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  • Antreas Hatzipolakis
    Let ABC be a triangle, L a line passing through H, intersecting BC,CA,AB at A ,B ,C , resp. Let La, Lb, Lc be the reflections of L in BC,CA,AB, resp.
    Message 1 of 22 , Jan 30, 2013
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      Let ABC be a triangle, L a line passing through H, intersecting
      BC,CA,AB at A',B',C', resp.

      Let La, Lb, Lc be the reflections of L in BC,CA,AB, resp. (concurrent
      on the circumcircle).

      Let Ab, Ac be the orth. projections of A' on Lb,Lc, resp. Similarly
      Bc,Ba and Ca,Cb.

      The circumcircles of A'AbAc, B'BcBa, C'CaCb are coaxial and their NPC
      centers are
      collinear. Which is this line ? Special case: L = Euler line.

      APH
    • Paul Yiu
      Dear Antreas, [APH] Let ABC be a triangle, L a line passing through H, intersecting BC,CA,AB at A ,B ,C , resp. Let La, Lb, Lc be the reflections of L in
      Message 2 of 22 , Jan 30, 2013
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        Dear Antreas,

        [APH] Let ABC be a triangle, L a line passing through H, intersecting
        BC,CA,AB at A',B',C', resp.

        Let La, Lb, Lc be the reflections of L in BC,CA,AB, resp. (concurrent
        on the circumcircle).

        Let Ab, Ac be the orth. projections of A' on Lb,Lc, resp. Similarly
        Bc,Ba and Ca,Cb.

        The circumcircles of A'AbAc, B'BcBa, C'CaCb are coaxial and their NPC
        centers are collinear. Which is this line ? Special case: L = Euler line.

        *** I think you mean circumcenters.
        Let P be the intersection of La, Lb, Lc on the circumcircle of ABC.
        These circumcenters are the midpoints of PA', PB', PC'.
        They lie on the Simson line of P.

        Best regards
        Sincerely
        Paul
      • Antreas
        Dear Paul The circumcircles are coaxial, but how about the NPCs ? Are their centers collinear for lines passing through H? APH ... [PY] I think you mean
        Message 3 of 22 , Jan 30, 2013
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          Dear Paul

          The circumcircles are coaxial, but how about the
          NPCs ? Are their centers collinear for lines passing through H?

          APH

          > [APH] Let ABC be a triangle, L a line passing through H, intersecting
          > BC,CA,AB at A',B',C', resp.
          >
          > Let La, Lb, Lc be the reflections of L in BC,CA,AB, resp. (concurrent
          > on the circumcircle).
          >
          > Let Ab, Ac be the orth. projections of A' on Lb,Lc, resp. Similarly
          > Bc,Ba and Ca,Cb.
          >
          > The circumcircles of A'AbAc, B'BcBa, C'CaCb are coaxial and their NPC
          > centers are collinear. Which is this line ? Special case: L = Euler line.
          >

          [PY] I think you mean circumcenters.
          > Let P be the intersection of La, Lb, Lc on the circumcircle of ABC.
          > These circumcenters are the midpoints of PA', PB', PC'.
          > They lie on the Simson line of P.
          >
          > Best regards
          > Sincerely
          > Paul
          >
        • Antreas Hatzipolakis
          A figure here: http://anthrakitis.blogspot.gr/2013/01/collinear-npc-centers.html ... [Non-text portions of this message have been removed]
          Message 4 of 22 , Jan 30, 2013
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            A figure here:

            http://anthrakitis.blogspot.gr/2013/01/collinear-npc-centers.html



            On Wed, Jan 30, 2013 at 9:41 PM, Antreas <anopolis72@...> wrote:

