ADGEOM 3945 * ADGEOM 3950

#3945

 

Dear geometers,

 

Let ABC be a triangle.

 

A'B'C' is cenvian triangle of incenter I.

 

A''B''C'' is pedal triangle of circumcenter O.

 

Then radical centers of the circles (A',A'A''), (B',B'B''), (C',C'C'') lies on OI line of ABC.

 

Which is this point ?

 

Best regards,

Tran Quang Hung. 

 

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#3950

 

Dear  Tran Quang Hung

The radical centers of the circles (A',A'A''), (B',B'B''), (C',C'C'') is:
W = (r^2+2rR+s^2) X(1)  - (r^2-14rR+s^2) X(3)

W =  (a (a^4 (b^2-6 b c+c^2)
            +a^3 (b^3+b^2 c+b c^2+c^3)
            -a^2 (b^4-7 b^3 c-7 b c^3+c^4)
            -a (b^5+b^4 c+b c^4+c^5)
            -b c (b^2-c^2)^2) : ... : ...).

W is the midpoint of X(3) and X(5482).
  Let  W- be the  radical  centers of the circles (A',A'B), )C',C'A) and (B',B'C), and W+  the  radical  centers of the circles (A',A'C), (B',B'A) and (C', C'B).
The points W+,W- form a bicentric pair, the midpoint of W+W- is a center of the triangle X(5482) (on the line IO,  see Hechos Geométricos en el Triángulo)

W lies on lines X(i)X(j) for these {i, j}:  {1,3}, {511,3530}, {3524,5752},....

 (6 - 9 - 13) - search numbers  of W: (1.09732354371282, 1.20066532197530, 2.30297762344172).
 
 Best regards,
 Angel Montesdeoca