## Κυριακή, 29 Ιανουαρίου 2012

### