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New circle related to a triangle

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  • amyakishev
    Dear friends! Last summer I ve found one new (as I hope) circle related to a triangle. The construction is as following: Let ABC be an arbitrary triangle and
    Message 1 of 14 , Nov 9 12:01 AM
      Dear friends!
      Last summer I've found one new (as I hope) circle related to a triangle.
      The construction is as following:
      Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle of ABC.
      Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine point K' of A'B'C'.
      By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots are with respect to ABC).
      Let Ab is the point of intersection of the parallel line to AB (trough A1) with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB - sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
      So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
      I can proof that fact by rather easy barycentric calculations but have failed to get synthetical (pure geometrical)
      proof.
      May be somebody be more successful in search of such proof?
      Best regards,
      Alexei Myakishev
    • Antreas Hatzipolakis
      ... Dear Alexei Is the center of the circle a known point? (ie already listed in ETC) Coordinates? APH [Non-text portions of this message have been removed]
      Message 2 of 14 , Nov 9 12:24 AM
        On Wed, Nov 9, 2011 at 10:01 AM, amyakishev <amyakishev@...> wrote:

        > **
        >
        >
        > Dear friends!
        > Last summer I've found one new (as I hope) circle related to a triangle.
        > The construction is as following:
        > Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle
        > of ABC.
        > Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine
        > point K' of A'B'C'.
        > By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots
        > are with respect to ABC).
        > Let Ab is the point of intersection of the parallel line to AB (trough A1)
        > with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB -
        > sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
        > So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
        > I can proof that fact by rather easy barycentric calculations but have
        > failed to get synthetical (pure geometrical)
        > proof.
        > May be somebody be more successful in search of such proof?
        > Best regards,
        > Alexei Myakishev
        >
        >
        >

        Dear Alexei

        Is the center of the circle a known point? (ie already listed in ETC)
        Coordinates?

        APH


        [Non-text portions of this message have been removed]
      • amyakishev
        Dear Antreas! The coordinates of the center are rather ugly, smth. like that (if Mathematica 5.1 tells the truth^)):
        Message 3 of 14 , Nov 9 1:06 AM
          Dear Antreas!
          The coordinates of the center are rather ugly, smth. like that (if Mathematica 5.1 tells the truth^)):
          8a^4b^2c^2(-a^2(b^2+c^2)(b^4-5b^2c^2+c^4)+a^4(b^4-4b^2c^2+c^4)+b^2c^2(b^4-4b^2c^2+c^4)):...:...
          This point is absent in ETC (was checked by magic triangle 6,9,13:))
          I've posted about it to Kimberling in ceptember but still have no reply:)
          Best regards,
          Alex

          --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
          >
          > On Wed, Nov 9, 2011 at 10:01 AM, amyakishev <amyakishev@...> wrote:
          >
          > > **
          > >
          > >
          > > Dear friends!
          > > Last summer I've found one new (as I hope) circle related to a triangle.
          > > The construction is as following:
          > > Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle
          > > of ABC.
          > > Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine
          > > point K' of A'B'C'.
          > > By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots
          > > are with respect to ABC).
          > > Let Ab is the point of intersection of the parallel line to AB (trough A1)
          > > with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB -
          > > sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
          > > So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
          > > I can proof that fact by rather easy barycentric calculations but have
          > > failed to get synthetical (pure geometrical)
          > > proof.
          > > May be somebody be more successful in search of such proof?
          > > Best regards,
          > > Alexei Myakishev
          > >
          > >
          > >
          >
          > Dear Alexei
          >
          > Is the center of the circle a known point? (ie already listed in ETC)
          > Coordinates?
          >
          > APH
          >
          >
          > [Non-text portions of this message have been removed]
          >
        • jpehrmfr
          Dear Alexei ... I don t know if the following can help you but If M,N are the isotomic conjugates (wrt ABC) of the Brocard points of A B C , it seems that your
          Message 4 of 14 , Nov 9 2:08 AM
            Dear Alexei

            > Last summer I've found one new (as I hope) circle related to a triangle.
            > The construction is as following:
            > Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle of ABC.
            > Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine point K' of A'B'C'.
            > By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots are with respect to ABC).
            > Let Ab is the point of intersection of the parallel line to AB (trough A1) with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB - sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
            > So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
            > I can proof that fact by rather easy barycentric calculations but have failed to get synthetical (pure geometrical)
            > proof.
            > May be somebody be more successful in search of such proof?

