'C' pass through the midpoint Ha of AH, since
      this midpoint is the circumcenter of both triangles. If Hb
      and Hc are the midpoints of BH and CH, we have analogous
      results. On the other hand, it is well-known that the
      points Ha, Hb, Hc lie on the nine-point circle of triangle
      ABC and the triangle HaHbHc is homothetic to triangle ABC.

      Jean-Pierre Ehrmann gave the following results in
      Hyacinthos messages #3693 and #3694:

      (2a) The Euler lines of triangles A'AbAc, B'BcBa, C'CaCb
      concur at one point, namely the
      complement-of-the-Schiffler point of triangle
      A'B'C'. It lies on the Euler line of triangle A'B'C'.

      (2b) The Euler lines of triangles AAbAc, BBcBa, CCaCb concur
      at one point.

      The point of concurrence in (2a) is called FIRST EHRMANN POINT
      of triangle ABC, included as X(973) in Kimberling's ETC, and
      the point of concurrence in (2b) is called SECOND EHRMANN
      POINT, included as X(974).

      Some days ago, I have found synthetic proofs of (2b) and a
      part of (2a). We begin with (2b):

      Let Ga, Gb, Gc be the centroids of triangles AAbAc, BBcBa,
      CCaCb. Since CbAb || CA, we have

      d(Cb; CA) = d(Ab; CA).

      [Here, d(P; g) means the distance from a point P to a line g.]

      Since the centroid of a triangle trisects the medians, we
      have

      d(Gc; CA) = d(Cb; CA)/3
      and d(Ga; CA) = d(Ab; CA)/3,

      hence d(Gc; CA) = d(Ga, CA), and GcGa || CA. Analogously,
      GaGb || AB and GbGc || BC. Hence, triangles GaGbGc and ABC
      are homothetic. Since triangle ABC is homothetic with triangle
      HaHbHc, triangles GaGbGc and HaHbHc are homothetic. But the
      Euler lines of triangles AAbAc, BBcBa, CCaCb pass through Ga,
      Gb, Gc, respectively (the centroid of a triangle lies on its
      Euler line!), while the Euler lines of triangles AB'C', BC'A',
      CA'B' pass through Ha, Hb, Hc, respectively. These Euler
      lines are respectively parallel (since triangles AB'C' and
      AAcAb are homothetic, and similarly for B and C). Hence, from
      the concurrence of the Euler lines of triangles AB'C', BC'A',
      CA'B' at a point on the nine-point circle of triangle ABC
      (i. e., on the circumcircle of triangle HaHbHc), we conclude
      the concurrence of the Euler lines of triangles AAbAc, BBcBa,
      CCaCb on the circumcircle of triangle GaGbGc. Herewith, we
      have proven even more than (2b)!

      Now to (2a).

      Call Wa, Wb, Wc the orthocenters of triangles A'AbAc, B'BcBa,
      C'CaCb. Since A'Ab is orthogonal to AB, while AcBc is parallel
      to AB, the line AcBc is an altitude of triangle A'AbAc and
      passes through its orthocenter Wa. Analogously, AcBc passes
      through Wb. Hence, the line WaWb coincides with AcBc and is
      therefore parallel to AB. Analogously, WbWc is parallel to
      BC and WcWa is parallel to CA. Hence, triangles WaWbWc and
      ABC are homothetic. Since triangles HaHbHc and ABC are also
      homothetic, we get the homothety of triangles WaWbWc and
      HaHbHc. But the Euler lines of triangles A'AbAc, B'BcBa,
      C'CaCb pass through Wa, Wb, Wc, respectively (the orthocenter
      of a triangle lies on its Euler line!), while the Euler
      lines of triangles HB'C', HC'A', HA'B' pass through Ha, Hb,
      Hc, respectively. These Euler lines are respectively
      parallel (since triangles HB'C' and A'AcAb are homothetic,
      and analogously for B and C). Hence, from the concurrence
      of the Euler lines of triangles HB'C', HC'A', HA'B', we
      deduce the concurrence of the Euler lines of triangles
      A'AbAc, B'BcBa, C'CaCb. This yields a part of (2a).

