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Re: Antreas' Feuerbach concurrence

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  • peter_mows
    Dear Antreas, [APH], (with P replacing O) ... If P is on the Keipert hyperbola, then, the Brocard axes of the triangles AAbAc, BBcBa and CCaCb concur at the
    Message 1 of 28 , Oct 3, 2004
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      Dear Antreas,

      [APH], (with P replacing O)

      >Let P be a point..
      >Let Ab and Ac be the orthogonal projections of the
      >point A on the lines BP and CP. Similarly, define
      >the points Bc and Ba, and the points Ca and Cb.

      If P is on the Keipert hyperbola, then, the Brocard axes of the
      triangles AAbAc, BBcBa and CCaCb concur at the the center of the
      hyperbola, X(115).
      As well as the Keipert hyperbola, it looks like other loci are the
      line at infinity, the sidelines and a degree 8.

      The Euler lines of the triangles AAbAc, BBcBa and CCaCb concur for P
      on the line at infinity and degrees 6 & 7.
      Some ETC points,

      1 -> 11
      4 -> 125
      13 -> 115
      14 -> 115
      74 -> 125
      80 -> 11

      Best regards
      Peter.
    • xpolakis
      Dear Peter ... Very nice ! Thanks. From these first points in ETC, we may conclude that the entire locus of P [a 1+6+7 deg. curve] is transformed in a three
      Message 2 of 28 , Oct 3, 2004
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        Dear Peter

        [APH]:
        > >Let P be a point..
        > >Let Ab and Ac be the orthogonal projections of the
        > >point A on the lines BP and CP. Similarly, define
        > >the points Bc and Ba, and the points Ca and Cb.

        [PM]:
        > If P is on the Keipert hyperbola, then, the Brocard axes of the
        > triangles AAbAc, BBcBa and CCaCb concur at the the center of the
        > hyperbola, X(115).
        > As well as the Keipert hyperbola, it looks like other loci are the
        > line at infinity, the sidelines and a degree 8.
        >
        > The Euler lines of the triangles AAbAc, BBcBa and CCaCb concur for P
        > on the line at infinity and degrees 6 & 7.
        > Some ETC points,
        >
        > 1 -> 11 Feuerbach Hyp. center
        > 4 -> 125 Jerabek
        > 13 -> 115 Kiepert
        > 14 -> 115 Kiepert
        > 74 -> 125 Jerabek
        > 80 -> 11 Feuerbach

        Very nice ! Thanks.

        From these first points in ETC, we may conclude that the entire
        locus of P [a 1+6+7 deg. curve] is transformed in a three point set:
        Centers of Feuerbach,Jerabek, and Kiepert Hyperbolae.

        But most likely the locus of the point of concurrence is the
        entire NPC (+ some other things ?)


        Greetings

        Antreas
      • Darij Grinberg
        ... Is it possible that antigonal conjugates are mapped to the same point? In other words, if a point P has the property that the Euler lines of triangles
        Message 3 of 28 , Oct 4, 2004
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          In Hyacinthos message #10593, Peter Moses wrote:

          >> The Euler lines of the triangles AAbAc, BBcBa
          >> and CCaCb concur for P on the line at infinity
          >> and degrees 6 & 7.
          >> Some ETC points,
          >>
          >> 1 -> 11 Feuerbach Hyp. center
          >> 4 -> 125 Jerabek
          >> 13 -> 115 Kiepert
          >> 14 -> 115 Kiepert
          >> 74 -> 125 Jerabek
          >> 80 -> 11 Feuerbach

          Is it possible that antigonal conjugates are mapped
          to the same point? In other words, if a point P has
          the property that the Euler lines of triangles
          AAbAc, BBcBa, CCaCb concur at one point Q, then the
          antigonal conjugate P' of P also has this property,
          and the Euler lines for P' concur at the same point
          Q ? This looks pretty plausible to me from the
          above data. (See Bernard Gibert's nice page

          http://perso.wanadoo.fr/bernard.gibert/gloss/pointsandmapping.html

          for a definition of antigonal conjugates.)

