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Re: X(5390)

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  • 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 1 of 26 , Jan 9, 2013
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      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 2 of 26 , Jan 16, 2013
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        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 3 of 26 , Jan 16, 2013
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          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 4 of 26 , Jan 16, 2013
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            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 5 of 26 , Mar 6, 2013
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              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 6 of 26 , Apr 6, 2013
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                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 7 of 26 , Apr 6, 2013
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                  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 8 of 26 , Apr 7, 2013
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                    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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