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Re: [EMHL] Re: X(5390)

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  • Bernard Gibert
    Dear Antreas, ... 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
    Message 1 of 26 , Jan 3, 2013
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      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]
    • Bernard Gibert
      Dear Antreas, ... The APH conjecture is unexpectedly FALSE ! Best regards Bernard [Non-text portions of this message have been removed]
      Message 2 of 26 , Jan 3, 2013
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        Dear Antreas,

        > I will try the APH conjecture but I'm pretty sure it's true also.

        The APH conjecture is unexpectedly FALSE !

        Best regards

        Bernard

        [Non-text portions of this message have been removed]
      • Antreas
        Dear Bernard Thank you! Naturally one may ask how about the case of the external Morley triangle ie the eq. triangle A B C we get by the external trisectors
        Message 3 of 26 , Jan 3, 2013
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          Dear Bernard

          Thank you!

          Naturally one may ask how about the case of the external
          Morley triangle ie the eq. triangle A'B'C' we get by
          the external trisectors of the triangle ABC.

          Is it true that the Euler lines of AB'C', BC'A', CA'B'
          are concurrent as well?

          If not, then I will not say that the God of Geometry made everything
          with sophia (wisdom) !!

          Antreas

          --- In Hyacinthos@yahoogroups.com, Bernard Gibert wrote:
          >
          > Dear Antreas,
          >
          > > I will try the APH conjecture but I'm pretty sure it's true also.
          >
          > The APH conjecture is unexpectedly FALSE !
          >
          > Best regards
          >
          > Bernard
        • Nikolaos Dergiades
          Dear Antreas, [APH] ... Yes. There is strong evidence that they are concurrent. ND
          Message 4 of 26 , Jan 3, 2013
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            Dear Antreas,

            [APH]
            > Naturally one may ask how about the case of the external
            > Morley triangle ie the eq. triangle A'B'C' we get by
            > the external trisectors of the triangle ABC.
            >
            > Is it true that the Euler lines of AB'C', BC'A', CA'B'
            > are concurrent as well?

            Yes. There is strong evidence that they are concurrent.

            ND
          • Antreas
            Dear Nikos, Thanks. Now, we know some equilateral triangles which are homothetic to Morley internal equilateral triangle A B C . One of these triangles is
            Message 5 of 26 , Jan 3, 2013
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              Dear Nikos,

              Thanks.

              Now, we know some equilateral triangles which are homothetic to Morley internal equilateral triangle A'B'C'. One of these triangles is Russel triangle A"B"C" (see here a figure http://www.cut-the-knot.org/Curriculum/Geometry/Roussel.shtml ).

              Conjecture:
              The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.

              I have put a not-so-good figure here
              http://anthrakitis.blogspot.gr/2013/01/morley-triangles-conjecture.html

              True?

              APH

              --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades wrote:
              >
              > Dear Antreas,
              >
              > [APH]
              > > Naturally one may ask how about the case of the external
              > > Morley triangle ie the eq. triangle A'B'C' we get by
              > > the external trisectors of the triangle ABC.
              > >
              > > Is it true that the Euler lines of AB'C', BC'A', CA'B'
              > > are concurrent as well?
              >
              > Yes. There is strong evidence that they are concurrent.
              >
              > ND
              >
            • Floor en Lyanne van Lamoen
              Dear all, First of all a Happy New Year to all Hyacinthians. Generalizating conjecture: Let t be a number. Consider the t-sector of an angle alpha the line
              Message 6 of 26 , Jan 3, 2013
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                Dear all,

                First of all a Happy New Year to all Hyacinthians.

                Generalizating conjecture:
                Let t be a number. Consider the t-sector of an angle alpha the line that
                cuts off an angle of t times alpha. The 6 t-sectors in a triangle
                constitute a triangle A'B'C' similar to the "standard" Morley triangle,
                which is found when t=1/3.
                Conjecture: the Euler lines of AB'C', A'BC' and A'B'C are concurrent.

                Kind regards,
                Floor van Lamoen.
              • Nikolaos Dergiades
                Dear Floor, ... For t=1/4 it is not correct. Best regards Nikos Dergiades
                Message 7 of 26 , Jan 4, 2013
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                  Dear Floor,

                  > Generalizating conjecture:
                  > Let t be a number. Consider the t-sector of an angle alpha
                  > the line that
                  > cuts off an angle of t times alpha. The 6 t-sectors in a
                  > triangle
                  > constitute a triangle A'B'C' similar to the "standard"
                  > Morley triangle,
                  > which is found when t=1/3.
                  > Conjecture: the Euler lines of AB'C', A'BC' and A'B'C are
                  > concurrent.

                  For t=1/4 it is not correct.

