HYACINTHOS 7461

• The Apollonius point X(181) of a triangle is defined as
  follows:
  
  Given a triangle ABC, the APOLLONIUS CIRCLE is the
  circle touching the three excircles externally. (This
  circle is the inverse of the nine-point circle in the
  radical circle of the excircles.)
  
  Now, if this Apollonius circle touches the excircles at
  Ja, Jb, Jc, then the lines AJa, BJb, CJc concur at the
  APOLLONIUS POINT X(181).
  
  In fact, this is very easy to prove. Consider the
  incircle, the A-excircle and the Apollonius circle of
  triangle ABC. After Monge's theorem,
  
  - the external center of similtude of the incircle and
  the A-excircle, i. e. the point A,
  - the external center of similtude of the A-excircle
  and the Apollonius circle, i. e. the point Ja,
  - the external center of similtude J of the incircle
  and the Apollonius circle
  
  are collinear. In other words, J lies on AJa.
  Analogously, J lies on BJb and CJc. Hence, the lines
  AJa, BJb, CJc concur at the external center of
  similtude J of the incircle and the Apollonius circle.
  
  Darij Grinberg