ADGEOM 3877 * ADGEOM 3880

#3877

 

Dear geometers,

 

Let ABC be a triangle with incenter I.

 

Ha is orthocenter of triangle IBC.

 

Euler line of triangle HaBC cuts the lines HaC,HaB at Ac,Ab.

 

Then Euler line da of triangle HaAbAc is parallel to line BC.

 

Similar, we have the lines db,dc.

 

Then the lines da,db,dc bound a triangle A'B'C' which is homothetic to ABC.

 

Triangle A'B'C' and ABC share the same OI line and the homothety center lies on this OI line. Which is this point ?

 

Best regards,

Tran Quang Hung.

 

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#3880

 

Dear Tran Quang Hung,

The homothety center of  triangles A'B'C' and ABC  is

((a (2 a-b-c) (a^2-a (b+c)-(b-c)^2+b c))/(b+c-a) : ... : ...),

with (6 - 9 - 13) - search numbers  (-0.106049542593117, 0.207008707141764, 3.54629670431383).
 Lies on lines X(i)X(j) for these {i, j}: {1,3}, {659, 6164},  {3911,4759}.
 
 Best regards,
 Angel Montesdeoca