            I don't know if the following can help you but
            If M,N are the isotomic conjugates (wrt ABC) of the Brocard points of A'B'C', it seems that your circle is the common cevian circle (wrt ABC) of M and N.
            Kind regards. Jean-Pierre
          • amyakishev
            Dear Jean-Pierre! Thank you for that wonderful observation. The next question is: Is this fact about isotomic Brocard s points and the corresponding circle
            Message 5 of 14 , Nov 9 3:00 AM
              Dear Jean-Pierre!
              Thank you for that wonderful observation.
              The next question is:
              Is this fact about isotomic Brocard's points and the corresponding circle well known?
              Or you have established it righn now?:)
              Best regards,
              Alex

              --- In Hyacinthos@yahoogroups.com, "jpehrmfr" <jean-pierre.ehrmann@...> wrote:
              >
              > Dear Alexei
              >
              > > Last summer I've found one new (as I hope) circle related to a triangle.
              > > The construction is as following:
              > > Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle of ABC.
              > > Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine point K' of A'B'C'.
              > > By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots are with respect to ABC).
              > > Let Ab is the point of intersection of the parallel line to AB (trough A1) with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB - sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
              > > So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
              > > I can proof that fact by rather easy barycentric calculations but have failed to get synthetical (pure geometrical)
              > > proof.
              > > May be somebody be more successful in search of such proof?
              >
              > I don't know if the following can help you but
              > If M,N are the isotomic conjugates (wrt ABC) of the Brocard points of A'B'C', it seems that your circle is the common cevian circle (wrt ABC) of M and N.
              > Kind regards. Jean-Pierre
              >
            • jpehrmfr
              Dear Alexei ... In my case, I didn t know this result before your message. In fact, if you put u=1/bb+1/cc-1/aa, v,w cyclically, we have
              Message 6 of 14 , Nov 9 6:21 AM
                Dear Alexei

                > Thank you for that wonderful observation.
                > The next question is:
                > Is this fact about isotomic Brocard's points and the corresponding circle well known?
                > Or you have established it righn now?:)

                In my case, I didn't know this result before your message.
                In fact, if you put u=1/bb+1/cc-1/aa, v,w cyclically, we have
                Ba=(0,u,w);Cb=(u,0,v);Ac=(w,v,0)
                Ca=(0,v,u);Ab=(v,0,w);Bc=(u,w,0)
                Thus BaCbAc is the cevian triangle of (1/v,1/w,1/u) and CaAbBc is the cevian triangle of (1/w,1/u,1/v)
                As (v,w,u) and (w,u,v) are the anticomplements of the Brocard points of ABC, my assertion follows.
                Now AAb.ACb = AAc.ABc = aabbccvw/4 and two similar identities shows that the six points are concyclic
                Kind regards. Jean-Pierre
              • yiuatfauedu
                Dear Alexei, Bernard has found this circle before in Hyacinthos 5710: [BG, 5710]: a new (?) circle Dear friends, P a point, PaPbPc its cevian triangle. the
                Message 7 of 14 , Nov 9 6:36 AM
                  Dear Alexei,

                  Bernard has found this circle before in Hyacinthos 5710:

                  [BG, 5710]: a new (?) circle

                  Dear friends,

                  P a point, PaPbPc its cevian triangle.
                  the parallel at Pa to AC meets AB at Pac.
                  the parallel at Pa to AB meets AC at Pab.
                  Similarly define Pba, Pbc, Pcb, Pca.

                  Those six points always lie on the same conic.

                  This conic is a circle iff P=X194 ie the perspector of the cevian triangle
                  of K=X6 and the antimedial triangle.

                  Its center is unknown.

                  Maybe there are some other interesting properties.

