HYACINTHOS 1190

• Hello Hyacinthos'members,
  it's my first contribution:
  do you know the following result?
  
  The perpendicular bisectors of the sides of the triangle ABC cut the
  other sides at 6
  points lying on 3 circles. (4 points on one circle)
  The centers ot these circles form a triangle homothetic to ABC.
  The center of homothety is point X(184) of the encyclopedia of
  triangle centers.
  The radical center of the 3 circles is the center of the circumcircle
  of ABC.
  
  The trilinear coordinates of the centers are:
  
  1)
  {-a^5 + (b^2 + c^2)*a^3 - 2*b^2*c^2*a , b^3*(a^2 - b^2 + c^2) ,
  c^3*(a^2 +
  b^2 - c^2)}
  =
  {-a^4 + (b^2 + c^2)*a^2 - 2*b^2*c^2, 2*b^3*c*cosB, 2*b*c^3*cosC}
  
  2)
  {a^3*(-a^2 + b^2 + c^2) , -b^5 + c^2*b^3 + a^2*(b^3 - 2*b*c^2) ,
  c^3*(a^2
  + b^2 - c^2)}
  =
  {2*a^3*c*cosA, -b^4 + c^2*b^2 + a^2*(b^2 - 2*c^2), 2*a*c^3*cosC}
  
  3)
  {a^3*(-a^2 + b^2 + c^2) , b^3*(a^2 - b^2 + c^2) , (b^2 - c^2)*c^3 +
  a^2*(
  c^3 - 2*b^2*c)}
  =
  {2*a^3*b*c*cosA, 2*a*b^3*c*cosB, (b^2 - c^2)*c^3 + a^2*(c^3 - 2*b^2*c)}
  
  The center or homothety is X(184) = {a^2 cosA, b^2 cosB, c^2 cosC}.
  
  Regards
  
  Fred Lang, Switzerland