          In Hyacinthos message #10594, Antreas P. Hatzipolakis
          wrote:

          >> From these first points in ETC, we may conclude
          >> that the entire locus of P [a 1+6+7 deg. curve]
          >> is transformed in a three point set:
          >> Centers of Feuerbach,Jerabek, and Kiepert
          >> Hyperbolae.

          Hardly. Anyway, extraversion shows that since the
          Feuerbach point X(11) serves as point of
          concurrence, the three external Feuerbach points
          must serve as points of concurrence, too. So we
          would have, at least, 6 different points. But I
          really suspect that the locus of the point of
          concurrence is the whole nine-point circle. If
          anybody could prove or disprove that...

          Darij Grinberg
        • Antreas P. Hatzipolakis
          Dear Darij ... Also, is it possible that the sextic and/or the septic of the locus is a self-antigonal curve? That is, if P lies on the curve, then the
          Message 4 of 28 , Oct 4, 2004
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            Dear Darij

            [PM]:
            >>> The Euler lines of the triangles AAbAc, BBcBa
            >>> and CCaCb concur for P on the line at infinity
            >>> and degrees 6 & 7.
            >>> Some ETC points,
            >>>
            >>> 1 -> 11 Feuerbach Hyp. center
            >>> 4 -> 125 Jerabek
            >>> 13 -> 115 Kiepert
            >>> 14 -> 115 Kiepert
            >>> 74 -> 125 Jerabek
            >>> 80 -> 11 Feuerbach

            [DG]:
            >Is it possible that antigonal conjugates are mapped
            >to the same point? In other words, if a point P has
            >the property that the Euler lines of triangles
            >AAbAc, BBcBa, CCaCb concur at one point Q, then the
            >antigonal conjugate P' of P also has this property,
            >and the Euler lines for P' concur at the same point
            >Q ? This looks pretty plausible to me from the
            >above data.


            Also, is it possible that the sextic and/or the septic
            of the locus is a self-antigonal curve?
            That is, if P lies on the curve, then the antigonal of
            P lies on the curve too ?


            [APH]:
            >> From these first points in ETC, we may conclude
            >> that the entire locus of P [a 1+6+7 deg. curve]
            >> is transformed in a three point set:
            >> Centers of Feuerbach,Jerabek, and Kiepert
            >> Hyperbolae.

            [DG]:
            >Hardly. Anyway, extraversion shows that since the

            Neither I believe that, since I wrote
            (but you misquoted me!):

            [APH]:
            >>But most likely the locus of the point of concurrence is the
            >>entire NPC (+ some other things ?)


            Antreas
            --
          • Antreas P. Hatzipolakis
            ... This is already contained in Darij s message, but I had not read it carefully! APH --
            Message 5 of 28 , Oct 4, 2004
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              >[DG]:
              >>Is it possible that antigonal conjugates are mapped
              >>to the same point? In other words, if a point P has
              >>the property that the Euler lines of triangles
              >>AAbAc, BBcBa, CCaCb concur at one point Q, then the
              >>antigonal conjugate P' of P also has this property,
              >>and the Euler lines for P' concur at the same point
              >>Q ? This looks pretty plausible to me from the
              >>above data.

              [APH]:
              >Also, is it possible that the sextic and/or the septic
              >of the locus is a self-antigonal curve?
              >That is, if P lies on the curve, then the antigonal of
              >P lies on the curve too ?

              This is already contained in Darij's message, but I had not
              read it carefully!

              APH


              --
            • Bernard Gibert
              Dear Antreas and Peter, ... ... which is bicircular and contains the points at infinity of the Steiner ellipse (double points) and (I think) only five singular
              Message 6 of 28 , Oct 4, 2004
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                Dear Antreas and Peter,

                > [APH]:
                > > >Let P be a point..
                > > >Let Ab and Ac be the orthogonal projections of the
                > > >point A on the lines BP and CP. Similarly, define
                > > >the points Bc and Ba, and the points Ca and Cb.
                >
                > [PM]:
                > > If P is on the Keipert hyperbola, then, the Brocard axes of the
                > > triangles AAbAc, BBcBa and CCaCb concur at the the center of the
                > > hyperbola, X(115).
                > > As well as the Kiepert hyperbola, it looks like other loci are the
                > > line at infinity, the sidelines and a degree 8.

