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[EMHL] Re: A concurrency at X1659

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  • Antreas
    Dear Francisco, Angel, Randy Let (Ma), (Mb), (Mc) be the three circles with diameters BC,CA,AB, resp. and (S),(T) the two circles touching (Ma),(Mb),(Mc)
    Message 1 of 7 , Feb 14, 2013
      Dear Francisco, Angel, Randy

      Let (Ma), (Mb), (Mc) be the three circles with diameters
      BC,CA,AB, resp. and (S),(T) the two circles touching
      (Ma),(Mb),(Mc) internally, externally, resp.

      1. Are the centers S, T interesting ? (are they lying on some
      interesting line / curve?)

      2. Let Sa, Sb, Sc the points of contact of (S) with (Ma),(Mb),(Mc),
      resp. and Ta,Tb,Tc of (T).
      Are these triangles perspective?
      2a. ABC, SaSbSc
      2b. ABC, TaTbTc
      2c. SaSbSc, TaTbTc

      APH

      --- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
      >
      > Dear Angel and Antreas,
      >
      > Non-ETC -1.2653434427 matches the isogonal conjugate of a center I recently submitted to ETC as:
      >
      > 3rd KENMOTU HOMOTHETIC CENTER
      >
      >     Trilinears    1 - sin A + cos A : :
      >
      >     Congruent squares U', V', W' can be constructed like U, V, W described at X(371), but with 2 vertices each on the extended sides of ABC, and having common vertex at X(372).
      >     Let La be the extended diagonal of U' that does not contain X(372), and define Lb, Lc cyclically.
      >     Let A' = Lb∩Lc, B' = Lc∩La, C' = La∩Lb.  The triangle A'B'C' is homothetic to the intangents triangle at this point.
      >
      >     Lies on hyperbola {A,B,C,X(1),X(3),...,X(2066)} and lines 1,372 3,1335 6,31.
      >     {X(6),X(55)}-harmonic conjugate of X(2066)
      >     {X(3),X(1335)}-harmonic conjugate of X(2067)
      >
      >     ETC search: 4.73298538410457
      >
      > Therefor, the trilinears of the perspector of ABC and KaKbKc should be 1/(1 - sin A + cos A) : :.
      >
      > Best regards,
      > Randy
      >
      >
      >
      >
      >
      >
      > >________________________________
      > > From: Angel
      > >To: Hyacinthos@yahoogroups.com
      > >Sent: Thursday, February 14, 2013 5:25 PM
      > >Subject: [EMHL] Re: A concurrency at X1659
      > >
      > >
      > > 
      > >Dear Antreas,
      > >
      > >ABC is perspective to K1K2K3 at X(1659) (As Francisco Javier says).
      > >
      > >ABC is perspective to KaKbKc at non-ETC -1.2653434427 (coordinates?).
      > >
      > >K1K2K3 is perspective to KaKbKc at the orthocenter.
      > >
      > >Best regards.
      > >Angel M.
      > >
      > >--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:
      > >>
      > >> If K1,K2,K3 are the centers of the circles and Ka,Kb,Kc their "brothers"
      > >> (ie the centers of the circles touching two of the original externally and
      > >> the third one internally),
      > >> are then K1K2K3, KaKbKc perspective?
      > >>
      > >> APH
      > >>
      > >> On Thu, Feb 14, 2013 at 10:34 PM, Francisco Javier
      > >> wrote:
      > >>
      > >> > **
      > >> >
      > >> >
      > >> > Dear friends:
      > >> >
      > >> > A concurrency at X1659.
      > >> > Given a triangle ABC, construct the circles with BC, CA, AB as diameters,
      > >> > then construct the three circles touching internally two of them and
      > >> > externally the third one. The lines joining the centers of these triangles
      > >> > and the corresponding vertices concur at X1659 (Yiu-Paasche point).
      > >> >
      > >> > See:
      > >> >
      > >> > http://garciacapitan.blogspot.com.es/2013/02/a-concurrency-at-x1659.html
      > >> >
      > >> > Love for everybody,
      > >> >
      > >> > Francisco Javier.
      > >> >
      > >> >
      > >> >
      > >>
      > >>
      > >>
      > >> --
      > >> http://anopolis72000.blogspot.com/
      > >>
      > >>
      > >> [Non-text portions of this message have been removed]
      > >>
      > >
      > >
      > >
      > >
      > >
      >
      > [Non-text portions of this message have been removed]
      >
    • rhutson2
      Also, V = {X(1),X(226)}-harmonic conjugate of X(1659) V = {X(4),X(278)}-harmonic conjugate of X(1659) And note the similarities in the trilinears for V and
      Message 2 of 7 , Feb 15, 2013
        Also,

        V = {X(1),X(226)}-harmonic conjugate of X(1659)
        V = {X(4),X(278)}-harmonic conjugate of X(1659)

        And note the similarities in the trilinears for V and U(=X(1659):

        V = 1/(1 - sin A + cos A) : :
        U = 1/(1 + sin A + cos A) : :

        Randy

        --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
        >
        > Correction (Antreas, you can delete message #21546)
        >
        > The coordinates of the ISOTOMIC CONJUGATE OF perspector V of KaKbKc are
        >
        > {(a^2 - b^2 - c^2) (a^3 - a b^2 - 2 a b c - a c^2 - 2 b S - 2 c S),
        > (a^2 - b^2 + c^2) (a^2 b - b^3 + 2 a b c + b c^2 + 2 a S + 2 c S),
        > (a^2 + b^2 - c^2) (a^2 c + 2 a b c + b^2 c - c^3 + 2 a S + 2 b S)}
        >
        > while those of the ISOTOMIC CONJUGATE OF perspector U of K1K2K3 (that is, U=X1659) are
        >
        > {(a^2 - b^2 - c^2) (a^3 - a b^2 - 2 a b c - a c^2 + 2 b S + 2 c S),
        > (a^2 - b^2 + c^2) (a^2 b - b^3 + 2 a b c + b c^2 - 2 a S - 2 c S),
        > (a^2 + b^2 - c^2) (a^2 c + 2 a b c + b^2 c - c^3 - 2 a S - 2 b S)}
        >
        > Both U and V lie on line IH and we have the cross ratio
        > (IHUV)=(2r^2+4rR-S)/(2r^2+4rR+S) where S is twice the area of ABC.
        >
        > Best regards,
        >
        > Francisco Javier.
        >
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