Σάββατο 13 Ιουλίου 2024

SAME CENTROID

Let ABC be a triangle and A'B'C' the cevian triangle of O

Denote

Ma, Mb, Mc = the midpoints of AA'. BB', CC', resp.

Ha, Hb, Hc = the orthocenters of OMbMc, OMcMa, OMaMb, resp,

The triangles ABC and HaHbHc share the same centroid G

APH

Francisco Javier García Capitán A triangle of orthocenters with centroid G
FJGC

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X(73160) = X(2)X(6120)∩X(15859)X(66554) Barycentrics    Sin[B/3]*Sin[(2*B)/3]*Sin[B] + (Sin[A]*Sin[B] + Sin[C/3]*Sin[(2*C)/3])*Sin[C] : : ...