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[EMHL] Re: X(5390)

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  • Chris Van Tienhoven
    Dear Randy, The outcome is in trigonometric functions of A/3, B/3, C/3. Just to give a rough idea here are the first 2 (out of 254) rules: (Sec[A/3] Sin[ A]
    Message 1 of 26 , Jan 4, 2013
      Dear Randy,

      The outcome is in trigonometric functions of A/3, B/3, C/3.
      Just to give a rough idea here are the first 2 (out of 254) rules:

      (Sec[A/3] Sin[
      A] (-(-b (3 c + 2 c Cos[(2 A)/3] + 2 a Cos[B/3] +
      4 Cos[A/3] (b + a Cos[C/3])) (-16 b^2 c^2 (SA + SB) SC Cos[A/
      3]^4 + 8 a Cos[A/
      3]^2 (-c SA (2 b^2 (SA + SB) + c^2 (SA + SC)) Cos[B/3] +
      b (2 c^2 SA^2 + b^2 (SA + SB)^2) Cos[C/3]) +
      b c (SA + SB) (b c SC - 2 a b SA Cos[B/3] +
      2 a c (SA + SC) Cos[C/3]) +
      2 Cos[A/3] (-b^3 c SB (SA + SB) +
      b c^3 (SB SC + SA (SB + 2 SC)) -
      2 a b (2 c^2 SA^2 + b^2 (SA + SB)^2) Cos[B/3] +
      2 a c SA (2 b^2 (SA + SB) + c^2 (SA + SC)) Cos[C/3]) +
      8 b c Cos[A/
      3]^3 (b^2 SB (SA + SB) - c^2 (SB SC + SA (SB + 2 SC)) +
      2 a (SA + SB) (-c (SA + SC) Cos[B/3] + b SA Cos[C/3]))) +

      Best regards,

      Chris van Tienhoven

      www.chrisvantienhoven.nl


      --- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
      >
      > Dear Chris,
      >
      > Were your coordinates in polynomial form?  If so, what degree?  I wonder if they could be more simply expressed in terms of trigonometric functions of A/3, B/3, C/3 (as in X(357), X(358), etc.).
      >
      > Best regards,
      > Randy
      >
      >
      >
      >
      >
      > >________________________________
      > > From: Chris Van Tienhoven
      > >To: Hyacinthos@yahoogroups.com
      > >Sent: Friday, January 4, 2013 8:11 AM
      > >Subject: [EMHL] Re: X(5390)
      > >
      > >
      > > 
      > >Dear friends,
      > >
      > >I managed to calculate the coordinates of X(5390).
      > >Indeed the outcome is horrendous.
      > >You need several pages to print it.
      > >Just like Bernard I found a SEARCH nr of: -0.3549673534054897101.
      > >I calculated in Mathematica and used my observation that X(5390) was lying on the line X(357).X(1136).X(3273).
      > >I also calculated that this line is the only connecting line of ETC-points where X(5390) is on.
      > >If someone wants the outcome to simplify it, please let me know.
      > >I have a strong feeling that this point should have simple coordinates.
      > >
      > >Best regards,
      > >
      > >Chris van Tienhoven
      > >
      > >www. chrisvantienhoven.nl
      > >
      > >--- In Hyacinthos@yahoogroups.com, Bernard Gibert wrote:
      > >>
      > >> Dear Antreas,
      > >>
      > >> > I am wondering if there is ANY known proof that they are concurrent.
      > >> > Drawing evidence is not a proof...!
      > >>
      > >> Indeed !...
      > >>
      > >> After some very ugly and painful computations, I did prove that the 3 Euler lines are concurrent.
      > >>
      > >> The SEARCH number of X(5390) is : -0.35496735340549.
      > >>
      > >> I expect the coordinates of X(5390) are horrendous...
      > >>
      > >> I will try the APH conjecture but I'm pretty sure it's true also.
      > >>
      > >> Best regards
      > >>
      > >> Bernard
      > >>
      > >> [Non-text portions of this message have been removed]
      > >>
      > >
      > >
      > >
      > >
      > >
      >
      > [Non-text portions of this message have been removed]
      >
    • Chris Van Tienhoven
      Dear Friends, I finally managed to find the coordinates of X(5390). I used a new method that –for as far as I know- never was used before. I already had
      Message 2 of 26 , Jan 8, 2013
        Dear Friends,

        I finally managed to find the coordinates of X(5390).
        I used a new method that –for as far as I know- never was used before.
        I already had found the coordinates and needed to scroll 16 times on my screen to reach the end of the coordinates.
        All Trig-commands in Mathematica could not help simplifying these coordinates.
        Still they were right, because when I substituted real values the outcome was perfect.