                  Best regards
                  Nikos Dergiades
                • Nikolaos Dergiades
                  Dear Antreas, Not only the Euler lines of triangles A B C , B C A , C A B are concurrent but also the Euler lines of A B C , B C A , C A B I am almost sure
                  Message 8 of 26 , Jan 4, 2013
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                    Dear Antreas,
                    Not only the Euler lines of triangles
                    A'B"C", B'C"A", C'A"B" are concurrent but also
                    the Euler lines of A"B'C', B"C'A', C"A'B'
                    I am almost sure that are concurrent.
                    Best regards
                    Nikos Dergiades

                    [APH]
                    > Now, we know some equilateral triangles which are homothetic
                    > to Morley internal equilateral triangle A'B'C'. One of these
                    > triangles is Russel triangle A"B"C" (see here a figure http://www.cut-the-knot.org/Curriculum/Geometry/Roussel.shtml
                    > ).
                    >
                    > Conjecture:
                    > The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
                    >
                    > I have put a not-so-good figure here
                    > http://anthrakitis.blogspot.gr/2013/01/morley-triangles-conjecture.html
                    >
                    > True?
                    >
                  • Antreas Hatzipolakis
                    Dear Nikos, Thanks! I had thought about that (of A B C , B C A , C A B ) and made a picture, but it seems that my picture was not good enough..... they were
                    Message 9 of 26 , Jan 4, 2013
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                      Dear Nikos,

                      Thanks!

                      I had thought about that (of A"B'C', B"C'A', C"A'B' ) and made a picture,
                      but it seems that my
                      picture was not good enough..... they were not concurrent.

                      Antreas


                      On Fri, Jan 4, 2013 at 11:39 AM, Nikolaos Dergiades <ndergiades@...>wrote:

                      > **
                      >
                      >
                      > Dear Antreas,
                      > Not only the Euler lines of triangles
                      > A'B"C", B'C"A", C'A"B" are concurrent but also
                      > the Euler lines of A"B'C', B"C'A', C"A'B'
                      > I am almost sure that are concurrent.
                      > Best regards
                      > Nikos Dergiades
                      >
                      > [APH]
                      >
                      > > Now, we know some equilateral triangles which are homothetic
                      > > to Morley internal equilateral triangle A'B'C'. One of these
                      > > triangles is Russel triangle A"B"C" (see here a figure
                      > http://www.cut-the-knot.org/Curriculum/Geometry/Roussel.shtml
                      > > ).
                      > >
                      > > Conjecture:
                      > > The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
                      > >
                      > > I have put a not-so-good figure here
                      > > http://anthrakitis.blogspot.gr/2013/01/morley-triangles-conjecture.html
                      > >
                      > > True?
                      > >
                      >
                      >
                      >


                      [Non-text portions of this message have been removed]
                    • Frank Jackson
                      Dear All Is it correct to assume that all these conjectured Euler line concurrences are colinear and lie on the Euler line of the original triangle ABC?
                      Message 10 of 26 , Jan 4, 2013
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                        Dear All
                        Is it correct to assume that all these conjectured Euler line
                        concurrences are colinear and lie on the Euler line of the original
                        triangle ABC?
                        Regards
                        Frank Jackson
                        > [ND]
                        >
                        > Now, we know some equilateral triangles which are homothetic to Morley internal equilateral triangle A'B'C'. One of these triangles is Russel triangle A"B"C" (see here a figurehttp://www.cut-the-knot.org/Curriculum/Geometry/Roussel.shtml ).
                        >
                        > Conjecture:
                        > The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
                        >
                        > I have put a not-so-good figure here
                        > http://anthrakitis.blogspot.gr/2013/01/morley-triangles-conjecture.html
                        >
                        > True?
                        >
                        > APH
                        >
                        > --- InHyacinthos@yahoogroups.com, Nikolaos Dergiades wrote:
                        >> Dear Antreas,
                        >>
                        >> [APH]
                        >>> Naturally one may ask how about the case of the external
                        >>> Morley triangle ie the eq. triangle A'B'C' we get by
                        >>> the external trisectors of the triangle ABC.
                        >>>
                        >>> Is it true that the Euler lines of AB'C', BC'A', CA'B'
                        >>> are concurrent as well?
                        >> Yes. There is strong evidence that they are concurrent.
                        >>
                        >> ND
                        >>
                        >
                        >
                        > ------------------------------------
                        >
                        > Yahoo! Groups Links
                        >
                        >
                        >
                      • Chris Van Tienhoven
                        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
                        Message 11 of 26 , Jan 4, 2013
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                          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]
                          >
                        • Randy Hutson
                          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
                          Message 12 of 26 , Jan 4, 2013
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                            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 <van10hoven@...>
                            >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 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 13 of 26 , Jan 4, 2013
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                              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 14 of 26 , Jan 8, 2013
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                                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 15 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 16 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 17 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 18 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 19 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 20 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 21 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 22 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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