Loading ...
Sorry, an error occurred while loading the content.

Re: A Nagel-Thebault configuration

Expand Messages
  • armpist
    Dear friends The posts on Mathlinks http://www.mathlinks.ro/Forum/viewtopic.php?t=114669 http://www.mathlinks.ro/Forum/viewtopic.php?t=113381 are related to
    Message 1 of 20 , Jan 7, 2008
    • 0 Attachment
      Dear friends

      The posts on Mathlinks

      http://www.mathlinks.ro/Forum/viewtopic.php?t=114669
      http://www.mathlinks.ro/Forum/viewtopic.php?t=113381

      are related to the current problem and show a
      connection of this problem to the equal inradii
      in partial triangles set-up: Nagel cevian extended
      is the equal inradii cevian in the triangle on the
      same base with vertex in the intersection of Nagel
      cevian and circumcircle. Another interesting fact:
      incenter is one of the limiting points in the
      coaxal system of two Thebault circles.

      Cosmin, the problem with JSTOR search is that
      the fresh stuff is not in the database yet,
      there is a delay of up to several years depending
      on a publication.


      Regards,
      Armpist.
    • jpehrmfr
      Dear Armpist and Cosmin ... Using this remark, it becomes obvious to find the second circle (O ) touching the 6 Thebault circles : The three radical axis of
      Message 2 of 20 , Jan 8, 2008
      • 0 Attachment
        Dear Armpist and Cosmin

        > Another interesting fact:
        > incenter is one of the limiting points in the
        > coaxal system of two Thebault circles.

        Using this remark, it becomes obvious to find the second circle (O')
        touching the 6 Thebault circles :
        The three radical axis of the three pairs of Thebault circle concur at
        the Cosmin point T; so, the circle (T) centered at T and going through
        I is orthogonal to the six Thebault circles which are invariant in the
        inversion wrt (T). Hence, the inverse of (O) in (T) touches the 6
        circles too.
        This is much more clear than my previous explanation.
        It remains to find a synthetic proof of the fac that the inverse of P
        in (T) lies on (O)
        Thank you again
        Jean-Pierre
      • Son Hong Ta
        ... touching ... (note ... point ... 2r) - ... Dear Jean-Pierre, Cosmin and all, Your first result above was mentioned and proved by Nguyen Minh Ha in a
        Message 3 of 20 , Jan 8, 2008
        • 0 Attachment
          --- In Hyacinthos@yahoogroups.com, "jpehrmfr" <jean-
          pierre.ehrmann@...> wrote:
          >
          > Dear Hyacinthists
          > A'B'C' is the cevian triangle of the Nagel point.
          > Consider the six corresponding Thébault circles (the circles
          touching
          > A'A, A'B and the circumcircle internally, ....)
          > Then :
          > 1) The six Thébault circles are congruent to the incircle
          > 2) Their 6 centers lie on the circle with center O,radius R-r
          (note
          > that this circle goes through X_1145, reflection of the Feuerbach
          point
          > in the Spieker center)
          > 3) The six circles touch externally the circle with center (O,R-
          2r) -
          > this circle goes through the Nagel point -
          > I suspect that this is not new. Have you head of this?
          > I'll upload soon a sketch in Hyacinthos file.
          > Friendly. Jean-Pierre
          >
          > PS : I apologize for a possible duplication
          >
          Dear Jean-Pierre, Cosmin and all,

          Your first result above was mentioned and proved by Nguyen Minh Ha
          in a Vietnamese Mathematics magazine, which was pblished in 2006.
          This paper also contains a nice synthetic proof of Sawayama-Thebault
          theorem. You can see it at:
          http://toantuoitho.nxbgd.com.vn/toantuoitho2/report.asp?
          soph=42&id=11&true=true ,
          http://toantuoitho.nxbgd.com.vn/toantuoitho2/report.asp?
          soph=43&id=11&true=true . But unfortunately, it's in Vietnamese.

          Fiendly, Son Hong Ta.
        • Cosmin Pohoata
          Dear Son and Armpist, Thank you both for the references to the particular cases. Son, does Nguyen Min Ha mentiones anything about the general configuration in
          Message 4 of 20 , Jan 9, 2008
          • 0 Attachment
            Dear Son and Armpist,

            Thank you both for the references to the particular cases.
            Son, does Nguyen Min Ha mentiones anything about the general
            configuration in his paper or by our first result you meant only the
            congruence of the Thebault circles in the Nagel case?


