HYACINTHOS 223

• Antreas P. Hatzipolakis
  
  

  I wrote:
  
  

  >Euler Midway Trinagles
  >
  >Definition:
  >Let P be a point on the Euler Line of a triangle ABC, and A', B', C'
  >the midpoints of AP, BP, CP, respectively. The triangle A'B'C'
  >is the P-Euler Midway Triangle (P-EMT).
  >Note: I found the term "midway triangle" in a posting of Paul Yiu:
  >>The perpendicular bisectors of QX, QY, QZ bound
  >>the ``midway'' triangle with Q as homothetic center.
  >(Subject: RE: [EMHL] point identification; Date: Tue, 11 Jan 2000)
  >
  >Notation:
  >Q(P) = The Point Q of the P-Euler Midway Triangle.
  >[For example: H(P) = The orthocenter of the P-EMT]
  >
  >Trilinear Coordinates of Q(P) (in respect to reference triangle ABC),
  >for Q, P belonging in {H (Orthocenter), N (Nine-Point Circle Center),
  >G (Barycenter), O (Circumcenter), L (de Longchamps Point)}.
  
  

  Of course we can calculate the coordinates of other points of these
  triangles (in respect to reference triangle), but these points are interesting
  if lie on some notable lines etc, or have simple algebraic expressions.
  The latter is true for I(G), that is the incenter of the G-Euler Mid. Triangle.
  
  [...]
  
  

  >
  > 3. G-Euler Midway Triangle.
  >
  >H(G) = (2cos(B-C) - cosA ::)
  >
  >N(G) = (5cos(B-C) + 2cosA ::)
  >
  >O(G) = (cos(B-C) + 4cosA ::)
  >
  >G(G) = G = (cos(B-C) + cosA) [= (sinBsinC ::) = (1/sinA ::) = (1/a ::)]
  >
  >L(G) = (-cos(B-C) + 5cosA ::)
  
  
  

  If my calculations were correct:
  
  4a + b + c
  I(G) = (---------- ::) = (bc(4a+b+c) ::)
  a
  
  
  Antreas