HYACINTHOS 29518

[Kadir Altintas]

Let ABC bea triangle.

Denote:

P = X(115) (the center of Kiepert hyperbola)
DEF = the medial triangle of ABC
Ka = the symmedian point of PEF. 

Define Kb, Kc cyclically

Prove:  AKa, BKb, CKc concurr at a point X

--------------------------------------------------------------------------------------------

 

 

[Ercole Suppa]
 
X = X(6)X(17500) ∩ X(53)X(6531) =

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

= (6 R^2 SB+6 R^2 SC+2 SB SC+4 R^2 SW-SB SW-SC SW)S^2 + 2 SB SC SW^2-SB SW^3-SC SW^3

= lies on these lines: {6,17500}, {53,6531}, {76,25322}, {83,597}, {115,804}, {141,308}, {251,1989}, {338,3124}, {524,20022}, {1086,4374}, {1211,18096}, {1799,13468}, {1990,21459}, {2872,6784}, {3051,30505}, {3589,18092}, {3613,8265}, {6543,18082}, {6748,10550}, {10130,11168}, {16889,23897}, {17056,18703}, {18088,23903}, {18091,23905}, {18104,23917}, {18105,31644}, {27376,32713}

= {X(i),X(j)}-harmonic conjugate of X(k) for these {i,j,k}: {115,1084,7668}, {308,16890,141}, {10549,32085,53}

= (6-9-13) search numbers [1.8157038322248938280, 3.3719244574583991513, 0.4682380887939979239]


Best regards,
Ercole Suppa