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

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• 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 1 of 9 , Jan 7, 2013
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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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