                ... which is bicircular and contains the points at infinity of the
                Steiner ellipse (double points) and (I think) only five singular real
                points A, B, C, X13, X14.

                > >> The Euler lines of the triangles AAbAc, BBcBa and CCaCb concur for P
                > > on the line at infinity and degrees 6 & 7.

                ... the septic is Q030 and the sextic is tricircular containing the
                same five singular real points.

                http://perso.wanadoo.fr/bernard.gibert/curves/q030.html

                Q030 is indeed self-antigonal since its isogonal conjugate is itself an
                inversible bicircular quintic which I will probably add to my web site.

                Very nice indeed !

                Best regards

                Bernard

                [Non-text portions of this message have been removed]
              • jpehrmfr
                Dear Antreas, Darij, Bernard, Peter and other Hyacinthists ... the ... the ... real ... concur for P ... the ... itself an ... site. The point is that the
                Message 7 of 28 , Oct 4, 2004
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                  Dear Antreas, Darij, Bernard, Peter and other Hyacinthists
                  > > [APH]:
                  > > > >Let P be a point..
                  > > > >Let Ab and Ac be the orthogonal projections of the
                  > > > >point A on the lines BP and CP. Similarly, define
                  > > > >the points Bc and Ba, and the points Ca and Cb.
                  > >
                  > > [PM]:
                  > > > If P is on the Keipert hyperbola, then, the Brocard axes of the
                  > > > triangles AAbAc, BBcBa and CCaCb concur at the the center of
                  the
                  > > > hyperbola, X(115).
                  > > > As well as the Kiepert hyperbola, it looks like other loci are
                  the
                  > > > line at infinity, the sidelines and a degree 8.
                  >
                  > ... which is bicircular and contains the points at infinity of the
                  > Steiner ellipse (double points) and (I think) only five singular
                  real
                  > points A, B, C, X13, X14.
                  >
                  > > >> The Euler lines of the triangles AAbAc, BBcBa and CCaCb
                  concur for P
                  > > > on the line at infinity and degrees 6 & 7.
                  >
                  > ... the septic is Q030 and the sextic is tricircular containing
                  the
                  > same five singular real points.
                  >
                  > http://perso.wanadoo.fr/bernard.gibert/curves/q030.html
                  >
                  > Q030 is indeed self-antigonal since its isogonal conjugate is
                  itself an
                  > inversible bicircular quintic which I will probably add to my web
                  site.

                  The point is that the sextic has probably no other finite real point
                  than A,B,C and the Fermat points. Hence,
                  YOU CAN REMOVE THE SEXTIC FROM THE LOCUS.
                  Here is an explanation :
                  If PaPbPc is the pedal triangle of P, the direct similitude with
                  center Pc mapping the midpoint of AP to the midpoint of BP will map
                  the triangle AAbAc to BaBBc. It follows that the three triangles
                  AAbAc, BBcBa, CCaCb are equilateral at the same time, and this can
                  occur only when P is one of the Fermat points. I presume that your
                  sextic is the condition for one of these triangles to be
                  equilateral - something like OaGa = 0 where Oa and Ga are the
                  circumcenter and the centroid of AAbAc -; but, in this case, the
                  three Euler lines don't exist and, of course, cannot concur.

