Τρίτη, 16 Απριλίου 2013

QUINTIC

Let ABC be a triangle, P a point and A1,B1,C1 are the NPC centers of PBC, PCA, PAB, resp.

Which is the locus of P such that the circumcenter O0 of A1B1C1 lies on the Euler line?

P : X1 = I, X4= H, X5 = N.......

For P = I we have O0 = N

Antreas P. Hatzipolakis, 15 April 2013

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It is a circular quintic through I, H and N. The other three intersection points with Euler line are X30 (the infinite point) and the instersections with the circumcircle X1113 and X1114.

Equation:

a^6 c^2 x^3 y^2 - 3 a^4 b^2 c^2 x^3 y^2 + 3 a^2 b^4 c^2 x^3 y^2 - b^6 c^2 x^3 y^2 - a^4 c^4 x^3 y^2 + a^2 b^2 c^4 x^3 y^2 - a^2 c^6 x^3 y^2 + c^8 x^3 y^2 + a^6 c^2 x^2 y^3 - 3 a^4 b^2 c^2 x^2 y^3 + 3 a^2 b^4 c^2 x^2 y^3 - b^6 c^2 x^2 y^3 - a^2 b^2 c^4 x^2 y^3 + b^4 c^4 x^2 y^3 + b^2 c^6 x^2 y^3 - c^8 x^2 y^3 + 3 a^2 b^4 c^2 x^3 y z - 3 b^6 c^2 x^3 y z - 3 a^2 b^2 c^4 x^3 y z + 3 b^2 c^6 x^3 y z + 2 a^6 c^2 x^2 y^2 z - 2 b^6 c^2 x^2 y^2 z - 4 a^4 c^4 x^2 y^2 z + 4 b^4 c^4 x^2 y^2 z + 2 a^2 c^6 x^2 y^2 z - 2 b^2 c^6 x^2 y^2 z + 3 a^6 c^2 x y^3 z - 3 a^4 b^2 c^2 x y^3 z + 3 a^2 b^2 c^4 x y^3 z - 3 a^2 c^6 x y^3 z - a^6 b^2 x^3 z^2 + a^4 b^4 x^3 z^2 + a^2 b^6 x^3 z^2 - b^8 x^3 z^2 + 3 a^4 b^2 c^2 x^3 z^2 - a^2 b^4 c^2 x^3 z^2 - 3 a^2 b^2 c^4 x^3 z^2 + b^2 c^6 x^3 z^2 - 2 a^6 b^2 x^2 y z^2 + 4 a^4 b^4 x^2 y z^2 - 2 a^2 b^6 x^2 y z^2 + 2 b^6 c^2 x^2 y z^2 - 4 b^4 c^4 x^2 y z^2 + 2 b^2 c^6 x^2 y z^2 + 2 a^6 b^2 x y^2 z^2 - 4 a^4 b^4 x y^2 z^2 + 2 a^2 b^6 x y^2 z^2 - 2 a^6 c^2 x y^2 z^2 + 4 a^4 c^4 x y^2 z^2 - 2 a^2 c^6 x y^2 z^2 + a^8 y^3 z^2 - a^6 b^2 y^3 z^2 - a^4 b^4 y^3 z^2 + a^2 b^6 y^3 z^2 + a^4 b^2 c^2 y^3 z^2 - 3 a^2 b^4 c^2 y^3 z^2 + 3 a^2 b^2 c^4 y^3 z^2 - a^2 c^6 y^3 z^2 - a^6 b^2 x^2 z^3 + b^8 x^2 z^3 + 3 a^4 b^2 c^2 x^2 z^3 + a^2 b^4 c^2 x^2 z^3 - b^6 c^2 x^2 z^3 - 3 a^2 b^2 c^4 x^2 z^3 - b^4 c^4 x^2 z^3 + b^2 c^6 x^2 z^3 - 3 a^6 b^2 x y z^3 + 3 a^2 b^6 x y z^3 + 3 a^4 b^2 c^2 x y z^3 - 3 a^2 b^4 c^2 x y z^3 - a^8 y^2 z^3 + a^2 b^6 y^2 z^3 + a^6 c^2 y^2 z^3 - a^4 b^2 c^2 y^2 z^3 - 3 a^2 b^4 c^2 y^2 z^3 + a^4 c^4 y^2 z^3 + 3 a^2 b^2 c^4 y^2 z^3 - a^2 c^6 y^2 z^3 = 0

Francisco Javier, Hyacinthos #21962

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