HYACINTHOS 15827

• Francisco Javier García Capitán

  Dear Cosmin, the first intersection point is X(399).
  
  The second one doesn't appear in ETC.
  Its homogeneus barycentric coordinates are shown below (in LaTeX
  format).
  
  Francisco Javier.
  
  ---------
  
  \left\{a^2 \left(a^{20}-6 b^2 a^{18}-6 c^2 a^{18}+14 b^4 a^{16}+14
  c^4 a^{16}+24 b^2 c^2 a^{16}-14 b^6 a^{14}-14 c^6 a^{14}-34
  b^2 c^4 a^{14}-34 b^4 c^2 a^{14}+18 b^2 c^6 a^{12}+25 b^4 c^4 a^
  {12}+18 b^6 c^2 a^{12}+14 b^{10} a^{10}+14 c^{10} a^{10}-b^2
  c^8 a^{10}-7 b^4 c^6 a^{10}-7 b^6 c^4 a^{10}-b^8 c^2 a^{10}-14 b^
  {12} a^8-14 c^{12} a^8+b^2 c^{10} a^8+5 b^4 c^8 a^8-2 b^6 c^6
  a^8+5 b^8 c^4 a^8+b^{10} c^2 a^8+6 b^{14} a^6+6 c^{14} a^6-4 b^2 c^
  {12} a^6-3 b^4 c^{10} a^6+b^6 c^8 a^6+b^8 c^6 a^6-3 b^{10}
  c^4 a^6-4 b^{12} c^2 a^6-b^{16} a^4-c^{16} a^4+4 b^2 c^{14} a^4-4
  b^4 c^{12} a^4+6 b^6 c^{10} a^4-10 b^8 c^8 a^4+6 b^{10} c^6
  a^4-4 b^{12} c^4 a^4+4 b^{14} c^2 a^4-3 b^2 c^{16} a^2+8 b^4 c^
  {14} a^2-6 b^6 c^{12} a^2+b^8 c^{10} a^2+b^{10} c^8 a^2-6
  b^{12} c^6 a^2+8 b^{14} c^4 a^2-3 b^{16} c^2 a^2+b^2 c^{18}-4 b^4
  c^{16}+4 b^6 c^{14}+4 b^8 c^{12}-10 b^{10} c^{10}+4 b^{12}
  c^8+4 b^{14} c^6-4 b^{16} c^4+b^{18} c^2\right),b^2 \left(b^{20}-6
  a^2 b^{18}-6 c^2 b^{18}+14 a^4 b^{16}+14 c^4 b^{16}+24 a^2
  c^2 b^{16}-14 a^6 b^{14}-14 c^6 b^{14}-34 a^2 c^4 b^{14}-34 a^4
  c^2 b^{14}+18 a^2 c^6 b^{12}+25 a^4 c^4 b^{12}+18 a^6 c^2
  b^{12}+14 a^{10} b^{10}+14 c^{10} b^{10}-a^2 c^8 b^{10}-7 a^4 c^6
  b^{10}-7 a^6 c^4 b^{10}-a^8 c^2 b^{10}-14 a^{12} b^8-14
  c^{12} b^8+a^2 c^{10} b^8+5 a^4 c^8 b^8-2 a^6 c^6 b^8+5 a^8 c^4
  b^8+a^{10} c^2 b^8+6 a^{14} b^6+6 c^{14} b^6-4 a^2 c^{12}
  b^6-3 a^4 c^{10} b^6+a^6 c^8 b^6+a^8 c^6 b^6-3 a^{10} c^4 b^6-4 a^
  {12} c^2 b^6-a^{16} b^4-c^{16} b^4+4 a^2 c^{14} b^4-4 a^4
  c^{12} b^4+6 a^6 c^{10} b^4-10 a^8 c^8 b^4+6 a^{10} c^6 b^4-4 a^
  {12} c^4 b^4+4 a^{14} c^2 b^4-3 a^2 c^{16} b^2+8 a^4 c^{14}
  b^2-6 a^6 c^{12} b^2+a^8 c^{10} b^2+a^{10} c^8 b^2-6 a^{12} c^6
  b^2+8 a^{14} c^4 b^2-3 a^{16} c^2 b^2+a^2 c^{18}-4 a^4
  c^{16}+4 a^6 c^{14}+4 a^8 c^{12}-10 a^{10} c^{10}+4 a^{12} c^8+4 a^
  {14} c^6-4 a^{16} c^4+a^{18} c^2\right),c^2 \left(c^{20}-6
  a^2 c^{18}-6 b^2 c^{18}+14 a^4 c^{16}+14 b^4 c^{16}+24 a^2 b^2 c^
  {16}-14 a^6 c^{14}-14 b^6 c^{14}-34 a^2 b^4 c^{14}-34 a^4 b^2
  c^{14}+18 a^2 b^6 c^{12}+25 a^4 b^4 c^{12}+18 a^6 b^2 c^{12}+14 a^
  {10} c^{10}+14 b^{10} c^{10}-a^2 b^8 c^{10}-7 a^4 b^6
  c^{10}-7 a^6 b^4 c^{10}-a^8 b^2 c^{10}-14 a^{12} c^8-14 b^{12}
  c^8+a^2 b^{10} c^8+5 a^4 b^8 c^8-2 a^6 b^6 c^8+5 a^8 b^4
  c^8+a^{10} b^2 c^8+6 a^{14} c^6+6 b^{14} c^6-4 a^2 b^{12} c^6-3
  a^4 b^{10} c^6+a^6 b^8 c^6+a^8 b^6 c^6-3 a^{10} b^4 c^6-4
  a^{12} b^2 c^6-a^{16} c^4-b^{16} c^4+4 a^2 b^{14} c^4-4 a^4 b^{12}
  c^4+6 a^6 b^{10} c^4-10 a^8 b^8 c^4+6 a^{10} b^6 c^4-4
  a^{12} b^4 c^4+4 a^{14} b^2 c^4-3 a^2 b^{16} c^2+8 a^4 b^{14} c^2-
  6 a^6 b^{12} c^2+a^8 b^{10} c^2+a^{10} b^8 c^2-6 a^{12} b^6
  c^2+8 a^{14} b^4 c^2-3 a^{16} b^2 c^2+a^2 b^{18}-4 a^4 b^{16}+4
  a^6 b^{14}+4 a^8 b^{12}-10 a^{10} b^{10}+4 a^{12} b^8+4 a^{14}
  b^6-4 a^{16} b^4+a^{18} b^2\right)\right\}
  
