HYACINTHOS 29557

[Tran Quang Hung]:

 

 

Let ABC be a triangle. 

 

The line d connecting two Brocard points of ABC meets BC, CA, AB at A', B', C', respectively. 

 

Let Oa, Ob, Oc be circumcenters of triangles AB'C', BC'A', CA'B'. 

 

Then orthocenter of triangle OaObOc lies on line d.

 

Which is this point?

 

 

[Ercole Suppa] 

 

 

Hi Tran Quang Hung, 

 

the orthocenter of the triangle OaObOc is the point:

 

 

Q = X(39)X(512) ∩ X(115)X(2971)

 

= a^4 (b-c) (b+c) (a^4 b^4-2 a^2 b^6+b^8-2 b^6 c^2+a^4 c^4+4 b^4 c^4-2 a^2 c^6-2 b^2 c^6+c^8) : : (barys)

 

= (2 R^2+SB-SC)S^4 + (6 R^2 SB SC+12 R^2 SC^2+2 SB SC^2-14 R^2 SB SW+2 R^2 SC SW-2 SB SC SW-2 SC^2 SW-2 R^2 SW^2+2 SB SW^2)S^2 + 2 R^2 SB SC SW^2+4 R^2 SC^2 SW^2+2 SB SC^2 SW^2-2 R^2 SB SW^3-2 R^2 SC SW^3-2 SB SC SW^3-2 SC^2 SW^3+SB SW^4+SC SW^4 : : (barys)

 

= lies on these lines: {39,512}, {115,2971}, {523,7697}, {3095,3566}

 

= ETC search numbers: [2.5751429252861840674, -0.3669096271828168137, 2.7061513352096288661]

 

 

Best regards 

Ercole Suppa