HYACINTHOS 28318

[Tran Quang Hung]:

 

Let ABC be a triangle.

O is the Circumcenter.

O* is the isotomic conjugate of O.

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

A''B''C'' is the antipedal triangle of O wrt A'B'C'.

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

Prove that O'' lies on theline OO*.

Please see AoPS

 

 

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[Ercole Suppa]

 

Dear Tran Quang Hung

 

O''=X(3)X(95) ∩ X(1154)X(10606)

 

=cos(B-C)*(-1-5*cos(4*A)-cos(6*A)+8*cos(A)cos(B-C)+6*cos(3*A)cos(B-C)+2*cos(5*A)*cos(B-C)-4*cos(2(B-C))-cos(2*A)*(3+2*cos(2(B-C)))) : : (trilinears)

 

=a^2 (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) (a^16-5 a^14 b^2+9 a^12 b^4-5 a^10 b^6-5 a^8 b^8+9 a^6 b^10-5 a^4 b^12+a^2 b^14-5 a^14 c^2 +21 a^12 b^2 c^2-33 a^10 b^4 c^2+23 a^8 b^6 c^2-7 a^6 b^8 c^2+3 a^4 b^10 c^2-3 a^2 b^12 c^2+b^14 c^2+9 a^12 c^4-33 a^10 b^2 c^4 +38 a^8 b^4 c^4-14 a^6 b^6 c^4+a^4 b^8 c^4-a^2 b^10 c^4-5 a^10 c^6+23 a^8 b^2 c^6-14 a^6 b^4 c^6+2 a^4 b^6 c^6+3 a^2 b^8 c^6-9 b^10 c^6

-5 a^8 c^8-7 a^6 b^2 c^8+a^4 b^4 c^8+3 a^2 b^6 c^8+16 b^8 c^8+9 a^6 c^10+3 a^4 b^2 c^10-a^2 b^4 c^10-9 b^6 c^10-5 a^4 c^12-3 a^2 b^2 c^12+a^2 c^14+b^2 c^14) : :  (barys)

 

=lies on these lines : {3,95},{1154,10606},{7395,12012},{11197,17928}

 

=(6-9-13) search numbers:  [-36.4490708229671888, 34.3019910430569337, -3.28421970653171859]

 

Best regards

Ercole Suppa