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Re: [EMHL] A question regarding table 14 of Bernard Gibert

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  • Hauke Reddmann
    ... Addendum: A numeric solution for the Kimberling triangle suggest the Fermat point #13 (can t be) and alternatively y=1.9094098548 (not in list, but it
    Message 1 of 16 , Dec 8, 2008
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      --- In Hyacinthos@yahoogroups.com, "Hauke Reddmann" <fc3a501@...> wrote:
      >
      > --- In Hyacinthos@yahoogroups.com, Bernard Gibert <bg42@> wrote:
      > >
      > >
      > > Could you please be more explicit ?
      > >
      > OK. Let ABC be a triangle, and A'B'C' an equilateral
      > triangle circumscribed around ABC, like in your
      > table 14. Now demand that AA',BB' and CC' concur in O.
      > O lies on your curve (it's also possible that a finite
      > number of O exist). But where, i.e., is the point
      > already on the Kimberlist?
      > As I said, the same construction with the *in*scribed
      > equilateral triangle gives the equilateral cevian center.
      >
      Addendum: A numeric solution for the Kimberling triangle
      suggest the Fermat point #13 (can't be) and alternatively
      y=1.9094098548 (not in list, but it somehow rings a bell).
      Since all other solutions are spurious (0/0, oo, corners,
      complex values) it should be possible to compute the unique trilinears.

      Hauke
    • Jeffrey Brooks
      Dear Hauke, What is it exactly you want? Do you demand I make a triangle circumscribe or inscribe a given triangle? If so, I have the similar problem.
      Message 2 of 16 , Dec 8, 2008
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        Dear Hauke,

        What is it exactly you want?
        Do you demand I make a triangle circumscribe or inscribe a given
        triangle? If so, I have the similar problem. Equilateral triangles are
        especially nice ones to begin with.

        Sincerely, Jeff
      • Jeffrey Brooks
        Dear Nikos, In message #16970 you gave a proof how to construct a triangle A B C inscrided in a given triangle ABC. But now, how to construct A B C
        Message 3 of 16 , Dec 8, 2008
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          Dear Nikos,

          In message #16970 you gave a proof how to construct a triangle A'B'C'
          inscrided in a given triangle ABC. But now, how to construct A'B'C'
          circumscribed to ABC?

          Sincerely, Jeff
        • Jeffrey Brooks
          Dear Nikos and friends, Sorry, I meant inscribed triangle A B C . Also, I meant similar triangles A B C . Sincerely, Jeff
          Message 4 of 16 , Dec 8, 2008
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            Dear Nikos and friends,

            Sorry, I meant 'inscribed' triangle A'B'C'. Also, I meant similar
            triangles A'B'C'.
            Sincerely, Jeff


            --- In Hyacinthos@yahoogroups.com, "Jeffrey Brooks" <cu1101@...> wrote:
            >
            > Dear Nikos,
            >
            > In message #16970 you gave a proof how to construct a triangle A'B'C'
            > inscrided in a given triangle ABC. But now, how to construct A'B'C'
            > circumscribed to ABC?
            >
            > Sincerely, Jeff
            >
          • Hauke Reddmann
            ... are ... Circumscribe an equilateral triangle A B C around ABC such that ABC is a Cevian triangle of A B C with respect to some point O (which coordinates
            Message 5 of 16 , Dec 9, 2008
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              --- In Hyacinthos@yahoogroups.com, "Jeffrey Brooks" <cu1101@...> wrote:
              >
              > Dear Hauke,
              >
              > What is it exactly you want?
              > Do you demand I make a triangle circumscribe or inscribe a given
              > triangle? If so, I have the similar problem. Equilateral triangles
              are
              > especially nice ones to begin with.
              >
              Circumscribe an equilateral triangle A'B'C' around ABC
              such that ABC is a Cevian triangle of A'B'C' with
              respect to some point O (which coordinates I desperately
              try to compute for the Kimberling triangle but for each
              of my NSolve[] tries I get different results. :-O )

              Hauke
            • Nikolaos Dergiades
              Dear Jeff and Hauke, In a triangle ABC we can inscribe infinitely many triangles A1B1C1 that are similar to a given triangle A B C . The method is the
              Message 6 of 16 , Dec 9, 2008
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                Dear Jeff and Hauke,
                In a triangle ABC we can inscribe infinitely
                many triangles A1B1C1 that are similar to
                a given triangle A'B'C'. The method is the following:
                In message #16970 we have seen how to construct the
                unique point P such that the pedal triangle XYZ of P
                is similar to A'B'C'.
                The line PX rotated by an angle w meets BC at the point A1.
                The line PY rotated by an angle w meets CA at the point B1.
                The line PZ rotated by an angle w meets AB at the point C1.
                The triangle A1B1C1 is similar to A'B'C' and w is arbitrary.
                If A1B1C1 is a cevian then tan(w) must be the root of
                a 3-degree equation. Hence there are 3 such points.

