Τρίτη 22 Οκτωβρίου 2019

HYACINTHOS 25186

[Antreas P. Hatzipolakis]:

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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