CONCURRENT CIRCUMCIRCLES - N1N2N3, NaNbNc

Let ABC be a triangle, A'B'C' the antipedal triangle of I (excentral triangle) 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 circumcircles of ABC, AN2N3, BN3N1, CN1N2 are concurrent.

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It is now the center X(5606) in ETC

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2. The circumcircles of ABC, ANbNc, BNcNa, CNaNb are concurent.

Note: The circumcenter of NaNbNc is the O.

Antreas P. Hatzipolakis, 2 June 2013

Addendum (26 - 11 - 2013)

1'. The circumcircles of N1BC, N2CA, N3AB are concurrent.

2'. The circumcircles of NaBC, NbCA, NcAB are concurrent.

APH, Anopolis #1119

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*** 1'. X(502)

*** 2'. ( (b+c) (a^6- a^4(b^2+c^2) - a^2(b^4-3b^2c^2+c^4) - 2a b c(b-c)^2(b+c) + (b-c)^4(b+c)^2 ) : ... : ... )

with (6-9-13)-search number 2.5719845710987353841258271936

Angel Montesdeoca, Anopolis #1120

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( (b+c) (a^6- a^4(b^2+c^2) - a^2(b^4-3b^2c^2+c^4) - 2a b c(b-c)^2(b+c) + (b-c)^4(b+c)^2 ) : ... : ... ) = R X[65]-(2r+R) X[1365], is on lines {{1,149},{10,1109},{36,759},{37,115},{65,1365},{162,1838},{267,3336},{897,1738},{1054,1247},{1737,2166},{2218,2915}}.

isogonal conjugate X(5127)

X(2245) cross-conjugate of X(226)

trilinear pole of line X(661) X(2294)

X(3) isoconjugate of X(2074)

X(21) isoconjugate of X(5172)

trilinear product of X(523) & X(1290)

barycentric product of X(1290) & X(1577)

antigonal of X(502)

Peter J.C. Moses, 2 Dec 2013

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It is now X(5620) = ISOGONAL CONJUGATE OF X(5127) in ETC