            > **
            >
            >
            > Dear Paul
            >
            > The circumcircles are coaxial, but how about the
            > NPCs ? Are their centers collinear for lines passing through H?
            >
            > APH
            >
            >
            > > [APH] Let ABC be a triangle, L a line passing through H, intersecting
            > > BC,CA,AB at A',B',C', resp.
            > >
            > > Let La, Lb, Lc be the reflections of L in BC,CA,AB, resp. (concurrent
            > > on the circumcircle).
            > >
            > > Let Ab, Ac be the orth. projections of A' on Lb,Lc, resp. Similarly
            > > Bc,Ba and Ca,Cb.
            > >
            > > The circumcircles of A'AbAc, B'BcBa, C'CaCb are coaxial and their NPC
            > > centers are collinear. Which is this line ? Special case: L = Euler line.
            > >
            >
            > [PY] I think you mean circumcenters.
            > > Let P be the intersection of La, Lb, Lc on the circumcircle of ABC.
            > > These circumcenters are the midpoints of PA', PB', PC'.
            > > They lie on the Simson line of P.
            > >
            > > Best regards
            > > Sincerely
            > > Paul
            > >
            >
            > __
            >


            [Non-text portions of this message have been removed]
          • Randy Hutson
            Dear Antreas, For L = Euler line, the NPC are indeed collinear, though I have not found the line.  The circumcenters lie on line X(30)X(113) (Simson line of
            Message 5 of 22 , Jan 30, 2013
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              Dear Antreas,

              For L = Euler line, the NPC are indeed collinear, though I have not found the line.  The circumcenters lie on line X(30)X(113) (Simson line of X(110)).

              For L = van Aubel line (HK), the NPCs are also collinear, though I have not found the line.  The circumcenters lie on line X(132)X^-1(1297) (Simson line of X(112)).

              For L = Brocard axis, though does not meet the condition of passing through H, the circumcenters are still collinear, though I have not found the line, but the NPCs are not.

              Randy





              >________________________________
              > From: Antreas <anopolis72@...>
              >To: Hyacinthos@yahoogroups.com
              >Sent: Wednesday, January 30, 2013 1:41 PM
              >Subject: Re: [EMHL] A line
              >
              >

              >Dear Paul
              >
              >The circumcircles are coaxial, but how about the
              >NPCs ? Are their centers collinear for lines passing through H?
              >
              >APH
              >
              >> [APH] Let ABC be a triangle, L a line passing through H, intersecting
              >> BC,CA,AB at A',B',C', resp.
              >>
              >> Let La, Lb, Lc be the reflections of L in BC,CA,AB, resp. (concurrent
              >> on the circumcircle).
              >>
              >> Let Ab, Ac be the orth. projections of A' on Lb,Lc, resp. Similarly
              >> Bc,Ba and Ca,Cb.
              >>
              >> The circumcircles of A'AbAc, B'BcBa, C'CaCb are coaxial and their NPC
              >> centers are collinear. Which is this line ? Special case: L = Euler line.
              >>
              >
              >[PY] I think you mean circumcenters.
              >> Let P be the intersection of La, Lb, Lc on the circumcircle of ABC.
              >> These circumcenters are the midpoints of PA', PB', PC'.
              >> They lie on the Simson line of P.
              >>
              >> Best regards
              >> Sincerely
              >> Paul
              >>
              >
              >
              >
              >
              >

              [Non-text portions of this message have been removed]
            • Antreas Hatzipolakis
              Dear Randy The case of the L = OK line is very interesting! So we have that for the lines L= OH,OK the circumcenters are collinear and the question is for
              Message 6 of 22 , Jan 30, 2013
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                Dear Randy

                The case of the L = OK line is very interesting!

                So we have that for the lines L= OH,OK the circumcenters are collinear and
                the question
                is for which other lines through O the circumcenters are collinear.

                APH


                On Wed, Jan 30, 2013 at 11:51 PM, Randy Hutson <rhutson2@...> wrote:

