HYACINTHOS 27207

• [Tran Quang Hung]:

   

  Let ABC be a triangle.

   

  A'B'C' is the medial triangle of ABC.

   

  L is the symmedian point of ABC.

   

  La, Lb, Lc are isogonal conjugate of L wrt triangles AB'C', BC'A', CA'B', resp.

   

  Then the orthocenter of the triangle LaLbLc lies on the Euler of ABC.

   

  Which is this point ?

   

   

  [César Lozada]:
  
  

   

  Q = EULER LINE INTERCEPT OF X(5139)X(6092)

  = SB*SC*(9*(12*R^2-SA-SW)*S^2-4* SW^3) : : (barys)

  = 9*S^2*(4*R^2-SW)*X(3)-(2*SW^3- 9*S^2*(8*R^2-SW))*X(4)

  = on lines: {2, 3}, {5139, 6092}

  = [ 5.7572482708648650, 4.8725405816600770, -2.3898243534871510 ]

   

   

  César Lozada