- Dear Antreas,

[APH]> Naturally one may ask how about the case of the external

Yes. There is strong evidence that they are concurrent.

> 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?

ND - 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]

> > >

> >

>