HYACINTHOS 23429

 

 

 

 

• Antreas Hatzipolakis

   

  
  

  [APH]:
  
  

  Let ABC be a triangle.
  
  

  Denote:
  
  

  O1,O2, O3 = the circumcenters of OBC,PCA,OAB, resp

  O12, O13 = the orthogonal projections of O1 on AC, AB, resp.

  O23, O21 = the orthogonal projections of O2 on BA, BC, resp.

  O31, O32 = the orthogonal projections of O3 on CB, CA, resp.
  
  

  M1, M2, M3 = the midpoints of O12O13, O23O21, O31O32, resp.
  
  

   

  The circumcenter of M1M2M3 lies on the Euler line of ABC

  (ie it is the intersection of the Euler lines of ABC and M1M2M3).

  Point?
  
  

  APH
  
  

  [Angel Montesdeoca]:

  
  

   

  Dear Antreas,
  

   

  The circumcenter of M1M2M3 lies on the Euler line of ABC
  
  

  
  (-2 a^10+5 a^8 (b^2+c^2)-2 a^6 (b^2+c^2)^2+a^4 (-4 b^6+2 b^4 c^2+2 b^2 c^4-4 c^6)+2 a^2 (b^2-c^2)^2 (2 b^4+b^2 c^2+2 c^4)-(b^2-c^2)^4 (b^2+c^2): ... : ... )
  
  
  Search number in ETC:
  {1.44283258574707, 0.569506434057196, 2.58046806490767}
  
  Best regards,
  Angel M.