      Maybe, using some special properties of the configuration
      (Taylor hexagon of triangle ABC and "Spieker hexagon" of
      triangle A'B'C') we can obtain a proof of the rest of (2a).

      Darij Grinberg
    • jpehrmfr
      Dear Darij in Hyacinthos 7408, you wrote ... Note that if an angle of ABC is obtuse, the incenter of A B C is not H, but the corresponding vertex of ABC. ...
      Message 2 of 14 , Aug 1, 2003
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        Dear Darij
        in Hyacinthos 7408, you wrote
        > Let ABC be a triangle and AA', BB', CC' its altitudes.
        > We have the following theorems:
        >
        > (1a) The Euler lines of triangles HB'C', HC'A', HA'B' concur
        > at one point, namely the Schiffler point of triangle
        > A'B'C'. It lies on the Euler line of triangle A'B'C'.
        >
        > (1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur
        > at one point on the nine-point circle of triangle ABC.
        >
        > Note that (1a) follows from the definition of the Schiffler
        > point (H is the incenter of triangle A'B'C')

        Note that if an angle of ABC is obtuse, the incenter of A'B'C' is
        not H, but the corresponding vertex of ABC.

        > and (1b) is a result of Victor Thebault.
        Notethat the common point is the center of Jerabek hyperbola.
        Friendly. Jean-Pierre
      • Darij Grinberg
        Dear Jean-Pierre Ehrmann, ... Yes, of course I know this, the excenters of a triangle also lie on the Neuberg cubic. ... I have met this before. The point M
        Message 3 of 14 , Aug 1, 2003
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          Dear Jean-Pierre Ehrmann,

          Thanks for the mail. You wrote:

          >> > (1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur
          >> > at one point on the nine-point circle of triangle ABC.
          >> >
          >> > Note that (1a) follows from the definition of the Schiffler
          >> > point (H is the incenter of triangle A'B'C')
          >>
          >> Note that if an angle of ABC is obtuse, the incenter of A'B'C' is
          >> not H, but the corresponding vertex of ABC.

          Yes, of course I know this, the excenters of a triangle
          also lie on the Neuberg cubic.

          >> > and (1b) is a result of Victor Thebault.
          >> Notethat the common point is the center of Jerabek hyperbola.

          I have met this before. The point M where the Euler lines
          of triangles AB'C', BC'A', CA'B' meet has the property
          that one of the equations MA' = MB' + MC' or cyclically
          holds. This was stated by Thebault. Can anybody find a
          SYNTHETIC proof?

          Thanks,
          Sincerely,
          Darij Grinberg
        • jpehrmfr
          Dear Darij [DG] ... [JPE] ... is ... [DG] ... I was just meaning that the common point is the Schiffler point of A B C only when ABC is acutangle ... I m
          Message 4 of 14 , Aug 2, 2003
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            Dear Darij

            [DG]
            > >> > (1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur
            > >> > at one point on the nine-point circle of triangle ABC.
            > >> >
            > >> > Note that (1a) follows from the definition of the Schiffler
            > >> > point (H is the incenter of triangle A'B'C')

            [JPE]
            > >> Note that if an angle of ABC is obtuse, the incenter of A'B'C'
            is
            > >> not H, but the corresponding vertex of ABC.

            [DG]
            > Yes, of course I know this, the excenters of a triangle
            > also lie on the Neuberg cubic.

            I was just meaning that the common point is the Schiffler point of
            A'B'C' only when ABC is acutangle

            > >> > and (1b) is a result of Victor Thebault.
            > >> Notethat the common point is the center of Jerabek hyperbola.

            > I have met this before. The point M where the Euler lines
            > of triangles AB'C', BC'A', CA'B' meet has the property
            > that one of the equations MA' = MB' + MC' or cyclically
            > holds. This was stated by Thebault. Can anybody find a
            > SYNTHETIC proof?

