HYACINTHOS 5437

• Dear Paul,
  
  

  > [AM]: A week ago I've found one property of 184-point in Clark List.
  > That point generates three equil triangles, similar to the base
  

  triangle ...
  
  [ND]
  

  >One solution is: If p, q, r are the angles of the isosceles
  >triangles in P then it is easy to prove that
  >p = 180 - 4A, q = 180 - 4B, r = 180 - 4C
  
  

  Correction --> p = 4A - 180, q = 4B - 180, r = 4C - 180
  
  

  >[PY]
  > Fred Lang, Hyacinthos message 1191, has found another interesting
  

  property
  

  > of X(184) that leads to a very easy construction of this point:
  >
  > Given a reference triangle ABC, let the perpendicular bisectors of
  

  BC, CA,
  

  > AC intersect the other pairs of sides at B1, C1; C2, A2; A3, B3
  > repsectively. The perpendicular bisectors of the segments B1C1,
  

  C2A2 and
  

  > A34B3 bound a triangle homothetic to ABC at X(184).
  >
  > Having constructed this point, it is now not difficult to construct
  

  three
  

  > congruent triangles similar to the reference triangle ABC.
  >
  > I wonder if the similarity ratio ZBaCa/ABC has a simple expression.
  
  

  The similarity ratio is
  k = S/(aasin(2A) + bbsin(2B) + ccsin(2C))
  = 1/(3 - ss) >= 1/3 where s = OH/R
  
  Other properties of X184 are
  X184 = X63 of the orthic triangle
  X184 = X226 of the tangential triangle
  and X184 is the homothety center of the orthic triangle
  and the medial triangle of the tangential triangle.
  
  Best regards
  Nikos Dergiades