22478CONJECTURE with PRIZE :-)
- Jun 29, 2014Antreasis proved true !If part of the locus is the Neuberg Cubic, then the conjecture(or, equivalently, Aa, Bb, Cc are collinear?)Which is the locus of P such that ABC, A'B'C' are perspective?A' = BAb /\ CAc. Similarly B',C'.Denote:resp.The perpendicular to La at Na intersects BC,CA,AB at Aa, Ab,Ac,Na, Nb, Nc = the NPC centers of PBC,PCA, PAB, resp.La, Lb, Lc = the Euler lines of PBC,PCA,PAB, resp.Denote:Let ABC be a triangle and P a point.That is:the property in question.A way to prove it is to find the locus of points withNow about the conjecture:library the original copy and other thing to have an electronic copy :-)Well ... it is available on-line, but it is an other thing to have in yourGeneral Conjecture:I offer an old book in German on Triliner Coordinates.
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