            I'm unable to find a synthetic proof of this fact but it is quite
            easy to check that
            MA' : MB' : MC' = |(b^2-c^2)SA| : |(c^2-a^2)SB| : |(a^2-b^2)SC| and
            (b^2-c^2)SA+(c^2-a^2)SB+(a^2-b^2)SC = 0
            Friendly. Jean-Pierre
          • Darij Grinberg
            Dear all, ... I think that a synthetic proof (in Russian) can be found at http://archive.1september.ru/mat/2000/no43_1.htm Alas, I cannot follow the author s
            Message 5 of 14 , Aug 3, 2003
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              Dear all,

              In Hyacinthos message #7410, I wrote:

              >> >> > (1b) The Euler lines of triangles AB'C', BC'A',
              >> >> > CA'B' concur at one point on the
              >> >> > nine-point circle of triangle ABC.

              >> >> > and (1b) is a result of Victor Thebault.
              >> >> Notethat the common point is the center of
              >> >> Jerabek hyperbola.

              >> I have met this before. The point M where the Euler lines
              >> of triangles AB'C', BC'A', CA'B' meet has the property
              >> that one of the equations MA' = MB' + MC' or cyclically
              >> holds. This was stated by Thebault. Can anybody find a
              >> SYNTHETIC proof?

              I think that a synthetic proof (in Russian) can be found at

              http://archive.1september.ru/mat/2000/no43_1.htm

              Alas, I cannot follow the author's observation. Anybody who
              can help?

              Darij Grinberg

              PS: Thanks Jean-Pierre, perhaps your equations can be
              helpful in finding a synthetic proof.

              Darij
            • Antreas P. Hatzipolakis
              ... This was once discussed here (Floor had proposed it as a problem in the AMM), and were generalizations by PY and APH I don t remember in what direction,
              Message 6 of 14 , Sep 11, 2003
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                in Hyacinthos 7408 Darij wrote:

                > Let ABC be a triangle and AA', BB', CC' its altitudes.
                > We have the following theorems:
                > (1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur
                > at one point on the nine-point circle of triangle ABC.

                This was once discussed here (Floor had proposed it
                as a problem in the AMM), and were generalizations
                by PY and APH

                I don't remember in what direction, but now let's
                parametrize it.

                Let ABC be a triangle and AHa, BHb, CHc its altitudes.

                Special case #1:

                Let Ab, Ac be the reflections of A in Hc, Hb, resp.
                Let Bc, Ba be the reflections of B in Ha, Hc, resp.
                Let Ca, Cb be the reflections of C in Hb, Ha, resp.

                The Euler Lines of the Triangles AHbHc, BHcHa, CHaHb
                concur (at H of ABC).

                Problem: Are their Nine Point Circles concurrent?

                Special Case #2:

                Let Mac, Mab be the midpoints of AHc, AHb, resp.
                Let Mba, Mbc be the midpoints of BHa, BHc, resp.
                Let Mcb, Mca be the midpoints of CHb, CHa, resp.

                The Euler Lines of the Triangles AMabMac, BMbcMba,
                CMcaMcb are concurrent (??)

                Generalization:

                Let Pac, Pab be points on AB, AC resp. such that:

                APac / AHc = APab / AHb = t

                Similarly we define the points Pba, Pbc; Pcb, Pca

                The Euler Lines of the Triangles APabPac, BPbcPba,
                CPcaPcb are concurrent (??) Locus??


