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Re: A question regarding table 14 of Bernard Gibert

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  • 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 1 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 2 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 3 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 4 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 5 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 6 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 7 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 8 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 9 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 10 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 11 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 12 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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