Σάββατο 14 Σεπτεμβρίου 2013

PRIZE (Re: ORTHOCENTER - REFLECTIONS - CONCURRENT CIRCLES)

[APH]: In fact we can take any point P (instead of H) and any points O1,O2,O3 on the circumcircles of PBC,PCA,PAB, resp.

Then the circumcircles of the triangles

AO2O3, BO3O1, CO1O2

are concurrent.

Anopolis #850

For a proof I offer the book:

R. G. SANGER: SYNTHETIC PROJECTIVE GEOMETRY (1939)

APH

1 σχόλιο:

z

X(72392) = X(110)X(3164)∩X(112)X(3168) Barycentrics    (a^10*b^2 - 2*a^8*b^4 + 2*a^6*b^6 - 2*a^4*b^8 + a^2*b^10 - a^10*c^2 - 2*a^8*b^2*c^...