HYACINTHOS 2346

• Dear Antreas,
  
  [APH]:
  The circles (A,R), (B,R), (C,R) intersect the extensions of the
  triangle sides as in the figure:
  
  Ac Ab
  \ /
  \ /
  \ /
  \ /
  \/ A
  /\
  / \
  / \
  / \
  / \
  / \
  Bc---------B------------C----------Cb
  / \
  / \
  / \
  / \
  Ba Ca
  
  
  ABa = c + R, ACa = b + R, BCb = a + R, BAb = c + R, CAc = b + R,
  CBc = a + R
  
  Which is the radical center of the circumcircles of the
  triangles ABaCa, BCbAb, CAcCBc ? Is it something interesting?
  
  --------
  
  I work with slightly different notations for the points.
  
  
  Ba Ca
  \ /
  \ /
  \ /
  \ /
  \/ A
  /\
  / \
  / \
  / \
  / \
  / \
  Ab---------B------------C----------Ac
  / \
  / \
  / \
  / \
  Cb Bc
  
  
  Homogeneous barycentric coordinates:
  
  Bc = (-R : 0 : R+b),
  Cb = (-R : R+c : 0).
  
  The circle ABcCb has equation
  
  a^2yz + b^2zx + c^2xy - (x+y+z)(qy+rz) = 0.
  
  Since it passes through Bc and Cb,
  
  -b^2R(R+b) - rb(R+b) = 0 ==> r = -bR;
  -c^2R(R+c) - qc(R+c) = 0 ==> q = -cR.
  
  The circle:
  
  a^2yz + b^2zx + c^2xy + R(x+y+z)(cy+bz) = 0.
  
  Similarly write down the equations of the other two circles.
  Radical center:
  
  cy + bz = az + cx = bx + ay.
  
  (1/a)(y/b+ z/c) = (1/b)(z/c + x/a) = (1/c)(x/a + y/b)
  
  m(x/a:y/b:z/c) ~ (a:b:c)
  
  (x/a:y/b:z/c) ~ d(a:b:c) = (b+c-a: ... : ...)
  
  (x:y:z) = (a(b+c-a):...:..),
  
  Mittenpunkt!
  
  Best regards.
  Sincerely,
  Paul Yiu