                  Now, an easy computation - starting, for instance, of the equation
                  xy=k for the rectangular hyperbola ABCP - gives the fact that, when
                  the three Euler lines concur, their common point is the center of
                  the hyperbola.
                  From this it follows that the locus of P for which the three Euler
                  lines concur is the union of the line at infinity (in which case,
                  the three lines are the same ones = GP) and of the curve of degree 7
                  above, in which case their common point, being the center of the
                  rectangular circumhyperbola going through P, lies on the NPCircle of
                  ABC.
                  Of course, we can now understand why the septic is self-antigonal
                  and why her isogonal conjugate - Bernard's quintic - is anallagmatic
                  wrt the circumcircle.
                  Friendly. Jean-Pierre
                • Antreas P. Hatzipolakis
                  ... We have seen the locus of P such that the Euler Lines of the triangles AAbAc, BBcBa and CCaCb are concurrent. Next natural question is: Which is the locus
                  Message 8 of 28 , Oct 4, 2004
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                    >[APH]
                    >
                    >>Let P be a point..
                    >>Let Ab and Ac be the orthogonal projections of the
                    >>point A on the lines BP and CP. Similarly, define
                    >>the points Bc and Ba, and the points Ca and Cb.

                    We have seen the locus of P such that the Euler Lines of
                    the triangles AAbAc, BBcBa and CCaCb are concurrent.

                    Next natural question is:

                    Which is the locus of P such that the Euler Lines (variations: OK lines, or...)
                    of the triangles ABcCb, BCaAc and CAbBa are concurrent.

                    It seems that this is more complicated than the first locus!


                    Antreas
                    --
                  • Antreas P. Hatzipolakis
                    ... And since it is too complicated, let s see a special case: Let ABC be a triangle, L a line, and La, Lb, Lc the parallels through A,B,C to the line L, resp.
                    Message 9 of 28 , Oct 4, 2004
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                      >>[APH]
                      >>
                      >>>Let P be a point..
                      >>>Let Ab and Ac be the orthogonal projections of the
                      >>>point A on the lines BP and CP. Similarly, define
                      >>>the points Bc and Ba, and the points Ca and Cb.
                      >
                      >We have seen the locus of P such that the Euler Lines of
                      >the triangles AAbAc, BBcBa and CCaCb are concurrent.
                      >
                      >Next natural question is:
                      >
                      >Which is the locus of P such that the Euler Lines (variations: OK
                      >lines, or...)
                      >of the triangles ABcCb, BCaAc and CAbBa are concurrent.
                      >
                      >It seems that this is more complicated than the first locus!


                      And since it is too complicated, let's see a special case:

                      Let ABC be a triangle, L a line, and La, Lb, Lc
                      the parallels through A,B,C to the line L, resp.

                      Bc := the orthogonal projection of B on Lc
                      Cb := the orthogonal projection of C on Lb

                      Similarly Ca,Ac; Ab, Ba

                      The Euler lines of ABcCb, BCaAc, CAbBa [and ABC ?] are concurrent (?)


                      Antreas
                      --
                    • peter_mows
                      Dear Antreas, [APH] ... Yes, I think so. I reckon that the Euler line locus is L.inf + degree 17 and Brocard a degree 18. I suspect X(4) lives on both locii.
                      Message 10 of 28 , Oct 4, 2004
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                        Dear Antreas,

                        [APH]
                        >>Let P be a point..
                        >>Let Ab and Ac be the orthogonal projections of the
                        >>point A on the lines BP and CP. Similarly, define
                        >>the points Bc and Ba, and the points Ca and Cb.

                        > Next natural question is:
                        >
                        > Which is the locus of P such that the Euler Lines (variations: OK
                        > lines, or...) of the triangles ABcCb, BCaAc and CAbBa are
                        > concurrent.
                        >
                        > It seems that this is more complicated than the first locus!

                        Yes, I think so.
                        I reckon that the Euler line locus is L.inf + degree 17 and Brocard
                        a degree 18.
                        I suspect X(4) lives on both locii.

