Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.
Denote:
A1, Ab, Ac = the midpoints of A'P, A'B, A'C, resp.
B2, Bc, Ba = the midpoints of B'P, B'C, B'A, resp.
C3, Ca, Cb = the midpoints of C'P, C'A, C'B, resp.
For P = O the Euler lines of the triangles A1AbAc, B2BcBa, C3CaCb are concurrent (trivial case)
How about for P = I ??
In general, which is the locus of P such that the Euler lines of the triangles A1AbAc, B2BcBa, C3CaCb are concurrent?
Antreas P. Hatzipolakis, 5 Dec. 2013, Anopolis #1143
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The locus is a degree-7 circum-excentral-curve which passes through ETC-centers I, O and X(1138)=Isogonal conjugate of X(399)
1) For P=X(3)=O, point of concurrence is Z=O
2) For P=X(1138), Z=X(30)
3) For P=X(1)=I
Z = (2*a^4-2*(b+c)*a^3-(3*b^2+4*b*c+3*c^2)*a^2+2*(c^2-3*b*c+b^2)*(b+c)*a+(b^2-c^2)^2)/a : : (Trilinears)
= midpoint of: (1,442), (21,3649)
= on lines (1,442), (7,21), (30,551), (78,3826), (79,5426), (191,3338), (497,2475), (758,942), (950,3838), (958,3487), (962,4428), (1387,3636), (2646,5249), (3035,3812), (3651,5603), (3671,4640), (3897,5434)
= [2.022889134016317502091, 1.585022278218526006039, 1.609700227440945328741]
César Lozada, 6 Dec. 2013, Anopolis #1144
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