                If the triangle A'B'C' is equilateral then P is not unique.
                There are two such points X(15), X(16).

                If Q is the isogonal conjugate of P then the antipedal triangle
                X'Y'Z' of Q wrt ABC is a triangle circumscribed to ABC and
                similar to A'B'C'. As previously the rotated QA meets Y'Z' at
                a point A2. Similarly we find B2, C2 and the triangle A2B2C2
                is a triangle similar to ABC and X'Y'Z' is circumscribed to A2B2C2.
                If A'B'C' is an equilateral triangle the points Q are the points
                X(13) and X(14).
                We can find that the triangles A2B2C2 and X'Y'Z' are
                perspective iff tan(w) is the root of a 3-degree equation.
                Hence there are 6 points such that Hauke wants to find
                and the barycentrics must be complicated.

                Best regards
                Nikos Dergiades


                > Dear Nikos and friends,
                >
                > Sorry, I meant 'inscribed' triangle
                > A'B'C'. Also, I meant similar
                > triangles A'B'C'.
                > Sincerely, Jeff

                > > Dear Nikos,
                > >
                > > In message #16970 you gave a proof how to construct a
                > triangle A'B'C'
                > > inscrided in a given triangle ABC. But now, how to
                > construct A'B'C'
                > > circumscribed to ABC?
                > >
                > > Sincerely, Jeff


                > Circumscribe an equilateral triangle A'B'C'
                > around ABC
                > such that ABC is a Cevian triangle of A'B'C'
                > with
                > respect to some point O (which coordinates I desperately
                > try to compute for the Kimberling triangle but for each
                > of my NSolve[] tries I get different results. :-O )
                >
                > Hauke





                >
                > Yahoo! Groups Links
                >
                >
                >


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              • van Tienhoven
                Dear Hauke, I had the same problem with constructing the equilateral circumscribed perspective triangle. Thanks to Jean Pierre Ehrmann who gave me a wonderful
                Message 7 of 16 , Dec 10, 2008
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                  Dear Hauke,

                  I had the same problem with constructing the equilateral circumscribed perspective triangle.
                  Thanks to Jean Pierre Ehrmann who gave me a wonderful solution I was able to construct the center at last.
                  See Hyancinthos message 16394.
                  With this message I calculated the ETC-coordinate being 1,9094098594 (does it match any of your trials?).
                  This coordinate does not match any ETC-center.
                  For as far as I can see there are hardly relations with any other center.
                  Except that this new point is on the line X16.X370.
                  It is also situated on the conic ABC-X14-X370.
                  This is good news for center X370 because sofar X370 did not have any relation with any other ETC-center.

                  This relationship with the point we are talking about is not strange,
                  because:X370 = the Equilateral Cevian Triangle Point
                  The center we are talking about is the Equilateral AntiCevian Triangle Point.

                  Best regards,

                  Chris van Tienhoven



                  ----- Original Message -----
                  From: Hauke Reddmann
                  To: Hyacinthos@yahoogroups.com
                  Sent: Tuesday, December 09, 2008 4:47 PM
                  Subject: [SPAM][EMHL] Re: A question regarding table 14 of Bernard Gibert


                  --- In Hyacinthos@yahoogroups.com, "Jeffrey Brooks" <cu1101@...> wrote:
                  >
                  > Dear Hauke,
                  >
                  > What is it exactly you want?
                  > Do you demand I make a triangle circumscribe or inscribe a given
                  > triangle? If so, I have the similar problem. Equilateral triangles
                  are
                  > especially nice ones to begin with.
                  >
                  Circumscribe an equilateral triangle A'B'C' around ABC
                  such that ABC is a Cevian triangle of A'B'C' with
                  respect to some point O (which coordinates I desperately
                  try to compute for the Kimberling triangle but for each
                  of my NSolve[] tries I get different results. :-O )

                  Hauke





                  [Non-text portions of this message have been removed]
                • jpehrmfr
                  Dear Hyacinthists two little remarks about the equilateral cevian or precevian triangles : If the cevian (or the precevian triangle) of P is equilateral with
                  Message 8 of 16 , Dec 10, 2008
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                    Dear Hyacinthists
                    two little remarks about the equilateral cevian or precevian triangles :
                    If the cevian (or the precevian triangle) of P is equilateral with
                    center W, then the line PW goes through the circumcenter.
                    There allways exist 3 equilateral cevian (or precevian) triangles with
                    the opposite orientation of ABC; there exists 1,2 or 3 cevian (or
                    precevian) triangles with the orientation of ABC. The centers of these
                    triangles (at least 8, at most 12) lie on the Napoleon cubic.
                    Best regards from France. Jean-Pierre
                  • jpehrmfr
                    Dear Hyacinthists I wrote ... I m sorry. This is true only for the equilateral cevian triangles. With my apologizes. Jean-Pierre
                    Message 9 of 16 , Dec 10, 2008
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                      Dear Hyacinthists
                      I wrote
                      > two little remarks about the equilateral cevian or precevian
                      triangles :
                      > If the cevian (or the precevian triangle) of P is equilateral with
                      > center W, then the line PW goes through the circumcenter.