                        Best regards
                        Peter.
                      • Antreas P. Hatzipolakis
                        Dear Peter ... Yes, X(4) = H, lives on both: If AA , BB , CC are the cevians of H, then Ab = Cb = B , etc and we know that the Euler Lines / Brocard axes of
                        Message 11 of 28 , Oct 4, 2004
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                          Dear Peter

                          >[APH]
                          >>>Let P be a point..
                          >>>Let Ab and Ac be the orthogonal projections of the
                          >>>point A on the lines BP and CP. Similarly, define
                          >>>the points Bc and Ba, and the points Ca and Cb.
                          >
                          >> Next natural question is:
                          >>
                          >> Which is the locus of P such that the Euler Lines (variations: OK
                          >> lines, or...) of the triangles ABcCb, BCaAc and CAbBa are
                          >> concurrent.
                          >>
                          >> It seems that this is more complicated than the first locus!

                          [PM]:
                          >Yes, I think so.
                          >I reckon that the Euler line locus is L.inf + degree 17 and Brocard
                          >a degree 18.
                          >I suspect X(4) lives on both locii.

                          Yes, X(4) = H, lives on both:

                          If AA', BB', CC' are the cevians of H, then Ab = Cb = B', etc
                          and we know that the Euler Lines / Brocard axes of
                          AB'C', BC'A', CA'B' are concurrent.

                          Greetings

                          Antreas
                          --
                        • Antreas P. Hatzipolakis
                          ... The lines BcCb, CaAc, AbBa concur at a point on the NPC of ABC. It is an old result I read in a Greek book and have posted in Hyacinthos (search the
                          Message 12 of 28 , Oct 4, 2004
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                            [APH]:
                            >Let ABC be a triangle, L a line, and La, Lb, Lc
                            >the parallels through A,B,C to the line L, resp.
                            >
                            >Bc := the orthogonal projection of B on Lc
                            >Cb := the orthogonal projection of C on Lb
                            >
                            >Similarly Ca,Ac; Ab, Ba
                            >The Euler lines of ABcCb, BCaAc, CAbBa [and ABC ?] are concurrent (?)

                            The lines BcCb, CaAc, AbBa concur at a point on the NPC of ABC.
                            It is an old result I read in a Greek book and have posted
                            in Hyacinthos (search the archive with keyword Panakis)

                            If the line L is the Euler line, or the Brocard axis or the OI
                            line of ABC, then which is the point of concurrence of the lines
                            BcCb, CaAc, AbBa ?

                            APH
                            --
                          • Antreas P. Hatzipolakis
                            ... Which is the locus of P such that ABC, 1. Triangle bounded by (AbAc, BcBa, CaCb) 2. Triangle bounded by (BcCb, CaAc, AbBa) are perspective? [Special Cases:
                            Message 13 of 28 , Oct 4, 2004
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                              >[APH]
                              >
                              >>Let P be a point..
                              >>Let Ab and Ac be the orthogonal projections of the
                              >>point A on the lines BP and CP. Similarly, define
                              >>the points Bc and Ba, and the points Ca and Cb.

                              Which is the locus of P such that ABC,

                              1. Triangle bounded by (AbAc, BcBa, CaCb)

                              2. Triangle bounded by (BcCb, CaAc, AbBa)

                              are perspective?

                              [Special Cases: (AbAc, BcBa, CaCb), (BcCb, CaAc, AbBa)
                              be concurrent]


                              APH
                              --
                            • peter_mows
                              Dear Antreas, ... 1/. DARBOUX CUBIC + L.inf + circumcircle + 3 ellipses. X(1) - X(2) X(3) - 1/((a^2 - SA) SA + 3 SB SC) : : X(4) - X(4) P on circumcircle -
                              Message 14 of 28 , Oct 5, 2004
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                                Dear Antreas,

                                > >[APH]
                                > >
                                > >>Let P be a point..
                                > >>Let Ab and Ac be the orthogonal projections of the
                                > >>point A on the lines BP and CP. Similarly, define
                                > >>the points Bc and Ba, and the points Ca and Cb.
                                >
                                > Which is the locus of P such that ABC,
                                >
                                > 1. Triangle bounded by (AbAc, BcBa, CaCb)
                                >
                                > 2. Triangle bounded by (BcCb, CaAc, AbBa)
                                >
                                > are perspective?

