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• [Hatzipolakis - Lozada]: Let ABC be a triangle and P a point. The circumcircle (Oa) of PBC intersects AB,AC at Ab, Ac, resp. other than B,C. The circumcircle
Message 1 of 1 , Jul 26
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Let ABC be a triangle and P a point. The circumcircle (Oa) of
PBC intersects AB,AC at Ab, Ac, resp. other than B,C.
The circumcircle (Ob) of PCA intersects BC, BA at Bc, Ba, resp.
other than C,A. The circumcircle (Oc) of PAB intersects CA, CB
at Ca, Cb, resp. other than A,B. Let (O1), (O2), (O3) be the
circumcircles of ABaCa, BCbAb, CAcBc,
resp. The triangles OaObOc, O1O2O3 are perspective.
For P = (u:v:w) in trilinears, Perspector:
2*a*b*c*u*(u*(w^2*b*cos(C)+c*cos(B)*v^2)+cos(A)*v*w*(b*v+c*w))+v*w*(2*v*w*a^2*b*c+(a^2*(-3*b^2+2*a^2-3*c^2)+(b^2-c^2)^2)*u^2) : :

P11:
For P = X3 the Perspector (in trilinears):
a*(2*a^8-(5*(b^2+c^2))*a^6+(3*c^4+4*b^2*c^2+3*b^4)*a^4+(b^2+c^2)*(b^2-c^2)^2*a^2-(b^4+c^4)^2+2*b^2*c^2*(b^4+c^4)): :

= Midpoint of (26,1147), (156,1658), (159,182) and others

P12:
For P = X7, the perspector (in trilinears):
(a*(a^2+2*b*c)-(b+c)*(b-c)^2)/a ::

=Midpoint of (7,2550), (9,4312) and others

Reference:

http://tech.groups.yahoo.com/group/Anopolis/message/682
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