## [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)

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• Dear Randy I am not sure what you mean by confirming computationally. If you used a coordinate system or other general algebraic method it is a proof! Not nice
Message 1 of 9 , Jan 4, 2013
Dear Randy

I am not sure what you mean by confirming computationally.
If you used a coordinate system or other general algebraic method
it is a proof! Not nice as a synthetic one, but a proof!

APH

PS I had quoted not the right message!!
The message that I should have quoted (with homothetic eq. triangles)
is #21349 ANOTHER CONJECTURE ([EMHL] Re: X(5390))

--- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
>
> Antreas,
>
> I was able to confirm this computationally.Â  Of course, not the same as a proof.
>
> Randy Hutson
>
>
>
>
>
> >________________________________
> > From: Antreas
> >To: Hyacinthos@yahoogroups.com
> >Sent: Friday, January 4, 2013 12:56 PM
> >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
> >
> >
> >Â
> >Let A'B'C', A"B"C" be two homothetic Equilateral triangles.
> >
> >Conjecture:
> >
> >The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
> >(and of A"B'C', B"C'A', C"A'B')
> >
> >True?
> >
> >APH
> >
> >--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
> >>
> >> This topic is very rich.............!!!!
> >>
> >> Let A'B'C', A"B"C" be the 1st Morley and 1st Morley Adjunct triangles, resp.
> >> [see definition http://mathworld.wolfram.com/FirstMorleyAdjunctTriangle.html ]
> >>
> >> Conjecture:
> >>
> >> The Euler Lines of A"B'C', B"C'A', C"A'B' are concurrent.
> >>
> >> If true, which point with respect A"B"C" is the point of concurrence?
> >>
> >> APH
• Dear Antreas, I constructed random homothetic equilateral triangles and the Euler lines of A B C , B C A , C A B (and of A B C , B C A , C A B ) are
Message 2 of 9 , Jan 4, 2013
Dear Antreas,

I constructed random homothetic equilateral triangles and the Euler lines of A'B"C", B'C"A", C'A"B" (and of A"B'C', B"C'A', C"A'B') are concurrent (the three intersections coincide to at least 15 decimal places).

Randy

>________________________________
> From: Antreas <anopolis72@...>
>To: Hyacinthos@yahoogroups.com
>Sent: Friday, January 4, 2013 5:41 PM
>Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
>
>

>Dear Randy
>
>I am not sure what you mean by confirming computationally.
>If you used a coordinate system or other general algebraic method
>it is a proof! Not nice as a synthetic one, but a proof!
>
>APH
>
>PS I had quoted not the right message!!
>The message that I should have quoted (with homothetic eq. triangles)
>is #21349 ANOTHER CONJECTURE ([EMHL] Re: X(5390))
>
>--- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
>>
>> Antreas,
>>
>> I was able to confirm this computationally.Â  Of course, not the same as a proof.
>>
>> Randy Hutson
>>
>>
>>
>>
>>
>> >________________________________
>> > From: Antreas
>> >To: Hyacinthos@yahoogroups.com
>> >Sent: Friday, January 4, 2013 12:56 PM
>> >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
>> >
>> >
>> >Â
>> >Let A'B'C', A"B"C" be two homothetic Equilateral triangles.
>> >
>> >Conjecture:
>> >
>> >The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
>> >(and of A"B'C', B"C'A', C"A'B')
>> >
>> >True?
>> >
>> >APH
>> >
>> >--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
>> >>
>> >> This topic is very rich.............!!!!
>> >>
>> >> Let A'B'C', A"B"C" be the 1st Morley and 1st Morley Adjunct triangles, resp.
>> >> [see definition http://mathworld.wolfram.com/FirstMorleyAdjunctTriangle.html ]
>> >>
>> >> Conjecture:
>> >>
>> >> The Euler Lines of A"B'C', B"C'A', C"A'B' are concurrent.
>> >>
>> >> If true, which point with respect A"B"C" is the point of concurrence?
>> >>
>> >> APH
>
>
>
>
>

[Non-text portions of this message have been removed]
• Dear Randy, in this case the concurrency can be completely proven by using computer algebra and suitable cartesian coordintates.
Message 3 of 9 , Jan 5, 2013
Dear Randy, in this case the concurrency can be completely proven by using computer algebra and suitable cartesian coordintates.

--- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
>
> Dear Antreas,
>
> I constructed random homothetic equilateral triangles and the Euler lines of A'B"C", B'C"A", C'A"B" (and of A"B'C', B"C'A', C"A'B') are concurrent (the three intersections coincide to at least 15 decimal places).
>
> Randy
>
• Dear Randy Note that the conjecture is not true if one of the six triangles A B C ,...., A B C is degenerated (if A ,B ,C etc are collinear). Figures here
Message 4 of 9 , Jan 6, 2013
Dear Randy

Note that the conjecture is not true if one of the six triangles A"B'C',...., A'B"C" is degenerated (if A",B',C' etc are collinear).

Figures here
http://tech.groups.yahoo.com/group/Hyacinthos/message/21357

APH

--- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
>
> Antreas,
>
> I was able to confirm this computationally.Â  Of course, not the same as a proof.
>
> Randy Hutson
>
>
>
>
>
> >________________________________
> > From: Antreas
> >To: Hyacinthos@yahoogroups.com
> >Sent: Friday, January 4, 2013 12:56 PM
> >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
> >
> >
> >Â
> >Let A'B'C', A"B"C" be two homothetic Equilateral triangles.
> >
> >Conjecture:
> >
> >The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
> >(and of A"B'C', B"C'A', C"A'B')
> >
> >True?
> >
> >APH
• Dear Antreas, The conjecture still holds if we extend the definition of Euler line to include degenerate triangles (e.g. A ,B ,C collinear) as: the line
Message 5 of 9 , Jan 6, 2013
Dear Antreas,

The conjecture still holds if we extend the definition of Euler line to include degenerate triangles (e.g. A",B',C' collinear) as: the line perpendicular to the line of vertices and passing through the centroid of the vertices.  This line meets the line at infinity at a point which is the 'circumcenter', 'orthocenter', and 'nine-point center' of the degenerate triangle.  In this case, the three Euler lines still concur.

Best regards,
Randy

>________________________________
> From: Antreas <anopolis72@...>
>To: Hyacinthos@yahoogroups.com
>Sent: Sunday, January 6, 2013 5:05 AM
>Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
>
>

>Dear Randy
>
>Note that the conjecture is not true if one of the six triangles A"B'C',...., A'B"C" is degenerated (if A",B',C' etc are collinear).
>
>Figures here
>http://tech.groups.yahoo.com/group/Hyacinthos/message/21357
>
>APH
>
>--- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
>>
>> Antreas,
>>
>> I was able to confirm this computationally.Â  Of course, not the same as a proof.
>>
>> Randy Hutson
>>
>>
>>
>>
>>
>> >________________________________
>> > From: Antreas
>> >To: Hyacinthos@yahoogroups.com
>> >Sent: Friday, January 4, 2013 12:56 PM
>> >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
>> >
>> >
>> >Â
>> >Let A'B'C', A"B"C" be two homothetic Equilateral triangles.
>> >
>> >Conjecture:
>> >
>> >The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
>> >(and of A"B'C', B"C'A', C"A'B')
>> >
>> >True?
>> >
>> >APH
>
>
>
>
>

