HYACINTHOS 28632

[Tran Quang Hung]:

 

 

Let ABC be a triangle with NPC center N.

 

The circle (NBC) meets AB, AC again at Ba,Ca, resp.

 

Define similarly points Ac, Bc, Cb, Ab.

 

Let A* be the orthocenter of the triangle AAbAc. 

Define similarly the points B* and C*.

 

Then X(186) of triangle the A*B*C* lies on the Euler line of ABC. 

 

Which is this point?

 

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[Ercole Suppa]

 

Dear Tran Quang Hung,

 

the center X(186) of triangle A*B*C* is 

 

W = X(2)X(3) ∩ X(6102)X(6798)

 

= (a^2+b^2-c^2) (a^2-b^2+c^2) (2 a^18-9 a^16 (b^2+c^2)+14 a^14 (b^2+c^2)^2+(b^2-c^2)^6 (b^6+b^4 c^2+b^2 c^4+c^6)-2 a^12 (3 b^6+14 b^4 c^2+14 b^2 c^4+3 c^6)+a^10 (-6 b^8+10 b^6 c^2+26 b^4 c^4+10 b^2 c^6-6 c^8)-2 a^2 (b^2-c^2)^4 (2 b^8-b^6 c^2-b^4 c^4-b^2 c^6+2 c^8)+a^4 (b^2-c^2)^2 (6 b^10-16 b^8 c^2-3 b^6 c^4-3 b^4 c^6-16 b^2 c^8+6 c^10)+a^8 (8 b^10-10 b^8 c^2-19 b^6 c^4-19 b^4 c^6-10 b^2 c^8+8 c^10)-2 a^6 (3 b^12-12 b^10 c^2+4 b^8 c^4-8 b^6 c^6+4 b^4 c^8-12 b^2 c^10+3 c^12)) : : (barys)

 

= (100 R^6+18 R^2 SB SC-105 R^4 SW-4 SB SC SW+36 R^2 SW^2-4 SW^3) S^2 -156 R^6 SB SC+175 R^4 SB SC SW-66 R^2 SB SC SW^2+8 SB SC SW^3 : : (barys)

=  As a point on the Euler line, X() has Shinagawa coefficients: {100 R^6 - 105 R^4 SW + 36 R^2 SW^2 - 4 SW^3, -156 R^6 + 18 R^2 S^2 + 175 R^4 SW - 4 S^2 SW - 66 R^2 SW^2 + 8 SW^3}.

 

= lies on these lines: {2,3}, {6102,6798}, {8146,20414}

= (6-8-13) search numbers [-7.52114848487373958, -8.37082742816174941, 12.9071520021150085]

 

 

Best regards

Ercole Suppa