Let ABC be triangle, P a point and A'B'C' the pedal triangle of P.
Denote:
Pa, Pb, Pc = the reflections of P in B'C', C'A', A'B', resp.
P1, P2, P3 = the reflections of P in BC, CA, AB, resp.
Which is the locus of P such that the Euler lines of P1PbPc, P2PcPa, P3PaPb are concurrent ?
ie (P1P2P3, PaPbPc) are Eulerologic triangles
[César Lozada]:
Locus = Napoleon-Feuerbach cubic
For P=N, Euler lines concur at:
cos(A)*((2*cos(A)-cos(3*A))* cos(B-C)+(cos(2*A)-1)*cos(2*( B-C))-1/2) : : (trilinears)
On lines: {5,6153}, {23,1173}, {30,143}, {51,567}, {52,3153}, {186,5462}, {265,1531}, {1216,2072}, {8254,10095}, {9969,11649}, {10110,11563}
= midpoint of X(52) and X(3153)
= reflection of X(i) in X(j) for these (i,j): (186,5462), (1216,2072), (10096,10095), (11563,10110)
= [ 0.068400064320402, 0.08827824699312, 3.547979511995110 ]
César Lozada
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