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

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ETC

X(72934) = (name pending) Barycentrics    a ((a^2+b^2-c^2) (a^2-b^2+c^2) (a^9 b^3-a^8 b^4-3 a^7 b^5+3 a^6 b^6+3 a^5 b^7-3 a^4 b^8-a^3 b^9+...