                APH


                --
              • Darij Grinberg
                Dear Antreas, ... ^^^^^^^^^^^^^^^^^^^ I guess you mean AAbAc, BBcBa, CCaCb. The Euler lines of these triangles concur since the orthocenter
                Message 7 of 14 , Sep 11, 2003
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                  Dear Antreas,

                  In Hyacinthos message #7858, you wrote (partly):

                  >> Let ABC be a triangle and AHa, BHb, CHc its altitudes.
                  >>
                  >> Special case #1:
                  >>
                  >> Let Ab, Ac be the reflections of A in Hc, Hb, resp.
                  >> Let Bc, Ba be the reflections of B in Ha, Hc, resp.
                  >> Let Ca, Cb be the reflections of C in Hb, Ha, resp.
                  >>
                  >> The Euler Lines of the Triangles AHbHc, BHcHa, CHaHb
                  ^^^^^^^^^^^^^^^^^^^
                  I guess you mean AAbAc, BBcBa, CCaCb.
                  The Euler lines of these triangles concur since the
                  orthocenter H of triangle ABC is their common
                  circumcenter.

                  >> concur (at H of ABC).

                  >> Problem: Are their Nine Point Circles concurrent?

                  Yes. In fact, the nine-point circle of a triangle
                  ABC is the reflection of the circumcircle of triangle
                  AB'C' in the line B'C', where A', B', C' are the
                  midpoints of the sides BC, CA, AB. This yields that
                  the nine-point circle of triangle AAbAc is the
                  reflection of the circumcircle of triangle AHbHc in
                  the line HbHc. Similarly for the other two nine-point
                  circles. Since the circles AHbHc, BHcHa, CHaHb concur
                  at H, the three nine-point circles concur at the
                  antigonal conjugate of H with respect to triangle
                  HaHbHc.

                  NOTE. Antigonal conjugates? If P is a point in the
                  plane of a triangle ABC, then the reflections of the
                  circles PBC, PCA, PAB in the sidelines BC, CA, AB
                  have a common point. This point is called antigonal
                  conjugate of P with respect to triangle ABC. But I
                  see (Hyacinthos message #7826) that you know this,
                  maybe with another name.

                  The point where your nine-point circles concur is
                  not in the ETC.

                  >> Generalization:
                  >>
                  >> Let Pac, Pab be points on AB, AC resp. such that:
                  >>
                  >> APac / AHc = APab / AHb = t
                  >>
                  >> Similarly we define the points Pba, Pbc; Pcb, Pca
                  >>
                  >> The Euler Lines of the Triangles APabPac, BPbcPba,
                  >> CPcaPcb are concurrent

                  Yes, they are: the proof is simple. Moreover, if
                  J = X(125) is the intersection of the Euler lines of
                  triangles AHbHc, BHcHa, CHaHb, then the Euler lines
                  of triangles APabPac, BPbcPba, CPcaPcb meet at the
                  point W lying on the line HJ and dividing the segment
                  HJ in the ratio

                  HW 2-t
                  -- = ---.
                  WJ t-1

                  Remark that the line HJ passes through X(74).

                  Sincerely,
                  Darij Grinberg
                • Antreas P. Hatzipolakis
                  Dear Darij [APH, carelessly!] ... I don t know if Paul and I have already discussed this point in the past (we had discussed a number of cases of concurrent
                  Message 8 of 14 , Sep 11, 2003
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                    Dear Darij

                    [APH, carelessly!]
                    >>> Let ABC be a triangle and AHa, BHb, CHc its altitudes.
                    >>>
                    >>> Special case #1:
                    >>>
                    >>> Let Ab, Ac be the reflections of A in Hc, Hb, resp.
                    >>> Let Bc, Ba be the reflections of B in Ha, Hc, resp.
                    >>> Let Ca, Cb be the reflections of C in Hb, Ha, resp.
                    >>>
                    >>> The Euler Lines of the Triangles AHbHc, BHcHa, CHaHb
                    > ^^^^^^^^^^^^^^^^^^^

                    [DG]:

                    >I guess you mean AAbAc, BBcBa, CCaCb.
                    >The Euler lines of these triangles concur since the
                    >orthocenter H of triangle ABC is their common
                    >circumcenter.

                    [APH]:
                    >>> concur (at H of ABC).
                    >
                    >>> Problem: Are their Nine Point Circles concurrent?