                                1/. DARBOUX CUBIC + L.inf + circumcircle + 3 ellipses.
                                X(1) -> X(2)
                                X(3) -> 1/((a^2 - SA) SA + 3 SB SC) : :
                                X(4) -> X(4)
                                P on circumcircle -> point on NPC midway between P and X(4)

                                2/. DARBOUX CUBIC + L.inf + circumcircle + sides.
                                X(1) -> X(7)
                                X(3) -> X(184)
                                X(4) -> X(4)
                                X(84) -> X(1857)

                                Best regards,
                                Peter.
                              • jpehrmfr
                                Dear Antreas and Peter ... I presume that your ellipses are the circles with diameter BC, CA and AB, corresponding to the case where two lines are the same
                                Message 15 of 28 , Oct 5, 2004
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                                  Dear Antreas and Peter
                                  > > >>Let P be a point..
                                  > > >>Let Ab and Ac be the orthogonal projections of the
                                  > > >>point A on the lines BP and CP. Similarly, define
                                  > > >>the points Bc and Ba, and the points Ca and Cb.
                                  > >
                                  > > Which is the locus of P such that ABC,
                                  > >
                                  > > 1. Triangle bounded by (AbAc, BcBa, CaCb)
                                  > >
                                  > > 2. Triangle bounded by (BcCb, CaAc, AbBa)
                                  > >
                                  > > are perspective?
                                  >
                                  > 1/. DARBOUX CUBIC + L.inf + circumcircle + 3 ellipses.
                                  > X(1) -> X(2)
                                  > X(3) -> 1/((a^2 - SA) SA + 3 SB SC) : :
                                  > X(4) -> X(4)
                                  > P on circumcircle -> point on NPC midway between P and X(4)

                                  I presume that your ellipses are the circles with diameter BC, CA
                                  and AB, corresponding to the case where two lines are the same ones.
                                  Friendly. Jean-Pierre
                                • peter_mows
                                  Dear Jean-Pierre & Antreas, [APH] ... [JPE] ... Yes!, I think so. I should have stared at my diagram a bit longer:) Best regards, Peter.
                                  Message 16 of 28 , Oct 5, 2004
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                                    Dear Jean-Pierre & Antreas,

                                    [APH]
                                    >>Let P be a point..
                                    >>Let Ab and Ac be the orthogonal projections of the
                                    >>point A on the lines BP and CP. Similarly, define
                                    >>the points Bc and Ba, and the points Ca and Cb.
                                    >>
                                    >> Which is the locus of P such that ABC,
                                    >>
                                    >> 1. Triangle bounded by (AbAc, BcBa, CaCb)
                                    >>
                                    >> 2. Triangle bounded by (BcCb, CaAc, AbBa)
                                    >>
                                    >> are perspective?
                                    >[PM]
                                    > 1/. DARBOUX CUBIC + L.inf + circumcircle + 3 ellipses.
                                    > X(1) -> X(2)
                                    > X(3) -> 1/((a^2 - SA) SA + 3 SB SC) : :
                                    > X(4) -> X(4)
                                    > P on circumcircle -> point on NPC midway between P and X(4)
                                    >
                                    [JPE]
                                    > I presume that your ellipses are the circles with diameter BC, CA
                                    > and AB, corresponding to the case where two lines are the same
                                    > ones.

                                    Yes!, I think so. I should have stared at my diagram a bit longer:)

                                    Best regards,
                                    Peter.
                                  • Antreas P. Hatzipolakis
                                    Dear Peter and Jean-Pierre ... [JPE] ... In other loci, we have seen these circles with the Neuberg cubic + Linf + (O) How about a locus problem where these
                                    Message 17 of 28 , Oct 5, 2004
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                                      Dear Peter and Jean-Pierre

                                      [PM]:
                                      >> 1/. DARBOUX CUBIC + L.inf + circumcircle + 3 ellipses.

                                      [JPE]
                                      >> I presume that your ellipses are the circles with diameter BC, CA
                                      >> and AB, corresponding to the case where two lines are the same
                                      >> ones.