                > **
                >
                >
                > Dear Antreas,
                >
                > For L = Euler line, the NPC are indeed collinear, though I have not found
                > the line. The circumcenters lie on line X(30)X(113) (Simson line of
                > X(110)).
                >
                > For L = van Aubel line (HK), the NPCs are also collinear, though I have
                > not found the line. The circumcenters lie on line X(132)X^-1(1297) (Simson
                > line of X(112)).
                >
                > For L = Brocard axis, though does not meet the condition of passing
                > through H, the circumcenters are still collinear, though I have not found
                > the line, but the NPCs are not.
                >
                > Randy
                >
                > >________________________________
                > > From: Antreas anopolis72@...>
                > >To: Hyacinthos@yahoogroups.com
                > >Sent: Wednesday, January 30, 2013 1:41 PM
                > >Subject: Re: [EMHL] A line
                >
                > >
                > >
                > >
                > >Dear Paul
                > >
                > >The circumcircles are coaxial, but how about the
                > >NPCs ? Are their centers collinear for lines passing through H?
                > >
                > >APH
                > >
                > >> [APH] Let ABC be a triangle, L a line passing through H, intersecting
                > >> BC,CA,AB at A',B',C', resp.
                > >>
                > >> Let La, Lb, Lc be the reflections of L in BC,CA,AB, resp. (concurrent
                > >> on the circumcircle).
                > >>
                > >> Let Ab, Ac be the orth. projections of A' on Lb,Lc, resp. Similarly
                > >> Bc,Ba and Ca,Cb.
                > >>
                > >> The circumcircles of A'AbAc, B'BcBa, C'CaCb are coaxial and their NPC
                > >> centers are collinear. Which is this line ? Special case: L = Euler
                > line.
                > >>
                > >
                > >[PY] I think you mean circumcenters.
                > >> Let P be the intersection of La, Lb, Lc on the circumcircle of ABC.
                > >> These circumcenters are the midpoints of PA', PB', PC'.
                > >> They lie on the Simson line of P.
                > >>
                > >> Best regards
                > >> Sincerely
                > >> Paul
                >


                [Non-text portions of this message have been removed]
              • Paul Yiu
                Dear Antreas and Randy, The three circumcircles are always coaxial. If L is the trilinear polar of (u:v:w), the line containing the circumcenters is
                Message 7 of 22 , Jan 30, 2013
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                  Dear Antreas and Randy,

                  The three circumcircles are always coaxial. If L is the trilinear polar of (u:v:w),
                  the line containing the circumcenters is

                  (S_A(S_Bv+S_Cw)/(v-w)) x + ... + ... = 0,

                  the trilinear polar of

                  ( (v-w)/(S_A(S_Bv+S_Cw)) : ... : ...).

                  Best regards
                  Sincerely
                  Paul
                • Paul Yiu
                  Dear Antreas, [APH] Let ABC be a triangle, L a line passing through H, intersecting BC,CA,AB at A ,B ,C , resp. Let La, Lb, Lc be the reflections of L in
                  Message 8 of 22 , Jan 30, 2013
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                    Dear Antreas,

                    [APH] Let ABC be a triangle, L a line passing through H, intersecting
                    BC,CA,AB at A',B',C', resp.

                    Let La, Lb, Lc be the reflections of L in BC,CA,AB, resp. (concurrent
                    on the circumcircle).

                    Let Ab, Ac be the orth. projections of A' on Lb,Lc, resp. Similarly
                    Bc,Ba and Ca,Cb.

                    The circumcircles of A'AbAc, B'BcBa, C'CaCb are coaxial and their NPC
                    centers are collinear. Which is this line ? Special case: L = Euler line.

                    [PY] I think you mean circumcenters.
                    Let P be the intersection of La, Lb, Lc on the circumcircle of ABC.
                    These circumcenters are the midpoints of PA', PB', PC'.
                    They lie on the Simson line of P.

                    ***
                    My apology. Your statement that the nine-point centers are collinear is indeed true,
                    though the equation of the line containing them is quite complicated, even for the Euler line.

                    Best regards
                    Sincerely
                    Paul
                  • Antreas
                    Dear Paul Naturally one can ask which is the envelope of the lines of the centers of the NPCs as the line L moves around H. Possibly the envelope is a simple
                    Message 9 of 22 , Jan 30, 2013
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                      Dear Paul

                      Naturally one can ask which is the envelope of the lines of the centers of the NPCs as the line L moves around H.

                      Possibly the envelope is a simple curve but with a very complicated equation (by taking as reference triangle the ABC)