                    [DG]:
                    >Yes. In fact, the nine-point circle of a triangle
                    >ABC is the reflection of the circumcircle of triangle
                    >AB'C' in the line B'C', where A', B', C' are the
                    >midpoints of the sides BC, CA, AB. This yields that
                    >the nine-point circle of triangle AAbAc is the
                    >reflection of the circumcircle of triangle AHbHc in
                    >the line HbHc. Similarly for the other two nine-point
                    >circles. Since the circles AHbHc, BHcHa, CHaHb concur
                    >at H, the three nine-point circles concur at the
                    >antigonal conjugate of H with respect to triangle
                    >HaHbHc.
                    >
                    >NOTE. Antigonal conjugates? If P is a point in the
                    >plane of a triangle ABC, then the reflections of the
                    >circles PBC, PCA, PAB in the sidelines BC, CA, AB
                    >have a common point. This point is called antigonal
                    >conjugate of P with respect to triangle ABC. But I
                    >see (Hyacinthos message #7826) that you know this,
                    >maybe with another name.
                    >
                    >The point where your nine-point circles concur is
                    >not in the ETC.

                    I don't know if Paul and I have already discussed this point
                    in the past (we had discussed a number of cases of
                    concurrent Nine point Circles)
                    Is it in your archive, Paul?

                    Anyway, it is a remarkable point and has place
                    in Kimberling's ETC, but what its coordinates are?

                    Of course, the locus problem is:

                    Which is the locus of the radical center of
                    the nine point circles in the parametrization below,
                    but I don't think that the curve is something
                    interesting.


                    [APH]:
                    >>> Generalization:
                    >>>
                    >>> Let Pac, Pab be points on AB, AC resp. such that:
                    >>>
                    >>> APac / AHc = APab / AHb = t
                    >>>
                    >>> Similarly we define the points Pba, Pbc; Pcb, Pca
                    >>>
                    >>> The Euler Lines of the Triangles APabPac, BPbcPba,
                    >>> CPcaPcb are concurrent

                    [DG]:

                    >Yes, they are: the proof is simple. Moreover, if
                    >J = X(125) is the intersection of the Euler lines of
                    >triangles AHbHc, BHcHa, CHaHb, then the Euler lines
                    >of triangles APabPac, BPbcPba, CPcaPcb meet at the
                    >point W lying on the line HJ and dividing the segment
                    >HJ in the ratio
                    >
                    > HW 2-t
                    > -- = ---.
                    > WJ t-1
                    >
                    >Remark that the line HJ passes through X(74).


                    Thanks

                    Greetings

                    Antreas
                    --
                  • Antreas P. Hatzipolakis
                    Let ABC be a triangle, HaHbHc its orthic triangle, and EaEbEc its Euler triangle (ie Ea = midpoint of AH etc). HbEa / AB = Hab HcEa / AC = Hac Similarly the
                    Message 9 of 14 , Sep 11, 2003
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                      Let ABC be a triangle, HaHbHc its orthic triangle, and
                      EaEbEc its Euler triangle (ie Ea = midpoint of AH etc).

                      HbEa /\ AB = Hab

                      HcEa /\ AC = Hac

                      Similarly the points Hbc, Hba; Hca, Hcb.

                      Are the Euler Lines of the Triangles AHabHac,
                      BHbcHba, CHcaHcb concurrent ?

                      Parametrization:

                      Ea, Eb, Ec points on AH, BH, CH, resp.
                      such that:

                      AEa / AH = BEb / BH = CEc / CH = t.

                      etc

                      Antreas

                      --
                    • Barry Wolk
                      ... Barycentrics (sin 3A)(1 + cos 2B + cos 2C)/(cos A) : ... : ... It was a routine computer calculation to get an answer. Reducing that answer to this short
                      Message 10 of 14 , Sep 12, 2003
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                        [APH]:
                        > Anyway, it is a remarkable point and has place
                        > in Kimberling's ETC, but what its coordinates are?

                        Barycentrics
                        (sin 3A)(1 + cos 2B + cos 2C)/(cos A) : ... : ...

