[Antreas P. Hatzipolakis]:
Let ABC be a triangle and A'B'C' the pedal triangle of H.
Denote: [Angel Montesdeoca]:
*** The circumcenter of N1N2N3 is X(3628) is the centroid of the set {A', B', C', X(5)}, where A'B'C' is the medial triangle.
[APH]:
Thanks, Angel !
And the radical center of the circles (N1), (N2), (N3) [ = reflections of the NPCs of A'AbAc, B'BcBa, C'CaCb in BC, CA, AB, resp.] lies on the Euler line.
[Angel Montesdeoca]:
The radical center of the circles (N1), (N2), (N3) [ = reflections of the NPCs of A'AbAc, B'BcBa, C'CaCb in BC, CA, AB, resp.] is W the midpoint of points on Euler line X(5133)=inverse-in-nine-point- circle of X(23) and X(7499) = {X(2),X(22)}-harmonic conjugate of X(5).
W = (2 a^6-3 a^4 (b^2+c^2)-2 a^2 (b^4+4 b^2 c^2+c^4)+3 (b^2-c^2)^2 (b^2+c^2): ...: ... ),
with (6,9,13) search numbers (1.40233422648944, 0.529114910493857, 2.62712297780120)
W lies on lines: {2,3}, {230,233}, {232,10184}, {590,8281}, {615,8280}, {3793,8878}, {3819,9969}, {6723,10219}.
Angel Montesdeoca
The radical center of the circles (N1), (N2), (N3) [ = reflections of the NPCs of A'AbAc, B'BcBa, C'CaCb in BC, CA, AB, resp.] is W the midpoint of points on Euler line X(5133)=inverse-in-nine-point- circle of X(23) and X(7499) = {X(2),X(22)}-harmonic conjugate of X(5).
W = (2 a^6-3 a^4 (b^2+c^2)-2 a^2 (b^4+4 b^2 c^2+c^4)+3 (b^2-c^2)^2 (b^2+c^2): ...: ... ),
with (6,9,13) search numbers (1.40233422648944, 0.529114910493857, 2.62712297780120)
W lies on lines: {2,3}, {230,233}, {232,10184}, {590,8281}, {615,8280}, {3793,8878}, {3819,9969}, {6723,10219}.
Angel Montesdeoca
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