HYACINTHOS 2123

• Dear Antreas,
  
  [FvL]
  

  > >Consider a triangle with squares erected on its sides:
  > >
  > > Bc
  > >
  > >
  > >Ba
  > > Cb
  > ^^
  > Ac
  > >
  > > C
  > > _.-' \
  > > _.-` \ Ab
  > > _.-' \
  > > A----------------B
  > > | |
  > > | |
  > > | |
  > > | |
  > > | |
  > > | |
  > > | |
  > > Ca---------------Cb
  > >Consider triangles ABaCa, AbBCb, AcBcC.
  

  <...>
  
  [APH]
  

  > I just recalled that I had studied the "opposite" figure (what I called
  > "orthial triangles"):
  >
  > Ca
  >
  > Cb
  > Ba
  >
  > A
  > _.-' \
  > _.-` \ Bc
  > _.-' \
  > B'a-----B----------------C--C'a--------
  > | |
  >
  > Let C'a = CaA /\ BC
  > B'a = BaA /\ BC
  >
  > The triangle AB'aC'a, I called "orthial" triangle, is "opposite"
  > to your ABaCa.
  >
  > Similarly we define the other two "orthial triangles" corresponding
  > to B, C.
  >
  > I am womdering if we have in these triangles a friendship as well.
  
  

  You mean my triangles, let's call them flanks, like Paul did, and your
  orthials?
  
  [APH]
  

  > What I remember is that the triangle formed by the G's of the three
  > orthials is in perspective with ABC, and the perspector is L.
  

  <...>
  

  > Paul had proved my conjecture that the perpendicular bisectors of
  > the bases of the orthials concur.
  > (at f(g+h)(g^2+h^2-f^2) : ... : ... in In P-perpendicularity)
  > http://www.egroups.com/message/Hyacinthos/76
  >
  > BTW, is the same true for your triangles's bases?
  > (namely: Are the perpendicular bisectors of BcAc, AbCb, CaBa concurrent?
  
  

  Yes, I checked that these perpendicular bisectors are concurrent at:
  
  ( ... : 2SS + 2SSB - bbS + 4SAC + bbSB - aaSA - ccSC : ... )
  
  Perhaps there is some simplification of these coordinates.
  
  I haven't tried the Euler lines yet.
  
  Kind regards,
  Sincerely,
  Floor van Lamoen