                  =============
                  Best regards
                  Sincerely
                  Paul

                  --- In Hyacinthos@yahoogroups.com, "amyakishev" <amyakishev@...> wrote:
                  >
                  > Dear friends!
                  > Last summer I've found one new (as I hope) circle related to a triangle.
                  > The construction is as following:
                  > Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle of ABC.
                  > Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine point K' of A'B'C'.
                  > By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots are with respect to ABC).
                  > Let Ab is the point of intersection of the parallel line to AB (trough A1) with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB - sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
                  > So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
                  > I can proof that fact by rather easy barycentric calculations but have failed to get synthetical (pure geometrical)
                  > proof.
                  > May be somebody be more successful in search of such proof?
                  > Best regards,
                  > Alexei Myakishev
                  >
                • Bernard Gibert
                  Dear Paul and friends, I had forgotten all this which also appears in §2.3 of http://bernard.gibert.pagesperso-orange.fr/files/tuckercubics.html Thank you and
                  Message 8 of 14 , Nov 9 6:45 AM
                    Dear Paul and friends,

                    I had forgotten all this which also appears in §2.3 of

                    http://bernard.gibert.pagesperso-orange.fr/files/tuckercubics.html

                    Thank you and best regards

                    Bernard



                    > [PY] Bernard has found this circle before in Hyacinthos 5710:



                    [Non-text portions of this message have been removed]
                  • jpehrmfr
                    Dear Paul, Bernard, Alexei and other Hyacinthists a good advice : before posting a message, we must read carefully the complete works of Bernard. Friendly.
                    Message 9 of 14 , Nov 9 9:14 AM
                      Dear Paul, Bernard, Alexei and other Hyacinthists
                      a good advice : before posting a message, we must read carefully the complete works of Bernard.
                      Friendly. Jean-Pierre

                      > Bernard has found this circle before in Hyacinthos 5710:
                      >
                      > [BG, 5710]: a new (?) circle
                      >
                      > Dear friends,
                      >
                      > P a point, PaPbPc its cevian triangle.
                      > the parallel at Pa to AC meets AB at Pac.
                      > the parallel at Pa to AB meets AC at Pab.
                      > Similarly define Pba, Pbc, Pcb, Pca.
                      >
                      > Those six points always lie on the same conic.
                      >
                      > This conic is a circle iff P=X194 ie the perspector of the cevian triangle
                      > of K=X6 and the antimedial triangle.
                      >
                      > Its center is unknown.
                      >
                      > Maybe there are some other interesting properties.
                    • Paul Yiu
                      Sound advice from Pope J.-P. Best regards Sincerely Paul ... From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] On Behalf Of jpehrmfr Sent:
                      Message 10 of 14 , Nov 9 9:53 AM
                        Sound advice from Pope J.-P.

                        Best regards
                        Sincerely
                        Paul

                        -----Original Message-----
                        From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] On Behalf Of jpehrmfr
                        Sent: Wednesday, November 09, 2011 12:15 PM
                        To: Hyacinthos@yahoogroups.com
                        Subject: [EMHL] Re: New circle related to a triangle



                        Dear Paul, Bernard, Alexei and other Hyacinthists a good advice : before posting a message, we must read carefully the complete works of Bernard.
                        Friendly. Jean-Pierre

                        > Bernard has found this circle before in Hyacinthos 5710:
                        >
                        > [BG, 5710]: a new (?) circle
                        >
                        > Dear friends,
                        >
                        > P a point, PaPbPc its cevian triangle.
                        > the parallel at Pa to AC meets AB at Pac.
                        > the parallel at Pa to AB meets AC at Pab.
                        > Similarly define Pba, Pbc, Pcb, Pca.
                        >
                        > Those six points always lie on the same conic.
                        >
                        > This conic is a circle iff P=X194 ie the perspector of the cevian
                        > triangle of K=X6 and the antimedial triangle.
                        >
                        > Its center is unknown.
                        >
                        > Maybe there are some other interesting properties.




