Κυριακή 2 Ιουνίου 2013

N1N2N3 - PERSPECTIVE

Let ABC be a triangle, A'B'C' the antipedal triangle of I (excentral tr.), N1,N2,N3 the NPC centers of IBC, ICA, IAB, resp. and Na,Nb,Nc the NPC centers of A'BC, B'CA, C'AB, resp.

1. The triangles A'B'C', N1N2N3 are perspective.

Perspector:

(aabc+a(a+b+c)(bb+4bc+cc-aa) , ... , ...)

Barry Wolk, Anopolis #347

ETC X(5506)

2. The triangles N1N2N3, NaNbNc are perspective (?).

Antreas P. Hatzipolakis, 2 June 2013

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ETC

X(72981) = X(2)X(1074)∩X(3)X(10) Barycentrics    (a^2-b^2-c^2)*(2*a^5-a^4*b-2*a^3*b^2+2*a^2*b^3-b^5-a^4*c+4*a^3*b*c-2*a^2*b^2*c-b^4*c-2*a^3...

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    Theorem 1. Let ABC be an equilateral triangle and P a point. The Euler lines of the triangles PBC,PCA,PAB are concurent.Denote the point ...
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    Let ABC be a triangle. Denote: (N1),(N2), (N3) = the NPCs of IBC, ICA, IAB, resp. (12), (13) = the reflections of (N1) in BI, CI, resp. ...