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

Expand Messages
• 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 1 of 9 , Jan 6, 2013
• 0 Attachment
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 2 of 9 , Jan 6, 2013
• 0 Attachment
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 3 of 9 , Jan 6, 2013
• 0 Attachment
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 4 of 9 , Jan 7, 2013
• 0 Attachment
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
>
Your message has been successfully submitted and would be delivered to recipients shortly.