                                      [PM]:
                                      >Yes!, I think so. I should have stared at my diagram a bit longer:)


                                      In other loci, we have seen these circles with the Neuberg cubic + Linf + (O)
                                      How about a locus problem where these circles come with an other isogonal
                                      cubic?


                                      APH
                                      --
                                    • peter_mows
                                      Dear Antreas, ... Call the triangle bounded by (AbAc, BcBa, CaCb), A1B1C1 Call the triangle bounded by (BcCb, CaAc, AbBa), A2B2C2 Circumcircles of A1B1C1 and
                                      Message 18 of 28 , Oct 6, 2004
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                                        Dear Antreas,

                                        >>[APH]
                                        >>
                                        >>Let P be a point..
                                        >>Let Ab and Ac be the orthogonal projections of the
                                        >>point A on the lines BP and CP. Similarly, define
                                        >>the points Bc and Ba, and the points Ca and Cb.

                                        Call the triangle bounded by (AbAc, BcBa, CaCb), A1B1C1

                                        Call the triangle bounded by (BcCb, CaAc, AbBa), A2B2C2

                                        Circumcircles of A1B1C1 and A2B2C2 meet the NPC of ABC at Q,
                                        x (a^2 y - b^2 y + c^2 y - a^2 z - b^2 z + c^2 z)
                                        (c^2 x y - b^2 x z - b^2 y z + c^2 y z) : :
                                        the center of the hyperbola ABCHP.

                                        Circumcircle of A2B2C2 & NPC of ABC are tangent.

                                        A3 = BBc /\ CCb = H of BCP
                                        B3 = CCa /\ AAc = H of CAP
                                        C3 = AAb /\ BBa = H of ABP

                                        A3,B3 & C3 are on the above hyperbola, opening the familiar flood
                                        gate of NP circles meeting at Q.

                                        NPC of A3BC passes through A1, Bc and Cb .. and Q
                                        NPC of AB3C3 passes through A1, AAc/\BcBa, AAb/\CaCb .. and Q
                                        NPC A3BC /\ NPC A3BC3 at midpoint BA3 on BBc .. and Q.

                                        Best regards,
                                        Peter.
                                      • xpolakis
                                        ... Let MaMbMc be the medial triangle of ABC. AbAc is the sideline MbMc of the medial triangle. So the problem in other words is: Let ABC be a triangle, A B C
                                        Message 19 of 28 , Apr 30, 2009
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                                          [APH #10485]:
                                          > Let ABC be a triangle and A'B'C' the cevian
                                          > triangle of I.
                                          >
                                          > Ab := the Orthogonal projection of A on BB'
                                          > Ac := the Orthogonal projection of A on CC'
                                          >
                                          > Similarly Bc,Ba; Ca,Cb
                                          >
                                          > The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent at F,
                                          > the Feuerbach point (??)

                                          Let MaMbMc be the medial triangle of ABC.

                                          AbAc is the sideline MbMc of the medial triangle.

                                          So the problem in other words is:

                                          Let ABC be a triangle, A'B'C' the cevian
                                          triangle of I and MaMbMc the cevian triangle of G.

                                          The perpendicular from A to BB' intersects
                                          the line MbMc at Ab.

                                          The perpendicular from A to CC' intersects
                                          the line MbMc at Ac.

                                          Similarly Bc,Ba; Ca,Cb

                                          The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent at F,
                                          the Feuerbach point.

                                          I was wondering what happens if we draw perpendiculars to
                                          the cevians of I, not from A,B,C, but from Ma,Mb,Mc.

                                          That is:

                                          Let ABC be a triangle, A'B'C' the cevian
                                          triangle of I and MaMbMc the cevian triangle of G.

                                          The perpendicular from Ma to BB' intersects
                                          the line MbMc at Ab.

                                          The perpendicular from Ma to CC' intersects
                                          the line MbMc at Ac.

                                          Similarly Bc,Ba; Ca,Cb

                                          The Euler Lines of MaAbAc, MbBcBa, McCaCb are concurrent
                                          (at some point on IG ???)


