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22166Fwd: [EGML] Re: Envelopes

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  • Antreas Hatzipolakis
    Apr 21, 2014
    • 0 Attachment


      ---------- Forwarded message ----------
      From: César Lozada 
      Date: Mon, Apr 21, 2014 at 4:36 PM
      Subject: RE: [EGML] Re: Envelopes
      To: Anopolis@yahoogroups.com


       

       

      Please forget that A’B’C’ may be the medial triangle of ABC in the last answer. In such case,  the reflected lines concur on a point on the NPC of ABC.

       

      Apologizes

      César Lozada


      De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com] En nombre de César Lozada
      Enviado el: Lunes, 21 de Abril de 2014 08:36 a.m.
      Para: Anopolis@yahoogroups.com
      Asunto: RE: [EGML] Re: Envelopes

       

       

      Special cases:

       

      If A’B’C’ is the medial or the antimedial triangle of ABC, the fixed point on Euler lines has trilinear coordinates:

       

       

      (2*f(6)+6*f(4)+14*f(2)+7)*g(1)

      -2*(2*f(5)+2*f(3)+5*f(1))*g(2)

      +(2*f(2)+2*f(4)+3)*g(3)

      -2*(f(3)+f(1))*g(4)

      -(f(7)+f(5)+6*f(3)+6*f(1)) : :

       

      Where f(n)=cos(n*A) and g(n)=cos(n*(B-C))

       

      This point is on line  (1141,3484) and has ETC-6-9-13 numbers:

      (4.923838044604385591130, 4.248841971282335390013, -1.573382134182338747412)

       

      Best regards

      César Lozada

       

       


      De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com] En nombre de Antreas Hatzipolakis
      Enviado el: Lunes, 21 de Abril de 2014 05:52 a.m.
      Para: anopolis@yahoogroups.com; Hyacinthos
      Asunto: [EGML] Re: Envelopes

       

       

      Conjecture:

      Let A'B'C' be the antimedial triangle of ABC. The reflections of the parallel lines L1,L2,L3 trough A,B,C, resp. in the sidelines of A'B'C' (instead of ABC) bound a triangle whose the Euler line passes through a fixed point.

      In general:

      Let ABC, A'B'C' be two homothetic triangles. Let L1,L2,L3 be three parallel lines through A,B,C, resp. The reflections of L1,L2,L3 in the sidelines B'C',C'A',A'B', resp. of A'B'C' bound a triangle whose the Euler line passes through a fixed point.

      http://anthrakitis.blogspot.gr/2014/04/euler-lines-of-triagles-bounded-by.html

       

      On Sun, Apr 20, 2014 at 10:59 PM, Antreas Hatzipolakis <anopolis72@...> wrote:

       

       

      On Sun, Apr 20, 2014 at 8:13 PM, Antreas Hatzipolakis <anopolis72@...> wrote:



      The problem is:
      Let ABC be a triangle and L1,L2,L3 three parallel lines through A,B,C, resp. and let A'B'C' be the triangle bounded by the reflections of L1,L2,L3, in BC,CA,AB, resp.
      As the lines L1,L2,L3 move around A,B,C, being parallel, which is the envelope of the Euler line of A'B'C'?
      Furthermore, which is the envelope of the circumcircle of A'B'C'?

      APH

       

       

      The problem was posed by the friend Dao Thanh Oai, in my wording:

      Let ABC be a triangle and L1,L2,L3 three parallel lines through A,B,C, resp. and let A'B'C' be the triangle bounded by the reflections of L1,L2,L3, in BC,CA,AB, resp.

      As the lines L1,L2,L3 move around A,B,C, the Euler line of the triangle
      A1B1C1 passes through a fixed point.

      At first glance I thought that if it was true, then the fixed point should be

      the Parry Point X(399), the point the reflections in BC,CA,AB of the parallels

      to Euler line through A,B,C, resp. concur at.

      But I was not right!!! If the lines L1,L2,L3 are parallels to Euler line,

      then the triangle A1B1C1 is degenerated in Parry point, and its Euler line

      is ANY line passing through Parry point!

      The fixed point has barycentrics (computed by Francisco Javier García Capitán):

      The point is (f(a,b,c):f(b,c,a):f(c,a,b)) where f(a,b,c) is
      a^22 - 8 a^20 b^2 + 28 a^18 b^4 - 56 a^16 b^6 + 70 a^14 b^8 -
      56 a^12 b^10 + 28 a^10 b^12 - 8 a^8 b^14 + a^6 b^16 - 8 a^20 c^2 +
      42 a^18 b^2 c^2 - 92 a^16 b^4 c^2 + 106 a^14 b^6 c^2 -
      62 a^12 b^8 c^2 + 7 a^10 b^10 c^2 + 13 a^8 b^12 c^2 -
      8 a^6 b^14 c^2 + 4 a^4 b^16 c^2 - 3 a^2 b^18 c^2 + b^20 c^2 +
      28 a^18 c^4 - 92 a^16 b^2 c^4 + 113 a^14 b^4 c^4 - 62 a^12 b^6 c^4 +
      17 a^10 b^8 c^4 - 9 a^8 b^10 c^4 + 5 a^6 b^12 c^4 - 6 a^4 b^14 c^4 +
      13 a^2 b^16 c^4 - 7 b^18 c^4 - 56 a^16 c^6 + 106 a^14 b^2 c^6 -
      62 a^12 b^4 c^6 + 4 a^10 b^6 c^6 + 4 a^8 b^8 c^6 + 8 a^6 b^10 c^6 -
      6 a^4 b^12 c^6 - 18 a^2 b^14 c^6 + 20 b^16 c^6 + 70 a^14 c^8 -
      62 a^12 b^2 c^8 + 17 a^10 b^4 c^8 + 4 a^8 b^6 c^8 - 12 a^6 b^8 c^8 +
      8 a^4 b^10 c^8 + 3 a^2 b^12 c^8 - 28 b^14 c^8 - 56 a^12 c^10 +
      7 a^10 b^2 c^10 - 9 a^8 b^4 c^10 + 8 a^6 b^6 c^10 + 8 a^4 b^8 c^10 +
      10 a^2 b^10 c^10 + 14 b^12 c^10 + 28 a^10 c^12 + 13 a^8 b^2 c^12 +
      5 a^6 b^4 c^12 - 6 a^4 b^6 c^12 + 3 a^2 b^8 c^12 + 14 b^10 c^12 -
      8 a^8 c^14 - 8 a^6 b^2 c^14 - 6 a^4 b^4 c^14 - 18 a^2 b^6 c^14 -
      28 b^8 c^14 + a^6 c^16 + 4 a^4 b^2 c^16 + 13 a^2 b^4 c^16 +
      20 b^6 c^16 - 3 a^2 b^2 c^18 - 7 b^4 c^18 + b^2 c^20

      https://www.facebook.com/photo.php?fbid=1498558183700588&set=gm.437719496364497&type=1&theater


       

       




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