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22025Fwd: [EGML] An image made from the midpoints of a triangle-derived hexagon

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  • Antreas Hatzipolakis
    Dec 9, 2013
    • 0 Attachment


      ---------- Forwarded message ----------
      From: César Lozada
      Date: Mon, Dec 9, 2013 at 5:37 PM
      Subject: RE: [EGML] An image made from the midpoints of a triangle-derived hexagon
      To: Anopolis@yahoogroups.com


       

      Dear Seichii:

       

      I see clearly why we didn’t coincide: I misread your construction and called Ab what you called Ba, and so on.

       

      Let’s assume your names for intersections of parallel lines with the sides of ABC. In fact, we have two different points where parallel lines concur:

       

      Yours:  Q = line(midpoint( Ab,Ac),midpoint(Bc,Cb)) /\ line(midpoint( Ba,Bc),midpoint(Ac,Ca)) /\ line(midpoint( Ca,Cb),midpoint(Ab,Ba))

       

      Mine:  Q’ = line(midpoint( Ba,Ca),midpoint(Bc,Cb)) /\ line(midpoint( Ab,Cb),midpoint(Ac,Ca)) /\ line(midpoint( Ac,Bc),midpoint(Ab,Ba))

       

      So, for P=[p : q : r ] (barycentrics)

       

      Q  = [ (2*p+q+r)(p^2+2*p*(q+r)+q*r : :  ]

       

      Q’ = [ (r+q)*(p^2+r*q)+2*p*q*r :  : ]

       

       

      If P’ = isotomic conjugate of P, then

      1)       Q and Q’ are on line ( P, P’ )

      2)       Q’ = midpoint(P,P’)

      3)       IF P lies on line at infinity, then Q=Q’=P

       

       

      ETC-pairs (P,Q):

      (2,2), (8,3696), (20,1350), (144,5223), (145,3243), (148,523), (192,3993), (193,1351), (4440,514)

       

      ETC-pairs (P,Q’):

       (2,2),  (4,1352), (7,2550), (8,2550), (13,623), (14,624),     

       (69,1352), (75,30002), (99,523), (190,514), (264,30003), (290,511), (298,623),     

       (299,624), (648,525), (664,522), (666,918), (668,513), (670,512), (671,524),        

       (886,888), (889,891), (892,690), (903,519), (1121,527), (1494,30), (2481,518),     

       (2966,2799), (3225,698), (3226,726), (3227,536), (3228,538), (4555,900), (4562,812),

       (4569,3900), (4577,826), (4586,824), (4597,4777)  

       

      Regards

      César Lozada

       

       

       

       

       

       

       

       


      De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com] En nombre de seiichi kirikami
      Enviado el: Lunes, 09 de Diciembre de 2013 02:53 a.m.
      Para: Anopolis@yahoogroups.com
      Asunto: Re: [EGML] An image made from the midpoints of a triangle-derived hexagon [1 Attachment]

       

       

      Dear Cesar,

       

      I mean the lines through P( example, X(1)) parallel to BC, CA and AB. See the attached picture. Originally I made Cabri picture of a=6, b=9 and c=13 and measured QR, which coincided with the computed value. Q is the concurrent point and R is the pedal point of Q on BC.

      (In the previous email, my P and Q are barycentric.)

       

      Best regards,

      Seiichi.

       

      On Sun, Dec 8, 2013 at 7:23 PM, César Lozada <cesar_e_lozada@...> wrote:

       

      Dear Seichii:

       

      >> we denote by Ba and Ca the intersections of the line parallel to BC with AB and BC respectively….

       

      What parallel line?

       

      I assume you didn’t mean “the line parallel through P” because, in such case I get comcurrence occurs always at:

        Q = ((b*q+c*r)*(p^2*a^2+c*b*q*r)+2*b*c*q*r*p*a)/a ::  for P=p:q:r  (trilinears)

       

      and, for P=X(1)

        Q(X(1))= ((b+c)*a^2+2*a*b*c+b*c*(b+c))/a ::

      also non-ETC with ETC-search: 2.698384066352449657972..

       

       

      Regards

      César Lozada

       

       

       


      De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com] En nombre de seiichi kirikami
      Enviado el: Domingo, 08 de Diciembre de 2013 01:54 a.m.
      Para: Anopolis@yahoogroups.com
      Asunto: [EGML] An image made from the midpoints of a triangle-derived hexagon

       

       

      Dear friends,

       

      Given a triangle ABC and a point P, we denote by Ba and Ca the intersections of the line parallel to BC with AB and BC respectively. The other intersections Cb, Ab and Ac, Bc are defined cyclically. The lines through the midpoints of AcAb and BcCb, BaBc and CaAc, CbCa and AbBa concur in a point Q.

      If P={p, q, r}, then Q={(2p+q+r)(p^2+2p(q+r)+qr,  ,  }.

      Example: If P=X(2), Q=X(2). If P=X(1), Q={(2a+b+c)(a^2+2a(b+c)+bc),  ,  }. The latter for a=6, b=9 and c=13 gives kx=1.93330058.. (non ETC).

       

      I confirmed that this image was not a quadrangle point.

       

      Best regards,

      Seiichi.

       







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