[Non-text portions of this message have been removed]
• Dear Antreas, If ABC is equilateral triangle P = (p : q : r) in ABC barycentrics and A B C is the homothetic of ABC in the homothety PA = t.PA then the Euler
Message 6 of 9 , Jan 6, 2013
Dear Antreas,
If ABC is equilateral triangle P = (p : q : r) in ABC barycentrics
and A'B'C' is the homothetic of ABC in the homothety PA' = t.PA
then the Euler lines of A'BC, B'CA, C'AB are concurrent at
Q = {2 p^3 q^2 + p^2 q^3 + 5 p^3 q r + 8 p^2 q^2 r + 2 p q^3 r +
2 p^3 r^2 + 8 p^2 q r^2 + 5 p q^2 r^2 + p^2 r^3 + 2 p q r^3 +
2 p^4 q t + 2 p^2 q^3 t + p q^4 t + 2 p^4 r t + 4 p^3 q r t +
4 p^2 q^2 r t + 4 p q^3 r t + 4 p^2 q r^2 t + 4 p q^2 r^2 t +
q^3 r^2 t + 2 p^2 r^3 t + 4 p q r^3 t + q^2 r^3 t + p r^4 t -
p^4 q t^2 + 2 p^3 q^2 t^2 - p^4 r t^2 - 2 p^3 q r t^2 +
p q^3 r t^2 + q^4 r t^2 + 2 p^3 r^2 t^2 + 3 p q^2 r^2 t^2 +
q^3 r^2 t^2 + p q r^3 t^2 + q^2 r^3 t^2 + q r^4 t^2 - p^3 q^2 t^3 -
p^3 r^2 t^3 + q^3 r^2 t^3 + q^2 r^3 t^3,
p^3 q^2 + 2 p^2 q^3 + 2 p^3 q r + 8 p^2 q^2 r + 5 p q^3 r +
5 p^2 q r^2 + 8 p q^2 r^2 + 2 q^3 r^2 + 2 p q r^3 + q^2 r^3 +
p^4 q t + 2 p^3 q^2 t + 2 p q^4 t + 4 p^3 q r t + 4 p^2 q^2 r t +
4 p q^3 r t + 2 q^4 r t + p^3 r^2 t + 4 p^2 q r^2 t +
4 p q^2 r^2 t + p^2 r^3 t + 4 p q r^3 t + 2 q^2 r^3 t + q r^4 t +
2 p^2 q^3 t^2 - p q^4 t^2 + p^4 r t^2 + p^3 q r t^2 -
2 p q^3 r t^2 - q^4 r t^2 + p^3 r^2 t^2 + 3 p^2 q r^2 t^2 +
2 q^3 r^2 t^2 + p^2 r^3 t^2 + p q r^3 t^2 + p r^4 t^2 -
p^2 q^3 t^3 + p^3 r^2 t^3 - q^3 r^2 t^3 + p^2 r^3 t^3,
2 p^3 q r + 5 p^2 q^2 r + 2 p q^3 r + p^3 r^2 + 8 p^2 q r^2 +
8 p q^2 r^2 + q^3 r^2 + 2 p^2 r^3 + 5 p q r^3 + 2 q^2 r^3 +
p^3 q^2 t + p^2 q^3 t + p^4 r t + 4 p^3 q r t + 4 p^2 q^2 r t +
4 p q^3 r t + q^4 r t + 2 p^3 r^2 t + 4 p^2 q r^2 t +
4 p q^2 r^2 t + 2 q^3 r^2 t + 4 p q r^3 t + 2 p r^4 t + 2 q r^4 t +
p^4 q t^2 + p^3 q^2 t^2 + p^2 q^3 t^2 + p q^4 t^2 + p^3 q r t^2 +
3 p^2 q^2 r t^2 + p q^3 r t^2 + 2 p^2 r^3 t^2 - 2 p q r^3 t^2 +
2 q^2 r^3 t^2 - p r^4 t^2 - q r^4 t^2 + p^3 q^2 t^3 + p^2 q^3 t^3 -
p^2 r^3 t^3 - q^2 r^3 t^3}

