HYACINTHOS 29090

[Tran Quang Hung]:

 

Let ABC be a triangle.

Euler line meets CA, AB at B0, C0, resp.

Euler line of AB0C0 meets CA, AB at B1, C1, resp.

Euler line of AB1C1 meets CA, AB at B2, C2, resp.

Then Euler line of AB2C2 is Ea2.

Define similarly we have Eb2, Ec2.

 

1)      Then lines Ea2, Eb2, Ec2 bound triangle which is homothetic with ABC. 

Which is the homothety center?

2)      If we define A1, A2,...,An, B1, B2, ... Bn, C1, C2, .. Cn, we have the Euler lines Ea2n, Eb2n, Ec2n. The lines Ea2n, Eb2n, Ec2n bound a triangle which is homothety with ABC. Which is the homothety center in term of n?

[César Lozada]:

 

1)      Q = X(2)X(15351) ∩ X(14401)X(15183)

= (2*a^4-(b^2+c^2)*a^2-(b^2-c^2)^2)^2*(a^8-(b^2+c^2)*a^6-((2*b^2-2*c^2)^2-b^2*c^2)*a^4+7*(b^4-c^4)*(b^2-c^2)*a^2-(3*b^4+7*b^2*c^2+3*c^4)*(b^2-c^2)^2)*(a^8-(b^2+c^2)*a^6-(2*b^2-c^2)*(b^2-2*c^2)*a^4+3*(b^4-c^4)*(b^2-c^2)*a^2-(b^4+3*b^2*c^2+c^4)*(b^2-c^2)^2) : : (barys)

= (4*S^2+3*(SB+SC)*(12*R^2-SA-3*SW))*(5*S^2-4*(6*SA-SW)*R^2+6*SA^2-4*SB*SC-SW^2)*(9*S^2-4*(12*SA+SW)*R^2+12*SA^2-8*SB*SC+SW^2) : : (barys)

= lies on these lines: {2, 15351}, {14401, 15183}

= (medial)-isotomic conjugate of-X(15184)

= complement of the isotomic conjugate of X(402)

= barycentric product X(402)*X(15184)

= [ 2.4666724743522280, 1.4934979756545030, 1.4682401259840800 ]

 

Best regards,

César Lozada