                        It was a routine computer calculation to get an answer. Reducing
                        that answer to this short formula was another problem.
                        --
                        Barry Wolk


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                      • Alexey.A.Zaslavsky
                        Dear colleagues! What is the number in ETC of Thebault point, i.e the common point of euler lines of triangles AB C , A BC and A B C where AA , BB , CC are
                        Message 11 of 14 , Dec 2, 2007
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                          Dear colleagues!
                          What is the number in ETC of Thebault point, i.e the common point of euler lines of triangles AB'C', A'BC' and A'B'C where AA', BB', CC' are the altitudes of ABC? Moscow scholar Fyodor Nilov asserts that this point is on Steiner inellipse . Is this correct?

                          Sincerely Alexey

                          Antivirus scanning: Symantec Mail Security for SMTP.

                          [Non-text portions of this message have been removed]
                        • jpehrmfr
                          Dear Alexey ... euler lines of triangles AB C , A BC and A B C where AA , BB , CC are the altitudes of ABC? Moscow scholar Fyodor Nilov asserts that this
                          Message 12 of 14 , Dec 3, 2007
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                            Dear Alexey

                            > What is the number in ETC of Thebault point, i.e the common point of
                            euler lines of triangles AB'C', A'BC' and A'B'C where AA', BB', CC' are
                            the altitudes of ABC? Moscow scholar Fyodor Nilov asserts that this
                            point is on Steiner inellipse . Is this correct?

                            I think that your point is the center of the Jerabek hyperbola (X_125)
                            As the four common points of the Steiner inellipse with the 9Pcircle
                            are the midpoints of the sides and the center of the Kiepert hyperbola,
                            I don't think that this point lies on the Steiner inellipse
                            Friendly. Jean-Pierre
                          • Paul Yiu
                            Dear Alexey, [AZ]: What is the number in ETC of Thebault point, i.e the common point of euler lines of triangles AB C , A BC and A B C where AA , BB , CC are
                            Message 13 of 14 , Dec 3, 2007
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                              Dear Alexey,

                              [AZ]: What is the number in ETC of Thebault point, i.e the common
                              point of euler lines of triangles AB'C', A'BC' and A'B'C where AA',
                              BB', CC' are the altitudes of ABC? Moscow scholar Fyodor Nilov asserts
                              that this point is on Steiner inellipse . Is this correct?


                              *** The Euler lines of the three triangles intersect at the center of
                              the Jerabek hyperbola. It is the point with barycentric coordinates

                              ((b^2+c^2-a^2)(b^2-c^2)^2 : ... : ...).

                              In ETC, this is X(125), called the Jerabek center.

                              It does not lie on the Steiner inellipse.

                              Best regards
                              Sincerely
                              Paul
                            • Francois Rideau
                              Dear Alexey Let T be the Thebault point and (L) the Euler line of ABC. The lines symmetric of (L) wrt the sides BC, CA, AB are on a same point U on the
                              Message 14 of 14 , Dec 3, 2007
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                                Dear Alexey
                                Let T be the Thebault point and (L) the Euler line of ABC.
                                The lines symmetric of (L) wrt the sides BC, CA, AB are on a same point U on
                                the ABC-circumcircle.
                                Then T is homothetic of U in the homothety of center G, the ABC-centroid and
                                ratio -1/2.
                                I think T is not on the ABC-Steiner inellipse.
                                Friendly
                                Francois

                                On Dec 3, 2007 8:24 AM, Alexey.A.Zaslavsky <zasl@...> wrote:

                                > Dear colleagues!
                                > What is the number in ETC of Thebault point, i.e the common point of euler
                                > lines of triangles AB'C', A'BC' and A'B'C where AA', BB', CC' are the
                                > altitudes of ABC? Moscow scholar Fyodor Nilov asserts that this point is on
                                > Steiner inellipse . Is this correct?
                                >
                                > Sincerely Alexey
                                >
                                > Antivirus scanning: Symantec Mail Security for SMTP.
                                >
                                > [Non-text portions of this message have been removed]
                                >
                                >
                                >


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