            Thank you in advance,
            Best regards,
            Cosmin
          • Son Hong Ta
            ... the ... Dear Cosmin, He only mentioned about the congurence of the Thebault circles in the Nagel case. I ve read this paper again, and in this paper, he
            Message 5 of 20 , Jan 10, 2008
            • 0 Attachment
              --- In Hyacinthos@yahoogroups.com, "Cosmin Pohoata"
              <pohoata_cosmin2000@...> wrote:
              >
              > Dear Son and Armpist,
              >
              > Thank you both for the references to the particular cases.
              > Son, does Nguyen Minh Ha mentiones anything about the general
              > configuration in his paper or by our first result you meant only
              the
              > congruence of the Thebault circles in the Nagel case?
              >
              >
              > Thank you in advance,
              > Best regards,
              > Cosmin
              >

              Dear Cosmin,

              He only mentioned about the congurence of the Thebault circles in
              the Nagel case. I've read this paper again, and in this paper, he
              said the author of the congurence is Ha Vu Anh, a teacher from Vinh
              Phuc, Vietnam. Anyway, your results and proofs are very nice. Thanks
              for sharing.

              Best regards, Son Hong Ta.
            • Luís Lopes
              Dear Hyacinthists, Both these problems have easy algebraic solutions/constructions. One can easily find and from b+c=s ; bc=u^2 and .
              Message 6 of 20 , Jan 18, 2008
              • 0 Attachment
                Dear Hyacinthists, Both these problems have easy algebraic
                solutions/constructions. One can easily find <b> and <c> from >b+c=s ; bc=u^2> and <b-c=d ; bc=v^2>. But... I am looking for an elegant "pure" geometrical approach. Is there one? Best regards, Luis
                _________________________________________________________________
                Veja mapas e encontre as melhores rotas para fugir do trânsito com o Live Search Maps!
                http://www.livemaps.com.br/index.aspx?tr=true

                [Non-text portions of this message have been removed]
              • Paul Yiu
                Dear Luis, [LL]: Both these problems have easy algebraic solutions/constructions. One can easily find b and c from b+c=s ; bc=u^2 and b-c=d ; bc=v^2. But... I
                Message 7 of 20 , Jan 19, 2008
                • 0 Attachment
                  Dear Luis,

                  [LL]: Both these problems have easy algebraic solutions/constructions.
                  One can easily find b and c from b+c=s ; bc=u^2 and b-c=d ; bc=v^2.
                  But... I am looking for an elegant "pure" geometrical approach. Is
                  there one?

                  *** Here is one. The relation (b+c)^2 + (b-c)^2 = 4m_a^2 + a^2
                  shows that the two problems can be solved simultaneously.

                  Given a segment $BC$ of length a, midpoint D, and a perpendicular
                  segment A_0D = m_a, construct
                  (i) a semicircle center D through A_0,
                  (ii) the circle with diameter A_0B (with perpendicular diameter MN, M
                  and C on opposite sides of A_0B),
                  (iii) the circle, center M, through A_0 and B.

                  Let P be a point on (ii) with BP = b + c for P on the semicircle A_0MB
                  and BP=b-c for P on A_0NB.

                  Extend BP to intersect the circle (iii) at Q.
                  Construct the circle through Q with center B to intersect the
                  semicircle (i) at A.

                  Then ABC is the required triangle.

                  Best regards
                  Sincerely
                  Paul
                • Paul Yiu
                  Dear Luis, My apology for omission of a factor 1/2 in the length of BP in the construction given in the prior posting (Hyacinthos 16048). [LL]: Both these
                  Message 8 of 20 , Jan 19, 2008
                  • 0 Attachment
                    Dear Luis,

                    My apology for omission of a factor 1/2 in the length of BP in the
                    construction given in the prior posting (Hyacinthos 16048).

                    [LL]: Both these problems have easy algebraic solutions/constructions.
                    One can easily find b and c from b+c=s ; bc=u^2 and b-c=d ; bc=v^2.
                    But... I am looking for an elegant "pure" geometrical approach. Is
                    there one?

                    *** Here is one. The relation (b+c)^2 + (b-c)^2 = 4m_a^2 + a^2
                    shows that the two problems can be solved simultaneously.

                    Given a segment $BC$ of length a, midpoint D, and a perpendicular
                    segment A_0D = m_a, construct
                    (i) a semicircle center D through A_0,
                    (ii) the circle with diameter A_0B (with perpendicular diameter MN, M
                    and C on opposite sides of A_0B),
                    (iii) the circle, center M, through A_0 and B.

                    Let P be a point on (ii) with BP = (b + c)/2 for P on the semicircle
                    A_0MB and BP = (b-c)/2 for P on A_0NB.

                    Extend BP to intersect the circle (iii) at Q.
                    Construct the circle through Q with center B to intersect the
                    semicircle (i) at A.

                    Then ABC is the required triangle.

                    Best regards
                    Sincerely
                    Paul
                  Your message has been successfully submitted and would be delivered to recipients shortly.