        The theory I used:
        Points that come forth from related subjects often are linearly related in the triangle field.
        I called this triangle field before a Perspective Field. It is a kind of coordinate system that can be used to describe all related points with real coordinates (often rational numbers).
        This relationship manifests itself in the property that each participating point in a Perspective Field can be described as the (weighted) sum of 3 other points in the field, whereas another 4th point is used to determine the standard counting weights, also called the compliance factors.
        So I reasoned that when I could find a Perspective Field and the location of X(5390) in this field, then it should be possible to calculate its coordinates from the 3+1 chosen reference points.

        I followed these steps:
        1. I gathered all Morley related points in ETC (31 points).
        2. I gathered some Euler line related points (13 points).
        3. I made a program in Mathematica in which I determined all permutations of 4 of these 44 points, no three of which were collinear.
        4. Per set of 4 points I calculated if X5390 could be the weighted sum of 3 of these points. The result would be 3 real / rational numbers (as coordinates), using the 4th point to determine the compliance factors.
        5. Of all 990 possible sets of 4 points only one set gave a result. It said:
        X5390 = 2.cf1.X2 + 1.cf2.X1136 + 2.cf3.X1137,
        the compliance factors cf1, cf2, cf3 being determined by the 4th point X(3273):
        cf1 = Det[X3273, X1136, X1137]
        cf2 = Det[ X2, X3273, X1137]
        cf3 = Det[ X2, X1136, X3273]

        I know this will appear pretty technically and it is of course.
        I think I will write a third document on this subject.
        The first two documents on Perspective Fields can be downloaded at:
        http://www.chrisvantienhoven.nl/index.php/mathematics/perspective-fields.html
        Especially the 2nd document explains the algebraic implications.

        The final result for the first BAYCENTRIC coordinate of X(5390) is:

        Sin[A] (Cos[ C/3 -Pi]/6] (-Sin[A] + Cos[1/3 (B + 4Pi])] Sec[1/3 (A + 4Pi])] Sin[B]) Sin[C] + Cos[B/3 -Pi]/6] Sin[B] (Sin[A] - Cos[1/3 (C + 4Pi])] Sec[1/3 (A + 4Pi])] Sin[C]) + Cos[A/3 -Pi]/ 6] (Cos[1/3 (C + 4Pi])] (-Sec[1/3 (A + 4Pi])] Sin[A] + 2 Sec[1/3 (B + 4Pi])] Sin[B]) Sin[C] + Cos[1/3 (B + 4Pi])] Sin[ B] (Sec[1/3 (A + 4Pi])] Sin[A] - 2 Sec[1/3 (C + 4Pi])] Sin[C]) + 2 Cos[1/3 (A + 4Pi])] Sin[ A] (-Sec[1/3 (B + 4Pi])] Sin[B] + Sec[1/3 (C + 4Pi])] Sin[C])))
        There still could be a simpler expression for this.

        There is also another way to describe the result.
        Let Det1 = Determinant:
        | Cos[1/3(A+4Pi)] Cos[1/3(B+4Pi)] Cos[1/3(C+4Pi)] |
        | Sec[1/3(A+4Pi)] Sec[1/3(B+4Pi)] Sec[1/3(C+4Pi)] |
        | Sin[1/3(A+4Pi)] Sin[1/3(B+4Pi)] Sin[1/3(C+4Pi)] |

        Let Det2 = Determinant:
        | 1 1 1 |
        | a Sec[1/3(A+4Pi)] b Sec[1/3(B+4Pi)] c Sec[1/3(C+4Pi)] |
        | a Sin[1/3(A+4Pi)] b Sin[1/3(B+4Pi)] c Sin[1/3(C+4Pi)] |

        Let Det3 = Determinant:
        | 1 1 1 |
        | a Cos[1/3(A+4Pi)] b Cos[1/3(B+4Pi)] c Cos[1/3(C+4Pi)] |
        | a Sin[1/3(A+4Pi)] b Sin[1/3(B+4Pi)] c Sin[1/3(C+4Pi)] |

        then the 1st TRILINEAR coordinate of X5390 is:
        2 b.c.Det1 - 2 Cos[1/3(A+4Pi)].Det2 + Sec[1/3(A+4Pi)].Det3

        A note from Peter Moses:
        X5390 is a point constructed from the first Morley Triangle.
        The equivalent points for the Second Morley Triangle can be derived by substituting 4Pi->2Pi.
        The equivalent points for the Third Morley Triangle can be derived by substituting 4Pi->0.

        Best regards,

        Chris van Tienhoven

        www.chris vantienhoven.nl



        --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" wrote:
        >
        > Dear friends,
        >
        > I managed to calculate the coordinates of X(5390).
        > Indeed the outcome is horrendous.
        > You need several pages to print it.
        > Just like Bernard I found a SEARCH nr of: -0.3549673534054897101.
        > I calculated in Mathematica and used my observation that X(5390) was lying on the line X(357).X(1136).X(3273).
        > I also calculated that this line is the only connecting line of ETC-points where X(5390) is on.
        > If someone wants the outcome to simplify it, please let me know.
        > I have a strong feeling that this point should have simple coordinates.
        >
        > Best regards,
        >
        > Chris van Tienhoven
        >
        > www. chrisvantienhoven.nl
        >
        > --- In Hyacinthos@yahoogroups.com, Bernard Gibert wrote:
        > >
        > > Dear Antreas,
        > >
        > > > I am wondering if there is ANY known proof that they are concurrent.
        > > > Drawing evidence is not a proof...!
        > >
        > > Indeed !...
        > >
        > > After some very ugly and painful computations, I did prove that the 3 Euler lines are concurrent.
        > >
        > > The SEARCH number of X(5390) is : -0.35496735340549.
        > >
        > > I expect the coordinates of X(5390) are horrendous...
        > >
        > > I will try the APH conjecture but I'm pretty sure it's true also.
        > >
        > > Best regards
        > >
        > > Bernard
        > >
        > > [Non-text portions of this message have been removed]
        > >
        >
      • Chris Van Tienhoven
        Dear Friends, I made a structured Mathematica-file with the calculation of the coordinates of X(5390). When you are interested please let me know so that I can
        Message 3 of 26 , Jan 9, 2013
          Dear Friends,

          I made a structured Mathematica-file with the calculation of the coordinates of X(5390).
          When you are interested please let me know so that I can send it to you.

          Best regards,

          Chris van Tienhoven

          www.chrisvantienhoven.nl


          --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" wrote:
          >
          > Dear Friends,
          >
          > I finally managed to find the coordinates of X(5390).
          > I used a new method that –for as far as I know- never was used before.
          > I already had found the coordinates and needed to scroll 16 times on my screen to reach the end of the coordinates.
          > All Trig-commands in Mathematica could not help simplifying these coordinates.
          > Still they were right, because when I substituted real values the outcome was perfect.
          >
          > The theory I used:
          > Points that come forth from related subjects often are linearly related in the triangle field.
          > I called this triangle field before a Perspective Field. It is a kind of coordinate system that can be used to describe all related points with real coordinates (often rational numbers).
          > This relationship manifests itself in the property that each participating point in a Perspective Field can be described as the (weighted) sum of 3 other points in the field, whereas another 4th point is used to determine the standard counting weights, also called the compliance factors.
          > So I reasoned that when I could find a Perspective Field and the location of X(5390) in this field, then it should be possible to calculate its coordinates from the 3+1 chosen reference points.
          >
          > I followed these steps:
          > 1. I gathered all Morley related points in ETC (31 points).
          > 2. I gathered some Euler line related points (13 points).
          > 3. I made a program in Mathematica in which I determined all permutations of 4 of these 44 points, no three of which were collinear.
          > 4. Per set of 4 points I calculated if X5390 could be the weighted sum of 3 of these points. The result would be 3 real / rational numbers (as coordinates), using the 4th point to determine the compliance factors.
          > 5. Of all 990 possible sets of 4 points only one set gave a result. It said:
          > X5390 = 2.cf1.X2 + 1.cf2.X1136 + 2.cf3.X1137,
          > the compliance factors cf1, cf2, cf3 being determined by the 4th point X(3273):
          > cf1 = Det[X3273, X1136, X1137]
          > cf2 = Det[ X2, X3273, X1137]
          > cf3 = Det[ X2, X1136, X3273]
          >
          > I know this will appear pretty technically and it is of course.
          > I think I will write a third document on this subject.
          > The first two documents on Perspective Fields can be downloaded at:
          > http://www.chrisvantienhoven.nl/index.php/mathematics/perspective-fields.html
          > Especially the 2nd document explains the algebraic implications.
          >
          > The final result for the first BAYCENTRIC coordinate of X(5390) is:
          >
          > Sin[A] (Cos[ C/3 -Pi]/6] (-Sin[A] + Cos[1/3 (B + 4Pi])] Sec[1/3 (A + 4Pi])] Sin[B]) Sin[C] + Cos[B/3 -Pi]/6] Sin[B] (Sin[A] - Cos[1/3 (C + 4Pi])] Sec[1/3 (A + 4Pi])] Sin[C]) + Cos[A/3 -Pi]/ 6] (Cos[1/3 (C + 4Pi])] (-Sec[1/3 (A + 4Pi])] Sin[A] + 2 Sec[1/3 (B + 4Pi])] Sin[B]) Sin[C] + Cos[1/3 (B + 4Pi])] Sin[ B] (Sec[1/3 (A + 4Pi])] Sin[A] - 2 Sec[1/3 (C + 4Pi])] Sin[C]) + 2 Cos[1/3 (A + 4Pi])] Sin[ A] (-Sec[1/3 (B + 4Pi])] Sin[B] + Sec[1/3 (C + 4Pi])] Sin[C])))
          > There still could be a simpler expression for this.
          >
          > There is also another way to describe the result.
          > Let Det1 = Determinant:
          > | Cos[1/3(A+4Pi)] Cos[1/3(B+4Pi)] Cos[1/3(C+4Pi)] |
          > | Sec[1/3(A+4Pi)] Sec[1/3(B+4Pi)] Sec[1/3(C+4Pi)] |
          > | Sin[1/3(A+4Pi)] Sin[1/3(B+4Pi)] Sin[1/3(C+4Pi)] |
          >
          > Let Det2 = Determinant:
          > | 1 1 1 |
          > | a Sec[1/3(A+4Pi)] b Sec[1/3(B+4Pi)] c Sec[1/3(C+4Pi)] |
          > | a Sin[1/3(A+4Pi)] b Sin[1/3(B+4Pi)] c Sin[1/3(C+4Pi)] |
          >
          > Let Det3 = Determinant:
          > | 1 1 1 |
          > | a Cos[1/3(A+4Pi)] b Cos[1/3(B+4Pi)] c Cos[1/3(C+4Pi)] |
          > | a Sin[1/3(A+4Pi)] b Sin[1/3(B+4Pi)] c Sin[1/3(C+4Pi)] |
          >
          > then the 1st TRILINEAR coordinate of X5390 is:
          > 2 b.c.Det1 - 2 Cos[1/3(A+4Pi)].Det2 + Sec[1/3(A+4Pi)].Det3
          >
          > A note from Peter Moses:
          > X5390 is a point constructed from the first Morley Triangle.
          > The equivalent points for the Second Morley Triangle can be derived by substituting 4Pi->2Pi.
          > The equivalent points for the Third Morley Triangle can be derived by substituting 4Pi->0.
          >
          > Best regards,
          >
          > Chris van Tienhoven
          >
          > www.chris vantienhoven.nl
          >
          >
          >
          > --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" wrote:
          > >
          > > Dear friends,
          > >
          > > I managed to calculate the coordinates of X(5390).
          > > Indeed the outcome is horrendous.
          > > You need several pages to print it.
          > > Just like Bernard I found a SEARCH nr of: -0.3549673534054897101.
          > > I calculated in Mathematica and used my observation that X(5390) was lying on the line X(357).X(1136).X(3273).
          > > I also calculated that this line is the only connecting line of ETC-points where X(5390) is on.
          > > If someone wants the outcome to simplify it, please let me know.
          > > I have a strong feeling that this point should have simple coordinates.
          > >
          > > Best regards,
          > >
          > > Chris van Tienhoven
          > >
          > > www. chrisvantienhoven.nl
          > >
          > > --- In Hyacinthos@yahoogroups.com, Bernard Gibert wrote:
          > > >
          > > > Dear Antreas,
          > > >
          > > > > I am wondering if there is ANY known proof that they are concurrent.
          > > > > Drawing evidence is not a proof...!
          > > >
          > > > Indeed !...
          > > >
          > > > After some very ugly and painful computations, I did prove that the 3 Euler lines are concurrent.
          > > >
          > > > The SEARCH number of X(5390) is : -0.35496735340549.
          > > >
          > > > I expect the coordinates of X(5390) are horrendous...
          > > >
          > > > I will try the APH conjecture but I'm pretty sure it's true also.
          > > >
          > > > Best regards
          > > >
          > > > Bernard
          > > >
          > > > [Non-text portions of this message have been removed]
          > > >
          > >
          >
        • Chris Van Tienhoven
          Dear Friends, Seiichi Kirikami did send me another property of X(5390). Construction: • Let A1,B1,C1 be the vertices of the 1st Morley Triangle. • Let La,
          Message 4 of 26 , Jan 16, 2013
            Dear Friends,

            Seiichi Kirikami did send me another property of X(5390).
            Construction:
            • Let A1,B1,C1 be the vertices of the 1st Morley Triangle.
            • Let La, Lb, Lc be the Euler Lines of triangles A.B1.C1, B.C1.A1, C.A1.B1.
            • Then La, Lb, Lc concur in X(5390).
            • The lines through A,B,C parallel to La, Lb, Lc also concur in a point.
            • This point happens to be X(1136) !
            When the parallel lines are drawn through the vertices of the Medial Triangle or the AntiComplementary Triangle, these lines also are concurrent, however not in ETC-points.
            This parallel-construction is similar to the construction method of points X(3647) to X(3652) in ETC.

            I suppose there will be more Morley related points that can be constructed this way.

            Best regards,

            Chris van Tienhoven


            --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
            >
            > X(5390) = EULER-MORLEY-ZHAO POINT
            >
            > Barycentrics (unknown)
            > Let DEF be the classical Morley triangle. The Euler lines of the three
            > triangles AEF, BFD, CDE
            > appear to concur in a point for which barycentric coordinates remain
            > to be discovered.
            > Construction by Zhao Yong of Anhui, China, October 2, 2012.
            >
            > http://faculty.evansville.edu/ck6/encyclopedia/ETCPart3.html
            >
          • Antreas
            Dear Chris Please check this whether it is true: The reflections of the Euler Lines La,Lb,Lc (concurrent in X5390) in the respective angle bisectors of ABC,
            Message 5 of 26 , Jan 16, 2013
              Dear Chris

              Please check this whether it is true:

              The reflections of the Euler Lines La,Lb,Lc (concurrent in X5390)
              in the respective angle bisectors of ABC, are concurrent.

              APH


              --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" wrote:
              >
              > Dear Friends,
              >
              > Seiichi Kirikami did send me another property of X(5390).
              > Construction:
              > • Let A1,B1,C1 be the vertices of the 1st Morley Triangle.
              > • Let La, Lb, Lc be the Euler Lines of triangles A.B1.C1, B.C1.A1, C.A1.B1.
              > • Then La, Lb, Lc concur in X(5390).
              > • The lines through A,B,C parallel to La, Lb, Lc also concur in a point.
              > • This point happens to be X(1136) !
              > When the parallel lines are drawn through the vertices of the Medial Triangle or the AntiComplementary Triangle, these lines also are concurrent, however not in ETC-points.
              > This parallel-construction is similar to the construction method of points X(3647) to X(3652) in ETC.
              >
              > I suppose there will be more Morley related points that can be constructed this way.
              >
              > Best regards,
              >
              > Chris van Tienhoven
              >
              >
              > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
              > >
              > > X(5390) = EULER-MORLEY-ZHAO POINT
              > >
              > > Barycentrics (unknown)
              > > Let DEF be the classical Morley triangle. The Euler lines of the three
              > > triangles AEF, BFD, CDE
              > > appear to concur in a point for which barycentric coordinates remain
              > > to be discovered.
              > > Construction by Zhao Yong of Anhui, China, October 2, 2012.
              > >
              > > http://faculty.evansville.edu/ck6/encyclopedia/ETCPart3.html
              > >
              >
            • Chris Van Tienhoven
              Dear Antreas, They are not concurrent. Chris
              Message 6 of 26 , Jan 16, 2013
                Dear Antreas,

                They are not concurrent.

                Chris

                --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
                >
                > Dear Chris
                >
                > Please check this whether it is true:
                >
                > The reflections of the Euler Lines La,Lb,Lc (concurrent in X5390)
                > in the respective angle bisectors of ABC, are concurrent.
                >
                > APH
                >
                >
                > --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" wrote:
                > >
                > > Dear Friends,
                > >
                > > Seiichi Kirikami did send me another property of X(5390).
                > > Construction:
                > > • Let A1,B1,C1 be the vertices of the 1st Morley Triangle.
                > > • Let La, Lb, Lc be the Euler Lines of triangles A.B1.C1, B.C1.A1, C.A1.B1.
                > > • Then La, Lb, Lc concur in X(5390).
                > > • The lines through A,B,C parallel to La, Lb, Lc also concur in a point.
                > > • This point happens to be X(1136) !
                > > When the parallel lines are drawn through the vertices of the Medial Triangle or the AntiComplementary Triangle, these lines also are concurrent, however not in ETC-points.
                > > This parallel-construction is similar to the construction method of points X(3647) to X(3652) in ETC.
                > >
                > > I suppose there will be more Morley related points that can be constructed this way.
                > >
                > > Best regards,
                > >
                > > Chris van Tienhoven
                > >
                > >
                > > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
                > > >
                > > > X(5390) = EULER-MORLEY-ZHAO POINT
                > > >
                > > > Barycentrics (unknown)
                > > > Let DEF be the classical Morley triangle. The Euler lines of the three
                > > > triangles AEF, BFD, CDE
                > > > appear to concur in a point for which barycentric coordinates remain
                > > > to be discovered.
                > > > Construction by Zhao Yong of Anhui, China, October 2, 2012.
                > > >
                > > > http://faculty.evansville.edu/ck6/encyclopedia/ETCPart3.html
                > > >
                > >
                >
              • Antreas
                If A*B*C* is the Morley triangle, the Euler Lines of AB*C*, BC*A*, CA*B* are concurrent [X(5390] A similar construction: I have constructed a central triangle
                Message 7 of 26 , Mar 6, 2013
                  If A*B*C* is the Morley triangle, the Euler Lines
                  of AB*C*, BC*A*, CA*B* are concurrent [X(5390]

                  A similar construction:

                  I have constructed a central triangle A*B*C* (using the bisectors), with the same property ie the Euler Lines of AB*C*, BC*A*, CA*B*
                  are concurrent.

                  See:
                  http://anthrakitis.blogspot.gr/2013/03/concurrent-euler-lines.html

                  Which is the point of concurrence?

                  APH


                  --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
                  >
                  > X(5390) = EULER-MORLEY-ZHAO POINT
                  >
                  > Barycentrics (unknown)
                  > Let DEF be the classical Morley triangle. The Euler lines of the three
                  > triangles AEF, BFD, CDE
                  > appear to concur in a point for which barycentric coordinates remain
                  > to be discovered.
                  > Construction by Zhao Yong of Anhui, China, October 2, 2012.
                  >
                  > http://faculty.evansville.edu/ck6/encyclopedia/ETCPart3.html
                  >
                • Barry Wolk
                  From ETC ... three triangles AEF, BFD, CDE to concur in X5390), ... of Anhui, China, October 2, 2012. ... barycentric coordinates ... Rewards. ... The
                  Message 8 of 26 , Apr 6, 2013
                    From ETC
                    >  X(5390) = EULER-MORLEY-ZHAO POINT
                    >   Barycentrics   (unknown)
                    >   Let DEF be the classical Morley triangle. The Euler lines
                    > of the
                    three triangles AEF, BFD, CDE to concur in X5390),
                    > as discovered by Zhao Yong
                    of Anhui, China, October 2, 2012.
                    > For a construction and derivation of
                    barycentric coordinates
                    > by Shi Yong, see Problem 20 at Unsolved Problems and
                    Rewards.
                    > For further developments type X(5390) into Search at Hyacinthos
                    The coordinates of X(5390) are listed as
                    unknown, but
                    some Hyacinthos messages a while ago shows that they
                    aren't really
                    unknown, just too long. See message 21367.
                    I used complex numbers to get fairly long
                    real-valued expressions
                    of complex parameters for these coordinates, and then
                    some ideas
                    to symmetrize and simplify them. I'll use the abbreviations
                       <m,n> =  sin(x+2my+2nz) +
                    sin(x+2ny+2mz)
                    and
                       <m> = <m,m>/2.
                    Define
                    f(x,y,z)=<-3,-1>+5*<-2,-1>+<-2,0>-5*<-1,1>-3*<-1,2>
                       -<-1,3>+<0,1>+2*<0,2>-<0,3>-2*<1,3>-2*<2,3>
                       +<-3>-2*<-2>+3*<-1>+3*<0>+3*<1>+3*<2>-2*<3>-<4>
                     
                    Then the barycentrics of X(5390) are
                       (
                    a*f(A/3,B/3,C/3) : b*f(B/3,C/3,A/3) : c*f(C/3,A/3,B/3) )
                    Antreas, can you forward this to Clark Kimberling? Everything
                    I've found from Clark hides his address.
                    --
                    Barry Wolk

                    [Non-text portions of this message have been removed]
                  • Antreas
                    OK, Barry, done. The email addresses are hiden for protection reasons. Clark s is: ck6(AT)evansville.edu where (AT) stands for @ Antreas
                    Message 9 of 26 , Apr 6, 2013
                      OK, Barry, done.

                      The email addresses are hiden for protection reasons.

                      Clark's is:

                      ck6(AT)evansville.edu

                      where (AT) stands for @

                      Antreas


                      --- In Hyacinthos@yahoogroups.com, Barry Wolk <wolkbarry@...> wrote:
                      >
                      > From ETC
                      > >  X(5390) = EULER-MORLEY-ZHAO POINT
                      > >   Barycentrics   (unknown)
                      > >   Let DEF be the classical Morley triangle. The Euler lines
                      > > of the
                      > three triangles AEF, BFD, CDE to concur in X5390),
                      > > as discovered by Zhao Yong
                      > of Anhui, China, October 2, 2012.
                      > > For a construction and derivation of
                      > barycentric coordinates
                      > > by Shi Yong, see Problem 20 at Unsolved Problems and
                      > Rewards.
                      > > For further developments type X(5390) into Search at Hyacinthos
                      > The coordinates of X(5390) are listed as
                      > unknown, but
                      > some Hyacinthos messages a while ago shows that they
                      > aren't really
                      > unknown, just too long. See message 21367.
                      > I used complex numbers to get fairly long
                      > real-valued expressions
                      > of complex parameters for these coordinates, and then
                      > some ideas
                      > to symmetrize and simplify them. I'll use the abbreviations
                      >    <m,n> =  sin(x+2my+2nz) +
                      > sin(x+2ny+2mz)
                      > and
                      >    <m> = <m,m>/2.
                      > Define
                      > f(x,y,z)=<-3,-1>+5*<-2,-1>+<-2,0>-5*<-1,1>-3*<-1,2>
                      >    -<-1,3>+<0,1>+2*<0,2>-<0,3>-2*<1,3>-2*<2,3>
                      >    +<-3>-2*<-2>+3*<-1>+3*<0>+3*<1>+3*<2>-2*<3>-<4>
                      >  
                      > Then the barycentrics of X(5390) are
                      >    (
                      > a*f(A/3,B/3,C/3) : b*f(B/3,C/3,A/3) : c*f(C/3,A/3,B/3) )
                      > Antreas, can you forward this to Clark Kimberling? Everything
                      > I've found from Clark hides his address.
                      > --
                      > Barry Wolk
                    • Chris Van Tienhoven
                      Dear Friends, I refined the coordinates of X(5390) that I mentioned earlier in message # 21379. The trilinear coordinates are: Cos[B - C] - Cos[B + C] -
                      Message 10 of 26 , Apr 7, 2013
                        Dear Friends,

                        I refined the coordinates of X(5390) that I mentioned earlier in message # 21379.
                        The trilinear coordinates are:

                        Cos[B - C] - Cos[B + C]
                        - Cos[B/3 + C/3] + Cos[5B/3 + 5C/3]
                        - Sin[C - B/3 - Pi/6] - Sin[B - C/3 - Pi/6]
                        + Sin[B + 5C/3 - Pi/6] + Sin[C + 5B/3 - Pi/6]