The locus of Q is

-p^2 (p - q) q^2 (p - r) (q - r) r^2 (p + q + r)^11 (p q + p r +
q r) x^3 (p^6 q^2 x^3 - 5 p^4 q^4 x^3 + 4 p^2 q^6 x^3 -
2 p^6 q r x^3 + 5 p^4 q^3 r x^3 + 2 p^3 q^4 r x^3 + 4 p q^6 r x^3 +
p^6 r^2 x^3 - 2 p^3 q^3 r^2 x^3 + q^6 r^2 x^3 + 5 p^4 q r^3 x^3 -
2 p^3 q^2 r^3 x^3 - 8 p^2 q^3 r^3 x^3 - 4 p q^4 r^3 x^3 -
5 p^4 r^4 x^3 + 2 p^3 q r^4 x^3 - 4 p q^3 r^4 x^3 -
2 q^4 r^4 x^3 + 4 p^2 r^6 x^3 + 4 p q r^6 x^3 + q^2 r^6 x^3 -
3 p^6 q^2 x^2 y + 9 p^5 q^3 x^2 y + 12 p^4 q^4 x^2 y -
18 p^3 q^5 x^2 y - 9 p^5 q^2 r x^2 y + 3 p^4 q^3 r x^2 y -
9 p^2 q^5 r x^2 y + 6 p q^6 r x^2 y + 3 p^6 r^2 x^2 y -
6 p^3 q^3 r^2 x^2 y + 3 q^6 r^2 x^2 y - 12 p^4 q r^3 x^2 y +
24 p^3 q^2 r^3 x^2 y - 9 p^2 q^3 r^3 x^2 y + 6 p q^4 r^3 x^2 y -
3 p^4 r^4 x^2 y + 18 p^2 q^2 r^4 x^2 y - 6 p q^3 r^4 x^2 y -
6 p q r^6 x^2 y - 3 q^2 r^6 x^2 y - 18 p^5 q^3 x y^2 +
12 p^4 q^4 x y^2 + 9 p^3 q^5 x y^2 - 3 p^2 q^6 x y^2 +
6 p^6 q r x y^2 - 9 p^5 q^2 r x y^2 + 3 p^3 q^4 r x y^2 -
9 p^2 q^5 r x y^2 + 3 p^6 r^2 x y^2 - 6 p^3 q^3 r^2 x y^2 +
3 q^6 r^2 x y^2 + 6 p^4 q r^3 x y^2 - 9 p^3 q^2 r^3 x y^2 +
24 p^2 q^3 r^3 x y^2 - 12 p q^4 r^3 x y^2 - 6 p^3 q r^4 x y^2 +
18 p^2 q^2 r^4 x y^2 - 3 q^4 r^4 x y^2 - 3 p^2 r^6 x y^2 -
6 p q r^6 x y^2 + 4 p^6 q^2 y^3 - 5 p^4 q^4 y^3 + p^2 q^6 y^3 +
4 p^6 q r y^3 + 2 p^4 q^3 r y^3 + 5 p^3 q^4 r y^3 - 2 p q^6 r y^3 +
p^6 r^2 y^3 - 2 p^3 q^3 r^2 y^3 + q^6 r^2 y^3 - 4 p^4 q r^3 y^3 -
8 p^3 q^2 r^3 y^3 - 2 p^2 q^3 r^3 y^3 + 5 p q^4 r^3 y^3 -
2 p^4 r^4 y^3 - 4 p^3 q r^4 y^3 + 2 p q^3 r^4 y^3 - 5 q^4 r^4 y^3 +
p^2 r^6 y^3 + 4 p q r^6 y^3 + 4 q^2 r^6 y^3 + 3 p^6 q^2 x^2 z -
3 p^4 q^4 x^2 z - 12 p^4 q^3 r x^2 z - 6 p q^6 r x^2 z -
3 p^6 r^2 x^2 z - 9 p^5 q r^2 x^2 z + 24 p^3 q^3 r^2 x^2 z +
18 p^2 q^4 r^2 x^2 z - 3 q^6 r^2 x^2 z + 9 p^5 r^3 x^2 z +
3 p^4 q r^3 x^2 z - 6 p^3 q^2 r^3 x^2 z - 9 p^2 q^3 r^3 x^2 z -
6 p q^4 r^3 x^2 z + 12 p^4 r^4 x^2 z + 6 p q^3 r^4 x^2 z -
18 p^3 r^5 x^2 z - 9 p^2 q r^5 x^2 z + 6 p q r^6 x^2 z +
3 q^2 r^6 x^2 z - 6 p^6 q^2 x y z + 9 p^5 q^3 x y z -
6 p^4 q^4 x y z + 9 p^3 q^5 x y z - 6 p^2 q^6 x y z -
6 p^6 q r x y z + 18 p^5 q^2 r x y z + 6 p^4 q^3 r x y z +
