HYACINTHOS 28720

 

[Tran Quang Hung]:

 

 

Let ABC be a triangle.

 

O is the circumcenter and H the orthocenter.

 

A', B', C' lie on BC, CA, AB, resp. such that HA'//OA, HB'//OB, HC'//OC.

 

HA meets B'C' at A", HB meets C'A' at B", HC meets A'B' at C".

 

Then the isogonal conjugate of H wrt A"B"C" also lies on Euler line of ABC.

 

Which is this point?

 

Also, I see that the triangles A'B'C' and A"B"C" are perspective and the perspector lies on the Euler line of ABC.

 

Which is this perspector?

 

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[Ercole Suppa]

 

Dear Tran Quang Hung,

 

*** the isogonal conjugate of H wrt A"B"C" is the point

 

Q = X(2)X(3) ∩ X(8537)X(14531)

 

= a^2 (a^2+b^2-c^2) (a^2-b^2+c^2) (a^10-3 a^8 b^2+2 a^6 b^4+2 a^4 b^6-3 a^2 b^8+b^10-3 a^8 c^2+7 a^6 b^2 c^2-a^4 b^4 c^2-7 a^2 b^6 c^2+4 b^8 c^2+2 a^6 c^4-a^4 b^2 c^4-4 a^2 b^4 c^4-5 b^6 c^4+2 a^4 c^6-7 a^2 b^2 c^6-5 b^4 c^6-3 a^2 c^8+4 b^2 c^8+c^10) :: (barys) 

 

= S^2 (8 R^4 - 6 R^2 SW + SW^2) + SB SC (-16 R^4 + 9 R^2 SW - SW^2) : : (barys)

 

= lies on these lines: {2,3}, {8537,14531}, {11425,15872}

 

= {X(i),X(j)}-harmonic conjugate of X(k) for these {i,j,k}: {4,3520,7512}, {378,3541,3520}, {427,12225,4}, {1593,7503,4}, {1885,5133,4},{9818,12084,3547}

 

= (6-8-13) search numbers [5.45654050946213040, 4.57262609492469488, -2.04340304971523908]

 

 

*** The perspector of triangles A'B'C' and A"B"C" is the point X(427)

 

 

Best regards

Ercole Suppa