18089Re: Problem from Vietnam Team Selection Tests 2009
- Aug 1, 2009Dear Antreas,
The locus is formed by the circumcircle and the quartic
-a^2 c^2 x^2 y^2 + b^2 c^2 x^2 y^2 + a^2 b^2 x^2 z^2 - b^2 c^2 x^2 z^2 - a^2 b^2 y^2 z^2 + a^2 c^2 y^2 z^2 = 0.
This quartic is the isogonal conjugate of the conic
b^4 c^2 x^2 - b^2 c^4 x^2 - a^4 c^2 y^2 + a^2 c^4 y^2 + a^4 b^2 z^2 - a^2 b^4 z^2 = 0
This conic, always an hyperbola, is centered at X110, the focus of the Kiepert parabola. Some points on this hyperbola include
Best regards from Almería, Spain.
--- In Hyacinthos@yahoogroups.com, "xpolakis" <anopolis72@...> wrote:
> Another problem with the "Vietnameze" triangle.
> Let ABC be a triangle, P a point, A1B1C1 the circumcevian
> triangle of P and A2B2C2 the cevian triangle of P.
> Let A3,B3,C3 be the second intersections of the circumcircle
> of ABC with the circumcircles of AB2C2, BC2A2, CA2B2, resp.
> Which is the locus of P such that A1B1C1, A3B3C3 are
> > Let an acute triangle ABC with curcumcircle (O). Call A_1,B_1,C_1
> > are foots of perpendicular line from A,B,C to opposite side.
> > A_2,B_2,C_2 are reflect points of A_1,B_1,C_1 over midpoints
> > of BC,CA,AB respectively.
> > Circle (AB_2C_2),(BC_2A_2),(CA_2B_2) cut (O) at A_3,B_3,C_3
> > respectively.
> > Prove that: A_1A_3,B_1B_3,C_1C_3 are concurent.
> > Vietnam Team Selection Tests 2009
> > http://www.mathlinks.ro/resources.php?c=186&cid=41&year=2009
> > In other words:
> > Let ABC be a triangle, A1B1C1 the cevian triangle of H, A2B2C2 the
> > cevian triangle of the isotomic conjugate of H, and A3,B3,C3
> > the second intersections of the circumcircle of ABC
> > with the circumcircles of AB2C2, BC2A2, CA2B2, resp.
> > The triangles A1B1C1, A3B3C3 are perspective.
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