[Tran Quang Hung]:
Let ABC be a triangle with circumcenter O.
Ka*,Kb*,Kc* are isogonal conjugate of Ka,Kb,Kc wrt triangle ABC.
Then orthocenter of triangle Ka*Kb*Kc* lies on Euler line of triangle ABC.
Which is this point ?
[César Lozada]:
Q = S^4+(SW^2-2*(4*R^2+SA)*SW+13* R^4+2*SA^2)*S^2+3*(2*R^2-SW)*( SA-SW)*R^2*SA : : (barycentrics)
= on line {2,3}
= [ -7.543499892698678, -8.39311987240848, 12.932901267128100 ]
Sorry. No other relation with ETC centers was found.
César Lozada
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