HYACINTHOS 2070

• Dear Antreas,
  
  [APH]: Let's make the things more complicated!
  
  Construct a circle touching the circumcircle of ABC at A,
  and the sideline BC.
  Let Ba, Ca be the points that the circle in question intersects
  the AB, AC, respectively.
  
  The triangle ABaCa is homothetic to ABC.
  ...
  
  If we similarly define the line segments CbAb, AcBc
  (corresponding two other two circles of the triad) then these three
  line segments (BaCa, CbAb, AcBc) either concur or bound a triangle
  homothetic to ABC.
  
  [PY]: ... the homothetic center is the point
  
  (a^2/(b+c)^2 : b^2/(c+a)^2 : c^2/(a+b)^2),
  
  a point not in ETC.
  
  Now, how about the radical center of the three circles?
  
  *******
  
  Let's do this.
  
  Write the equation of the A-circle as
  
  aayz + bbzx+ ccxy - k(x+y+z)(y/bb + z/cc) = 0.
  
  Since this circle passes through (0:b:c),
  
  k = aabc/((b+c)(1/b+1/c)) = aabbcc/(b+c)^2.
  
  Similarly, we find the equations of the other two circles.
  Their radical center is the point (x:y:z) satisfying
  
  1/(b+c)^2(y/bb+z/cc) = 1/(c+a)^2(z/cc+x/aa) = 1/(a+b)^2(y/bb+z/cc)
  
  From these, it is clear that
  
  m(x/aa:y/bb:z/cc) = ((b+c)^2:(c+a)^2:(a+b)^2)
  
  and
  
  (x/aa:y/bb:z/cc) = d((b+c)^2:...:...)
  ~((c+a)^2+(a+b)^2-(b+c)^2: ... : ...)
  ~(a^2+ca+ab-bc : ... : ...)
  ~((c+a)(a+b)-2bc : ... : ...).
  
  This gives, as the radical center of the three circles,
  the point
  
  (a^2(a(a+b+c)-bc) : ... : ...)
  
  This point, again, is not in ETC.
  
  Best regards.
  Sincerely,
  Paul Yiu