6 p^3 q^4 r x y z + 18 p^2 q^5 r x y z - 6 p q^6 r x y z -
6 p^6 r^2 x y z + 18 p^5 q r^2 x y z - 36 p^4 q^2 r^2 x y z -
6 p^3 q^3 r^2 x y z - 36 p^2 q^4 r^2 x y z + 18 p q^5 r^2 x y z -
6 q^6 r^2 x y z + 9 p^5 r^3 x y z + 6 p^4 q r^3 x y z -
6 p^3 q^2 r^3 x y z - 6 p^2 q^3 r^3 x y z + 6 p q^4 r^3 x y z +
9 q^5 r^3 x y z - 6 p^4 r^4 x y z + 6 p^3 q r^4 x y z -
36 p^2 q^2 r^4 x y z + 6 p q^3 r^4 x y z - 6 q^4 r^4 x y z +
9 p^3 r^5 x y z + 18 p^2 q r^5 x y z + 18 p q^2 r^5 x y z +
9 q^3 r^5 x y z - 6 p^2 r^6 x y z - 6 p q r^6 x y z -
6 q^2 r^6 x y z - 3 p^4 q^4 y^2 z + 3 p^2 q^6 y^2 z -
6 p^6 q r y^2 z - 12 p^3 q^4 r y^2 z - 3 p^6 r^2 y^2 z +
18 p^4 q^2 r^2 y^2 z + 24 p^3 q^3 r^2 y^2 z - 9 p q^5 r^2 y^2 z -
3 q^6 r^2 y^2 z - 6 p^4 q r^3 y^2 z - 9 p^3 q^2 r^3 y^2 z -
6 p^2 q^3 r^3 y^2 z + 3 p q^4 r^3 y^2 z + 9 q^5 r^3 y^2 z +
6 p^3 q r^4 y^2 z + 12 q^4 r^4 y^2 z - 9 p q^2 r^5 y^2 z -
18 q^3 r^5 y^2 z + 3 p^2 r^6 y^2 z + 6 p q r^6 y^2 z +
3 p^6 q^2 x z^2 - 3 p^2 q^6 x z^2 + 6 p^6 q r x z^2 +
6 p^4 q^3 r x z^2 - 6 p^3 q^4 r x z^2 - 6 p q^6 r x z^2 -
9 p^5 q r^2 x z^2 - 9 p^3 q^3 r^2 x z^2 + 18 p^2 q^4 r^2 x z^2 -
18 p^5 r^3 x z^2 - 6 p^3 q^2 r^3 x z^2 + 24 p^2 q^3 r^3 x z^2 +
12 p^4 r^4 x z^2 + 3 p^3 q r^4 x z^2 - 12 p q^3 r^4 x z^2 -
3 q^4 r^4 x z^2 + 9 p^3 r^5 x z^2 - 9 p^2 q r^5 x z^2 -
3 p^2 r^6 x z^2 + 3 q^2 r^6 x z^2 - 3 p^6 q^2 y z^2 +
3 p^2 q^6 y z^2 - 6 p^6 q r y z^2 - 6 p^4 q^3 r y z^2 +
6 p^3 q^4 r y z^2 + 6 p q^6 r y z^2 + 18 p^4 q^2 r^2 y z^2 -
9 p^3 q^3 r^2 y z^2 - 9 p q^5 r^2 y z^2 + 24 p^3 q^2 r^3 y z^2 -
6 p^2 q^3 r^3 y z^2 - 18 q^5 r^3 y z^2 - 3 p^4 r^4 y z^2 -
12 p^3 q r^4 y z^2 + 3 p q^3 r^4 y z^2 + 12 q^4 r^4 y z^2 -
9 p q^2 r^5 y z^2 + 9 q^3 r^5 y z^2 + 3 p^2 r^6 y z^2 -
3 q^2 r^6 y z^2 + p^6 q^2 z^3 - 2 p^4 q^4 z^3 + p^2 q^6 z^3 +
4 p^6 q r z^3 - 4 p^4 q^3 r z^3 - 4 p^3 q^4 r z^3 +
4 p q^6 r z^3 + 4 p^6 r^2 z^3 - 8 p^3 q^3 r^2 z^3 +
4 q^6 r^2 z^3 + 2 p^4 q r^3 z^3 - 2 p^3 q^2 r^3 z^3 -
2 p^2 q^3 r^3 z^3 + 2 p q^4 r^3 z^3 - 5 p^4 r^4 z^3 +
5 p^3 q r^4 z^3 + 5 p q^3 r^4 z^3 - 5 q^4 r^4 z^3 + p^2 r^6 z^3 -
2 p q r^6 z^3 + q^2 r^6 z^3)