                        Best regards,

                        Chris van Tienhoven
                        www.chrisvantienhoven.nl


                        --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@...> wrote:
                        >
                        > Dear Friends,
                        >
                        > I finally managed to find the coordinates of X(5390).
                        > I used a new method that –for as far as I know- never was used before.
                        > I already had found the coordinates and needed to scroll 16 times on my screen to reach the end of the coordinates.
                        > All Trig-commands in Mathematica could not help simplifying these coordinates.
                        > Still they were right, because when I substituted real values the outcome was perfect.
                        >
                        > The theory I used:
                        > Points that come forth from related subjects often are linearly related in the triangle field.
                        > I called this triangle field before a Perspective Field. It is a kind of coordinate system that can be used to describe all related points with real coordinates (often rational numbers).
                        > This relationship manifests itself in the property that each participating point in a Perspective Field can be described as the (weighted) sum of 3 other points in the field, whereas another 4th point is used to determine the standard counting weights, also called the compliance factors.
                        > So I reasoned that when I could find a Perspective Field and the location of X(5390) in this field, then it should be possible to calculate its coordinates from the 3+1 chosen reference points.
                        >
                        > I followed these steps:
                        > 1. I gathered all Morley related points in ETC (31 points).
                        > 2. I gathered some Euler line related points (13 points).
                        > 3. I made a program in Mathematica in which I determined all permutations of 4 of these 44 points, no three of which were collinear.
                        > 4. Per set of 4 points I calculated if X5390 could be the weighted sum of 3 of these points. The result would be 3 real / rational numbers (as coordinates), using the 4th point to determine the compliance factors.
                        > 5. Of all 990 possible sets of 4 points only one set gave a result. It said:
                        > X5390 = 2.cf1.X2 + 1.cf2.X1136 + 2.cf3.X1137,
                        > the compliance factors cf1, cf2, cf3 being determined by the 4th point X(3273):
                        > cf1 = Det[X3273, X1136, X1137]
                        > cf2 = Det[ X2, X3273, X1137]
                        > cf3 = Det[ X2, X1136, X3273]
                        >
                        > I know this will appear pretty technically and it is of course.
                        > I think I will write a third document on this subject.
                        > The first two documents on Perspective Fields can be downloaded at:
                        > http://www.chrisvantienhoven.nl/index.php/mathematics/perspective-fields.html
                        > Especially the 2nd document explains the algebraic implications.
                        >
                        > The final result for the first BAYCENTRIC coordinate of X(5390) is:
                        >
                        > Sin[A] (Cos[ C/3 -Pi]/6] (-Sin[A] + Cos[1/3 (B + 4Pi])] Sec[1/3 (A + 4Pi])] Sin[B]) Sin[C] + Cos[B/3 -Pi]/6] Sin[B] (Sin[A] - Cos[1/3 (C + 4Pi])] Sec[1/3 (A + 4Pi])] Sin[C]) + Cos[A/3 -Pi]/ 6] (Cos[1/3 (C + 4Pi])] (-Sec[1/3 (A + 4Pi])] Sin[A] + 2 Sec[1/3 (B + 4Pi])] Sin[B]) Sin[C] + Cos[1/3 (B + 4Pi])] Sin[ B] (Sec[1/3 (A + 4Pi])] Sin[A] - 2 Sec[1/3 (C + 4Pi])] Sin[C]) + 2 Cos[1/3 (A + 4Pi])] Sin[ A] (-Sec[1/3 (B + 4Pi])] Sin[B] + Sec[1/3 (C + 4Pi])] Sin[C])))
                        > There still could be a simpler expression for this.
                        >
                        > There is also another way to describe the result.
                        > Let Det1 = Determinant:
                        > | Cos[1/3(A+4Pi)] Cos[1/3(B+4Pi)] Cos[1/3(C+4Pi)] |
                        > | Sec[1/3(A+4Pi)] Sec[1/3(B+4Pi)] Sec[1/3(C+4Pi)] |
                        > | Sin[1/3(A+4Pi)] Sin[1/3(B+4Pi)] Sin[1/3(C+4Pi)] |
                        >
                        > Let Det2 = Determinant:
                        > | 1 1 1 |
                        > | a Sec[1/3(A+4Pi)] b Sec[1/3(B+4Pi)] c Sec[1/3(C+4Pi)] |
                        > | a Sin[1/3(A+4Pi)] b Sin[1/3(B+4Pi)] c Sin[1/3(C+4Pi)] |
                        >
                        > Let Det3 = Determinant:
                        > | 1 1 1 |
                        > | a Cos[1/3(A+4Pi)] b Cos[1/3(B+4Pi)] c Cos[1/3(C+4Pi)] |
                        > | a Sin[1/3(A+4Pi)] b Sin[1/3(B+4Pi)] c Sin[1/3(C+4Pi)] |
                        >
                        > then the 1st TRILINEAR coordinate of X5390 is:
                        > 2 b.c.Det1 - 2 Cos[1/3(A+4Pi)].Det2 + Sec[1/3(A+4Pi)].Det3
                        >
                        > A note from Peter Moses:
                        > X5390 is a point constructed from the first Morley Triangle.
                        > The equivalent points for the Second Morley Triangle can be derived by substituting 4Pi->2Pi.
                        > The equivalent points for the Third Morley Triangle can be derived by substituting 4Pi->0.
                        >
                        > Best regards,
                        >
                        > Chris van Tienhoven
                        >
                        > www.chris vantienhoven.nl
                        >
                        >
                        >
                        > --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" wrote:
                        > >
                        > > Dear friends,
                        > >
                        > > I managed to calculate the coordinates of X(5390).
                        > > Indeed the outcome is horrendous.
                        > > You need several pages to print it.
                        > > Just like Bernard I found a SEARCH nr of: -0.3549673534054897101.
                        > > I calculated in Mathematica and used my observation that X(5390) was lying on the line X(357).X(1136).X(3273).
                        > > I also calculated that this line is the only connecting line of ETC-points where X(5390) is on.
                        > > If someone wants the outcome to simplify it, please let me know.
                        > > I have a strong feeling that this point should have simple coordinates.
                        > >
                        > > Best regards,
                        > >
                        > > Chris van Tienhoven
                        > >
                        > > www. chrisvantienhoven.nl
                        > >
                        > > --- In Hyacinthos@yahoogroups.com, Bernard Gibert wrote:
                        > > >
                        > > > Dear Antreas,
                        > > >
                        > > > > I am wondering if there is ANY known proof that they are concurrent.
                        > > > > Drawing evidence is not a proof...!
                        > > >
                        > > > Indeed !...
                        > > >
                        > > > After some very ugly and painful computations, I did prove that the 3 Euler lines are concurrent.
                        > > >
                        > > > The SEARCH number of X(5390) is : -0.35496735340549.
                        > > >
                        > > > I expect the coordinates of X(5390) are horrendous...
                        > > >
                        > > > I will try the APH conjecture but I'm pretty sure it's true also.
                        > > >
                        > > > Best regards
                        > > >
                        > > > Bernard
                        > > >
                        > > > [Non-text portions of this message have been removed]
                        > > >
                        > >
                        >
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