• ## ANOTHER CONJECTURE ([EMHL] Re: X(5390))

(26)
• NextPrevious
• 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 1 of 26 , Jan 3, 2013
View Source
• 0 Attachment
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
>
• 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 2 of 26 , Jan 3, 2013
View Source
• 0 Attachment
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.
• Dear Floor, ... For t=1/4 it is not correct. Best regards Nikos Dergiades
Message 3 of 26 , Jan 4, 2013
View Source
• 0 Attachment
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
• 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 4 of 26 , Jan 4, 2013
View Source
• 0 Attachment
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

[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?
>
• 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 5 of 26 , Jan 4, 2013
View Source
• 0 Attachment
Dear Nikos,

Thanks!

but it seems that my
picture was not good enough..... they were not concurrent.

Antreas

> **
>
>
> 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
>
> [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]
• 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 6 of 26 , Jan 4, 2013
View Source
• 0 Attachment
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
>>
>
>
> ------------------------------------
>
>
>
>
• 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 7 of 26 , Jan 4, 2013
View Source
• 0 Attachment
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]
>
• 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 8 of 26 , Jan 4, 2013
View Source
• 0 Attachment
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]
• 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 9 of 26 , Jan 4, 2013
View Source
• 0 Attachment
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?Â  If so, what degree?Â  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)
> >
> >
> >Â
> >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]
>
• 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 10 of 26 , Jan 8, 2013
View Source
• 0 Attachment
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.
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]
> >
>
• 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 11 of 26 , Jan 9, 2013
View Source
• 0 Attachment
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]
> > >
> >
>
• 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 12 of 26 , Jan 16, 2013
View Source
• 0 Attachment
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
>
• 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 13 of 26 , Jan 16, 2013
View Source
• 0 Attachment
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
> >
>
• Dear Antreas, They are not concurrent. Chris
Message 14 of 26 , Jan 16, 2013
View Source
• 0 Attachment
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
> > >
> >
>
• 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 15 of 26 , Mar 6 10:40 PM
View Source
• 0 Attachment
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
>
• From ETC ... three triangles AEF, BFD, CDE to concur in X5390), ... of Anhui, China, October 2, 2012. ... barycentric coordinates ... Rewards. ... The
Message 16 of 26 , Apr 6, 2013
View Source
• 0 Attachment
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]
• OK, Barry, done. The email addresses are hiden for protection reasons. Clark s is: ck6(AT)evansville.edu where (AT) stands for @ Antreas
Message 17 of 26 , Apr 6, 2013
View Source
• 0 Attachment
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
• 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 18 of 26 , Apr 7, 2013
View Source
• 0 Attachment
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]
> > >
> >
>
Your message has been successfully submitted and would be delivered to recipients shortly.