and the Euler lines of AB'C', BC'A', CA'B' are concurrent at

{2 p^3 q^2 + 2 p^3 r^2 + 3 p q^2 r^2 + q^3 r^2 + q^2 r^3 + 2 p^4 q t -
p^3 q^2 t + 3 p^2 q^3 t + 2 p^4 r t + 4 p^3 q r t +
6 p^2 q^2 r t + 4 p q^3 r t + q^4 r t - p^3 r^2 t + 6 p^2 q r^2 t +
q^3 r^2 t + 3 p^2 r^3 t + 4 p q r^3 t + q^2 r^3 t + q r^4 t -
p^4 q t^2 + 3 p^3 q^2 t^2 - p^2 q^3 t^2 + p q^4 t^2 - p^4 r t^2 +
7 p^3 q r t^2 + 7 p^2 q^2 r t^2 + p q^3 r t^2 + 3 p^3 r^2 t^2 +
7 p^2 q r^2 t^2 + 7 p q^2 r^2 t^2 + q^3 r^2 t^2 - p^2 r^3 t^2 +
p q r^3 t^2 + q^2 r^3 t^2 + p r^4 t^2 - p^3 q^2 t^3 + p^2 q^3 t^3 -
4 p^3 q r t^3 - p^2 q^2 r t^3 + 2 p q^3 r t^3 - p^3 r^2 t^3 -
p^2 q r^2 t^3 + 2 p q^2 r^2 t^3 + p^2 r^3 t^3 + 2 p q r^3 t^3,
2 p^2 q^3 + p^3 r^2 + 3 p^2 q r^2 + 2 q^3 r^2 + p^2 r^3 +
3 p^3 q^2 t - p^2 q^3 t + 2 p q^4 t + p^4 r t + 4 p^3 q r t +
6 p^2 q^2 r t + 4 p q^3 r t + 2 q^4 r t + p^3 r^2 t +
6 p q^2 r^2 t - q^3 r^2 t + p^2 r^3 t + 4 p q r^3 t + 3 q^2 r^3 t +
p r^4 t + p^4 q t^2 - p^3 q^2 t^2 + 3 p^2 q^3 t^2 - p q^4 t^2 +
p^3 q r t^2 + 7 p^2 q^2 r t^2 + 7 p q^3 r t^2 - q^4 r t^2 +
p^3 r^2 t^2 + 7 p^2 q r^2 t^2 + 7 p q^2 r^2 t^2 + 3 q^3 r^2 t^2 +
p^2 r^3 t^2 + p q r^3 t^2 - q^2 r^3 t^2 + q r^4 t^2 + p^3 q^2 t^3 -
p^2 q^3 t^3 + 2 p^3 q r t^3 - p^2 q^2 r t^3 - 4 p q^3 r t^3 +
2 p^2 q r^2 t^3 - p q^2 r^2 t^3 - q^3 r^2 t^3 + 2 p q r^3 t^3 +
q^2 r^3 t^3,
p^3 q^2 + p^2 q^3 + 3 p^2 q^2 r + 2 p^2 r^3 + 2 q^2 r^3 + p^4 q t +
p^3 q^2 t + p^2 q^3 t + p q^4 t + 4 p^3 q r t + 4 p q^3 r t +
3 p^3 r^2 t + 6 p^2 q r^2 t + 6 p q^2 r^2 t + 3 q^3 r^2 t -
p^2 r^3 t + 4 p q r^3 t - q^2 r^3 t + 2 p r^4 t + 2 q r^4 t +
p^3 q^2 t^2 + p^2 q^3 t^2 + p^4 r t^2 + p^3 q r t^2 +
7 p^2 q^2 r t^2 + p q^3 r t^2 + q^4 r t^2 - p^3 r^2 t^2 +
7 p^2 q r^2 t^2 + 7 p q^2 r^2 t^2 - q^3 r^2 t^2 + 3 p^2 r^3 t^2 +
7 p q r^3 t^2 + 3 q^2 r^3 t^2 - p r^4 t^2 - q r^4 t^2 +
2 p^3 q r t^3 + 2 p^2 q^2 r t^3 + 2 p q^3 r t^3 + p^3 r^2 t^3 -
p^2 q r^2 t^3 - p q^2 r^2 t^3 + q^3 r^2 t^3 - p^2 r^3 t^3 -
4 p q r^3 t^3 - q^2 r^3 t^3}

ND

> Let A'B'C', A"B"C" be two homothetic
> Equilateral triangles.
>
> Conjecture:
>
> The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
> (and of A"B'C', B"C'A', C"A'B')
>
> True?
>
> APH
• Dear Nikos Thanks!! Using this Therorem we can find new Triangle centers, but with very complicated coordinates, I guess! A family of triangle equilateral
Message 7 of 9 , Jan 7, 2013
Dear Nikos

Thanks!!

Using this Therorem we can find new Triangle centers, but with very
complicated coordinates, I guess!
A family of triangle equilateral triangles is Morley triangle and its homothetics. We have already discussed the case of the 1st Morley and Russel triangles. It will be interesting if the two centers we get are on the Euler line of the reference triangle (remarked by Frank Jackson, Hyacinthos #21354)

APH

--- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades wrote:
>
> Dear Antreas,
> If ABC is equilateral triangle P = (p : q : r) in ABC barycentrics
> and A'B'C' is the homothetic of ABC in the homothety PA' = t.PA
> then the Euler lines of A'BC, B'CA, C'AB are concurrent at
> Q = ......

>
> The locus of Q is ...........

>
> and the Euler lines of AB'C', BC'A', CA'B' are concurrent at......
>
> ND
>
> > Let A'B'C', A"B"C" be two homothetic
> > Equilateral triangles.
> >
> > Conjecture:
> >
> > The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
> > (and of A"B'C', B"C'A', C"A'B')
> >
> > True?
> >
> > APH
>
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