  
  
  
  
  --- In Hyacinthos@yahoogroups.com, "Cosmin Pohoata"
  <pohoata_cosmin2000@...> wrote:
  

  >
  > Dear friends,
  >
  > If A', B', C' are the reflections of the vertices A, B, C in the
  > sidelines BC, CA, AB and XYZ is the tangential triangle of ABC,
  

  then
  

  > the circles
  > a) B'C'X, C'A'Y, A'B'Z (*)
  > b) A'YZ, B'ZX, C'XY (**)
  > are concurrent.
  >
  > I didn't finish the computations in homogenous barycentric
  > coordinates. (I stopped at finding the p1, q1, r1 from the equation
  > a^2xy+b^2yz+c^2zx+(x+y+z)(p1x+q1y+r1z)=0 of the circle B'C'X; for
  

  the
  

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  > ones who want to continue them by hand, the p1, q1, r1 verify the
  > system of equations
  > (1) 2S_C p1 - b^2(2S_B+q1) + 2S_A r1 = 0
  > (2) 2S_B p1 + 2S_A q1 - c^2(2S_C+r1) = 0
  > (3) a^2b^2c^2+(b^2+c^2)(q1b^2+r1c^2) = 0...)
  >
  > I wonder if the concurrences from (*), (**) can be related to the
  > nine-point circle as in the quite similar problem involving the
  > inversive image of the Kosnita point in the circumcircle. [P. Yiu,
  > Hyacinthos message #4533].
  >
  > Best regards,
  > Cosmin
  >