                      Greetings

                      Antreas


                      --- In Hyacinthos@yahoogroups.com, Paul Yiu wrote:
                      >
                      > Dear Antreas,
                      >
                      > [APH] Let ABC be a triangle, L a line passing through H, intersecting
                      > BC,CA,AB at A',B',C', resp.
                      >
                      > Let La, Lb, Lc be the reflections of L in BC,CA,AB, resp. (concurrent
                      > on the circumcircle).
                      >
                      > Let Ab, Ac be the orth. projections of A' on Lb,Lc, resp. Similarly
                      > Bc,Ba and Ca,Cb.
                      >
                      > The circumcircles of A'AbAc, B'BcBa, C'CaCb are coaxial and their NPC
                      > centers are collinear. Which is this line ? Special case: L = Euler line.
                      >
                      > [PY] I think you mean circumcenters.
                      > Let P be the intersection of La, Lb, Lc on the circumcircle of ABC.
                      > These circumcenters are the midpoints of PA', PB', PC'.
                      > They lie on the Simson line of P.
                      >
                      > ***
                      > My apology. Your statement that the nine-point centers are collinear is indeed true,
                      > though the equation of the line containing them is quite complicated, even for the Euler line.
                      >
                      > Best regards
                      > Sincerely
                      > Paul
                      >
                    • Paul Yiu
                      Dear Antreas, [APH]: Naturally one can ask which is the envelope of the lines of the centers of the NPCs as the line L moves around H. Possibly the envelope is
                      Message 10 of 22 , Jan 30, 2013
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                        Dear Antreas,

                        [APH]: Naturally one can ask which is the envelope of the lines of the centers of the NPCs as the line L moves around H. Possibly the envelope is a simple curve but with a very complicated equation (by taking as reference triangle the ABC).

                        *** A ``simple" case (with complicated equation) occurs if we fix the line L and consider the line through the E_t points of the three triangles. (By the E_t point I mean the point dividing the circumcenter and orthocenter in the ratio t:1-t). In this case the envelope is a conic.

                        Best regards
                        Sincerely
                        Paul
                      • Paul Yiu
                        Dear Antreas, [APH]: Naturally one can ask which is the envelope of the lines of the centers of the NPCs as the line L moves around H. Possibly the envelope is
                        Message 11 of 22 , Jan 30, 2013
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                          Dear Antreas,

                          [APH]: Naturally one can ask which is the envelope of the lines of the centers of the NPCs as the line L moves around H. Possibly the envelope is a simple curve but with a very complicated equation (by taking as reference triangle the ABC).

                          *** A ``simple" case (with complicated equation) occurs if we fix the line L

                          *** through the orthocenter

                          and consider the line through the E_t points of the three triangles. (By the E_t point I mean the point dividing the circumcenter and orthocenter in the ratio t:1-t). In this case the envelope is a conic.

                          Best regards
                          Sincerely
                          Paul
                        • Antreas
                          Dear Paul I think it is true for the centroids as well (ie they are collinear) And, in general, how about for any point on the Euler line??? That is, let P be
                          Message 12 of 22 , Jan 31, 2013
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                            Dear Paul

                            I think it is true for the centroids as well (ie they are collinear)

                            And, in general, how about for any point on the Euler line???

                            That is, let P be a point on the Euler line of ABC and Pa,Pb,Pc the
                            respective points on the Euler lines of the triangles A'AbAc,
                            B'BcBa, C'CaCb.
                            Conjecture: The points Pa,Pb,Pc are collinear.

                            APH

                            --- In Hyacinthos@yahoogroups.com, Paul Yiu wrote:
                            >
                            > Dear Antreas,
                            >
                            > [APH] Let ABC be a triangle, L a line passing through H, intersecting
                            > BC,CA,AB at A',B',C', resp.
                            >
                            > Let La, Lb, Lc be the reflections of L in BC,CA,AB, resp. (concurrent
                            > on the circumcircle).
                            >
                            > Let Ab, Ac be the orth. projections of A' on Lb,Lc, resp. Similarly
                            > Bc,Ba and Ca,Cb.
                            >
                            > The circumcircles of A'AbAc, B'BcBa, C'CaCb are coaxial and their NPC
                            > centers are collinear. Which is this line ? Special case: L = Euler line.
                            >
                            > [PY] I think you mean circumcenters.
                            > Let P be the intersection of La, Lb, Lc on the circumcircle of ABC.
                            > These circumcenters are the midpoints of PA', PB', PC'.
                            > They lie on the Simson line of P.
                            >
                            > ***
                            > My apology. Your statement that the nine-point centers are collinear is indeed true,
                            > though the equation of the line containing them is quite complicated, even for the Euler line.
                            >
                            > Best regards
                            > Sincerely
                            > Paul
                            >
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