ADGEOM 3956 * ADGEOM 3960 * ADGEOM 3962

#3956

 

Dear geometers,

 

Let ABC be a triangle with incenter I.

 

A'B'C' is cyclocevian triangle of I wrt ABC.

 

Then orthocenter of A'B'C' lies on OI line of ABC.

 

Is this a new point ?

 

Best regards,

Tran Quang Hung.

 

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#3960

 

The  orthocenter of A'B'C' is the intersection of  IO and X(5948)X(5164), the line that passes through the orthocenter of Feuerbach triangle and the inverse in circumcircle of the isogonal conjugate of the cyclocevian conjugate of I.

Barycentric coordinates:

a (a^3 (b+c)-a (b-c)^2 (b+c)-(b^2-c^2)^2+a^2 (b^2+c^2)) (a^5+a^4 (b+c)+(b-c)^2 (b+c)^3-a^3 (2 b^2+b c+2 c^2)-2 a^2 (b^3+b^2 c+b c^2+c^3)+a (b^4+b^3 c+2 b^2 c^2+b c^3+c^4)):... : ...),

with  (6 - 9 - 13) - search numbers (-0.532738560673522, -0.145319569198488, 3.98714967320188)

Angel Montesdeoca

 

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#3962

 

 

Dear geometers,

 

I see the general problem.

 

Let ABC be a triangle with P,Q are two isogonal conjugate point and circumcenter O.

 

A'B'C' is cyclocevian triangle of O wrt ABC.

 

Then orthocenter of A'B'C' lies on line OQ.

 

Thank you for your interest.

 

Best regards,

Tran Quang Hung.