HYACINTHOS 25630

[Antreas P. Hatzipolaks]:

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of O. .

Denote:

A"B"C" = the midway triangle of P

(ie A", B", C" = the midpoints of AP, BP, CP, resp.)

Which is the locus of P such that the circumcircles of PA'A", PB'B", PC'C" are coaxial?

H lies on the locus. 2nd intersection on the Euler line.

O lies on the locus.

 

[César Lozada]:

Locus = {Gibert’s curve Q050, through ETCs 2, 3, 4, 13, 14, 67, 1113, 1114, 11058 and vertices of triangles ABC, ANTICOMPLEMENTARY, MEDIAL} ∪ {L∞}

 

2nd point Q2(P):

Q2(O) = X(6150)

Q2(H) = X(2072)

 

Q2(X(67)) = (a^6+(b^2+c^2)*a^4-(b^4+5*b^2* c^2+c^4)*a^2-(b^2+c^2)*(b^4-4* b^2*c^2+c^4))*((b^2+c^2)*a^6-( 3*b^4-2*b^2*c^2+3*c^4)*a^4-(b^ 2+c^2)*(3*b^4-7*b^2*c^2+3*c^4) *a^2+(b^4-c^4)^2) :: (baryc.)

= On line: {67,524}

= complementary conjugate of X(111)

= [ 1.577347832955612, 3.95984308251163, 0.171227963419889 ]

 

César Lozada