                                          Antreas
                                        • xpolakis
                                          [APH] ... In general: For MaMbMc = a triangle homothetic to ABC with homothety (G,t). APH
                                          Message 20 of 28 , Apr 30, 2009
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                                            [APH]
                                            > Let ABC be a triangle, A'B'C' the cevian
                                            > triangle of I and MaMbMc the cevian triangle of G.
                                            >
                                            > The perpendicular from Ma to BB' intersects
                                            > the line MbMc at Ab.
                                            >
                                            > The perpendicular from Ma to CC' intersects
                                            > the line MbMc at Ac.
                                            >
                                            > Similarly Bc,Ba; Ca,Cb
                                            >
                                            > The Euler Lines of MaAbAc, MbBcBa, McCaCb are concurrent
                                            > (at some point on IG ???)

                                            In general: For MaMbMc = a triangle homothetic to
                                            ABC with homothety (G,t).

                                            APH
                                          • jan.vonk
                                            Dear Antreas, ... Yes, you are right! In fact, we can say more: The three triangles share the same circumcenter. It s the Spieker point Sp of ABC. ... Here s a
                                            Message 21 of 28 , May 1, 2009
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                                              Dear Antreas,

                                              > Let ABC be a triangle, A'B'C' the cevian
                                              > triangle of I and MaMbMc the cevian triangle of G.
                                              >
                                              > The perpendicular from Ma to BB' intersects
                                              > the line MbMc at Ab.
                                              >
                                              > The perpendicular from Ma to CC' intersects
                                              > the line MbMc at Ac.
                                              >
                                              > Similarly Bc,Ba; Ca,Cb
                                              >
                                              > The Euler Lines of MaAbAc, MbBcBa, McCaCb are concurrent
                                              > (at some point on IG ???)

                                              Yes, you are right! In fact, we can say more: The three triangles
                                              share the same circumcenter. It's the Spieker point Sp of ABC.

                                              -----------
                                              Here's a proof:
                                              Note that MaAbAc is taken to a familiar triangle by a homothecy g
                                              with factor -1 and center the midpoint of MbMc. It's the triangle of
                                              the initial problem, which has his circumcenter at the midpoint of
                                              AI. Now consider the homothecy h with center A and factor 2, so we
                                              know that the circumcenter of MaAbAc is mapped to I by gh.

                                              On the other hand, gh is a homothecy with center G and factor -2, so
                                              it maps the incenter of MaMbMc to the incenter of ABC, which is I.
                                              This shows that Sp, the incenter of MaMbMc coincides with the
                                              circumcenter of MaAbAc.
                                              ------------

                                              Unfortunately, I think this proof is worthless for your general
                                              theorem, which seems to be true! Amazing...

                                              All the best,
                                              Jan
                                            • xpolakis
                                              [APH] ... Let La, Lb, Lc be the Reflections of the Euler Lines of AAbAc, BBcBa, CCaCb in the angle bisectors of A, B, C, resp. and Ma, Mb, Mc the reflections
                                              Message 22 of 28 , Jun 27, 2009
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                                                [APH]
                                                > Let ABC be a triangle and A'B'C' the cevian
                                                > triangle of I.
                                                >
                                                > Ab := the Orthogonal projection of A on BB'
                                                > Ac := the Orthogonal projection of A on CC'
                                                >
                                                > Similarly Bc,Ba; Ca,Cb
                                                >
                                                > The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent at F,
                                                > the Feuerbach point (??)

                                                Let La, Lb, Lc be the Reflections of the Euler Lines of
                                                AAbAc, BBcBa, CCaCb in the angle bisectors of A, B, C, resp.
                                                and Ma, Mb, Mc the reflections of La, Lb, Lc in the
                                                corresponding angle bisectors of the Excentral triangle
                                                of ABC.

                                                I think that:

                                                1. La, Lb, Lc

                                                2. Ma, Mb, Mc

                                                are concurrent.

                                                Points?

                                                Antreas
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