Let ABC be a triangle, and (Na), (Nb), (Nc) the NPCs of the triangles IBC, ICA, IAB, resp., concurrent at the Feuerbach point F. I think that the Reflections...
Dear Francisco Thanks Another one: Let ABC be a triangle, and A'B'C' its orthic triangle. The Euler Lines E1,E2,E3 of AB'C',BC'A',CA'B', resp. are concurrent....
Let ABC be a triangle and (Oa), (Ob),(Oc) the circumcircles of IBC, ICA, IAB, resp. Let (Qa), (Qb), (Qc) be the reflections of (Oa), (Ob), (Oc) in BC, CA, AB,...
The intersection point L has coordinates {a (a^2 b - b^3 + a^2 c - a b c + b^2 c + b c^2 - c^3), b (-a^3 + a b^2 + a^2 c - a b c + b^2 c + a c^2 - c^3), c...
Let ABC be a triangle and P = (x:y:z) a point. Denote: A* :=(Perpendicular from B to CP) /\ (Perpendicular from C in BP) B* :=(Perpendicular from C to AP) /\...
Let ABC be a triangle, A'B'C' its medial triangle and P a point. Let La, Lb, Lc be the reflections of PA', PB', PC' in the bisectors AI, BI, CI, resp. Which is...
Dear Antreas, ... The locus is the linf + the conic with center X(1112) that passes through the vertices of the cevian triangles of X(4) Orthocenter and X(648)...
[APH] ... [ND] ... Dear Nikos It is nice! Thanks. Probably there are also nice results for pedal triangles of other points. For the pedal triangle of I, for...
Dear Antreas, [APH] ... If A', B', C' are arbitrary not collinear points with barycentrics A'=(0:q1:r1), B'=(p2:0:r2), C'=(p3:q3:0) then the locus is the linf...
Dear Francisco [APH] ... [FGC] ... A natural question is which other than I points P have that property. Anyway, I think that this variation is true as well: ...
For F* the intersection point has coordinates {a (a^5 b - a^4 b^2 - 2 a^3 b^3 + 2 a^2 b^4 + a b^5 - b^6 + a^5 c - 6 a^4 b c + 8 a^3 b^2 c + 4 a^2 b^3 c - 9 a...
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