                        ------------------------------------

                        Yahoo! Groups Links
                      • amyakishev
                        Dear Paul, thank you. It s really so hard tj invent something new. But the question about geometrical proof is still remain? Best regards, Sincerely yours,
                        Message 11 of 14 , Nov 9 12:28 PM
                          Dear Paul,
                          thank you.
                          It's really so hard tj invent something new.
                          But the question about geometrical proof is still remain?
                          Best regards,
                          Sincerely yours,
                          Alex

                          --- In Hyacinthos@yahoogroups.com, "yiuatfauedu" <yiu@...> wrote:
                          >
                          > Dear Alexei,
                          >
                          > Bernard has found this circle before in Hyacinthos 5710:
                          >
                          > [BG, 5710]: a new (?) circle
                          >
                          > Dear friends,
                          >
                          > P a point, PaPbPc its cevian triangle.
                          > the parallel at Pa to AC meets AB at Pac.
                          > the parallel at Pa to AB meets AC at Pab.
                          > Similarly define Pba, Pbc, Pcb, Pca.
                          >
                          > Those six points always lie on the same conic.
                          >
                          > This conic is a circle iff P=X194 ie the perspector of the cevian triangle
                          > of K=X6 and the antimedial triangle.
                          >
                          > Its center is unknown.
                          >
                          > Maybe there are some other interesting properties.
                          >
                          > =============
                          > Best regards
                          > Sincerely
                          > Paul
                          >
                          > --- In Hyacinthos@yahoogroups.com, "amyakishev" <amyakishev@> wrote:
                          > >
                          > > Dear friends!
                          > > Last summer I've found one new (as I hope) circle related to a triangle.
                          > > The construction is as following:
                          > > Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle of ABC.
                          > > Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine point K' of A'B'C'.
                          > > By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots are with respect to ABC).
                          > > Let Ab is the point of intersection of the parallel line to AB (trough A1) with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB - sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
                          > > So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
                          > > I can proof that fact by rather easy barycentric calculations but have failed to get synthetical (pure geometrical)
                          > > proof.
                          > > May be somebody be more successful in search of such proof?
                          > > Best regards,
                          > > Alexei Myakishev
                          > >
                          >
                        • amyakishev
                          Dear Bernard, sorry for the new circle. Newertheless, it would be interesting to know (for me) how you have proved it. Best regards, Alex
                          Message 12 of 14 , Nov 9 12:58 PM
                            Dear Bernard, sorry for the "new" circle.
                            Newertheless, it would be interesting to know (for me) how you have proved it.
                            Best regards,
                            Alex

                            --- In Hyacinthos@yahoogroups.com, Bernard Gibert <bg42@...> wrote:
                            >
                            > Dear Paul and friends,
                            >
                            > I had forgotten all this which also appears in §2.3 of
                            >
                            > http://bernard.gibert.pagesperso-orange.fr/files/tuckercubics.html
                            >
                            > Thank you and best regards
                            >
                            > Bernard
                            >
                            >
                            >
                            > > [PY] Bernard has found this circle before in Hyacinthos 5710:
                            >
                            >
                            >
                            > [Non-text portions of this message have been removed]
                            >
                          • Bernard Gibert
                            Dear Alex, ... That s OK. I m pretty sure one can find it in older literature ! ... through barycentric calculations. Best regards Bernard [Non-text portions
                            Message 13 of 14 , Nov 9 10:37 PM
                              Dear Alex,

                              > Dear Bernard, sorry for the "new" circle.

                              That's OK. I'm pretty sure one can find it in older literature !

                              > Newertheless, it would be interesting to know (for me) how you have proved it.

                              through barycentric calculations.

                              Best regards

                              Bernard



                              [Non-text portions of this message have been removed]
                            • amyakishev
                              Dear Bernard, in fact, nominal by barycentric, I ve get it:) Now I m prepareing the article about this circle. Of course, you would be menthion in it.:) But
                              Message 14 of 14 , Nov 10 12:41 AM
                                Dear Bernard,
                                in fact, nominal by barycentric, I've get it:)
                                Now I'm prepareing the article about this circle.
                                Of course, you would be menthion in it.:)
                                But can you help me with "older literature", i.e. some sourses?:)
                                Best regards,
                                Alex

                                --- In Hyacinthos@yahoogroups.com, Bernard Gibert <bg42@...> wrote:
                                >
                                > Dear Alex,
                                >
                                > > Dear Bernard, sorry for the "new" circle.
                                >
                                > That's OK. I'm pretty sure one can find it in older literature !
                                >
                                > > Newertheless, it would be interesting to know (for me) how you have proved it.
                                >
                                > through barycentric calculations.
                                >
                                > Best regards
                                >
                                > Bernard
                                >
                                >
                                >
                                > [Non-text portions of this message have been removed]
                                >
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