HYACINTHOS 25668

[Tran Quang Hung]:

Let ABC be a triangle.

 

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

 

Circle (A'B'C') intersects BC,CA,AB again at A'',B'',C''.

 

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

 

Which is this point ?

 

[Peter Moses]:

Hi Antreas,

 

(2 a^4-3 a^2 b^2+b^4-3 a^2 c^2-2 b^2 c^2+c^4) (a^12-5 a^10 b^2+9 a^8 b^4-6 a^6 b^6-a^4 b^8+3 a^2 b^10-b^12-5 a^10 c^2+11 a^8 b^2 c^2-2 a^6 b^4 c^2-2 a^4 b^6 c^2-9 a^2 b^8 c^2+7 b^10 c^2+9 a^8 c^4-2 a^6 b^2 c^4+6 a^4 b^4 c^4+6 a^2 b^6 c^4-19 b^8 c^4-6 a^6 c^6-2 a^4 b^2 c^6+6 a^2 b^4 c^6+26 b^6 c^6-a^4 c^8-9 a^2 b^2 c^8-19 b^4 c^8+3 a^2 c^10+7 b^2 c^10-c^12)::
on line {2,3}.

 

Generalisation:

A'B'C' is cevian triangle of a point P on the Jerabek circumhyperbola.

Circle (A'B'C') intersects BC,CA,AB again at A'',B'',C''.

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

 

Best regards

Peter.