                      I'm sorry. This is true only for the equilateral cevian triangles.
                      With my apologizes. Jean-Pierre
                    • Jeffrey Brooks
                      Dear Jean-Pierre and Nikos, [ND] Hence there are 6 points such that Hauke wants to find [JPE] There allways exist 3 equilateral cevian (or precevian) triangles
                      Message 10 of 16 , Dec 11, 2008
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                        Dear Jean-Pierre and Nikos,

                        [ND]
                        Hence there are 6 points such that Hauke wants to find

                        [JPE]
                        There allways exist 3 equilateral cevian (or precevian) triangles with
                        the opposite orientation of ABC; there exists 1,2 or 3 cevian (or
                        precevian) triangles with the orientation of ABC. The centers of these
                        triangles (at least 8, at most 12) lie on the Napoleon cubic.

                        Okay, is it 6, 8 or 12 and what determines this number?

                        Sincerely, Jeff
                      • Hauke Reddmann
                        ... (does it match any of your trials?). Yes. That s the one. Thread closed :-) Hauke
                        Message 11 of 16 , Dec 11, 2008
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                          --- In Hyacinthos@yahoogroups.com, "van Tienhoven" <van10hoven@...> wrote:
                          >
                          > Dear Hauke,
                          >
                          > See Hyancinthos message 16394.
                          > With this message I calculated the ETC-coordinate being 1,9094098594
                          (does it match any of your trials?).

                          Yes. That's the one. Thread closed :-)

                          Hauke
                        • Nikolaos Dergiades
                          Dear Jeff, when I spoke, about 6 I was speaking only for the inscribed or the circumscribed and this number is the maximum possible. Jean Pierre spoke about
                          Message 12 of 16 , Dec 11, 2008
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                            Dear Jeff,
                            when I spoke, about 6 I was speaking only
                            for the inscribed or the circumscribed
                            and this number is the maximum possible.
                            Jean Pierre spoke about all and thats why
                            he gave the number 6+6=12.
                            More precisely he gave the number 8 as
                            the minimum.
                            For example if we want the cevian inscribed
                            triangles that are similar to ABC,
                            the maximum number is 3 because there is
                            a unique point with pedal triangle directly similar
                            to ABC that is the circumcenter O. If we rotate
                            around O the perpendiculars from O to the sides
                            of ABC by an angle w we get as intersections
                            triangles A'B'C' similar to ABC and these
                            triangles are cevian triangles iff d=tan(w)
                            is the zero of a 3-degree equation.
                            This equation has a real root d = 0.
                            (because O lies on the Darboux cubic and its
                            pedal triangle is also a cevian triangle)
                            The other 2 roots are real only when
                            the triangle ABC has an obtuse angle.
                            If I remember well
                            dd = -S_A.S_B.S_C/(S_A + S_B + S_C).

                            I don't know in the general case what
                            determines this number of cevian triangles.

                            Best regards
                            Nikos Dergiades


                            > Dear Jean-Pierre and Nikos,
                            >
                            > [ND]
                            > Hence there are 6 points such that Hauke wants to find
                            >
                            > [JPE]
                            > There allways exist 3 equilateral cevian (or precevian)
                            > triangles with
                            > the opposite orientation of ABC; there exists 1,2 or 3
                            > cevian (or
                            > precevian) triangles with the orientation of ABC. The
                            > centers of these
                            > triangles (at least 8, at most 12) lie on the Napoleon
                            > cubic.
                            >
                            > Okay, is it 6, 8 or 12 and what determines this number?
                            >
                            > Sincerely, Jeff
                            >




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                          • Nikolaos Dergiades
                            Dear Jeff, I wrote ... Sorry. The correct equation is d.d.S_A.S_B.S_C + S.S.(S_A + S_B + S_C) = 0 Best regards Nikos Dergiades
                            Message 13 of 16 , Dec 11, 2008
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                              Dear Jeff,
                              I wrote

                              > If I remember well
                              > dd = -S_A.S_B.S_C/(S_A + S_B + S_C).

                              Sorry. The correct equation is

                              d.d.S_A.S_B.S_C + S.S.(S_A + S_B + S_C) = 0

                              Best regards
                              Nikos Dergiades




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