HYACINTHOS 28548

 

[Antreas P. Hatzipolakis]:

 

 

 

Variation:

 

Let ABC be a triangle and A'B'C' the pedal triangle of I.

 

Denote:

 

Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.

 

(Oa), (Ob), (Oc) = the circles (A', A'Na), (B', B'Nb), (C', C'Nc), resp.

 

Ra = the radical axis of (Ob), (Oc)

Rb = the radical axis of (Oc), (Oa)

Rc = the radical axis of (Oa), (Ob)

 

The point of concurrence of Ra, Rb, Rc (radical center of the circles) U lies on the OI line.

The reflections of Ra, Rb, Rc in AI, BI, CI, resp. are concurrent at a point W on the OI line such that W and U are symmetric in I.

 

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[Ercole Suppa]

 

*** Ra, Rb, Rc are concurrent at a point U = X(11567)

 

*** The reflections of Ra, Rb, Rc in AI, BI, CI, resp. are concurrent at a point W 

 

W = X(1)X(3) ∩ X(10)X(19907) 

 

= a (2 a^6-4 a^5 b-2 a^4 b^2+8 a^3 b^3-2 a^2 b^4-4 a b^5+2 b^6-4 a^5 c+12 a^4 b c-9 a^3 b^2 c-8 a^2 b^3 c+13 a b^4 c-4 b^5 c-2 a^4 c^2-9 a^3 b c^2+22 a^2 b^2 c^2-9 a b^3 c^2-2 b^4 c^2+8 a^3 c^3-8 a^2 b c^3-9 a b^2 c^3+8 b^3 c^3-2 a^2 c^4+13 a b c^4-2 b^2 c^4-4 a c^5-4 b c^5+2 c^6)  : : (barys)

 

= X[952]-X[24387], 3*X[7967]+X[10525], X[10526]-5*X[10595]

 

= lies on these lines: {1,3}, {10,19907}, {952,24387}, {1317,6842}, {3825,5901}, {4861,6265}, {5154,5886}, {5882,21630}, {7967,10525}, {10526,10595}, {10914,22935}, {11230,17619}, {12737,21740}

 

= reflection of X(11567) in X(1)

 

= {X(i),X(j)}-harmonic conjugate of X(k) for these {i,j,k}: {1385,10222,3057},{13145,15178,1385},{21842,25413,23961}

 

= (6-8-13) search numbers [7.27565566271796642, 6.30227566090485772, -4.08044435843496844]

 

 

Best regards

Ercole Suppa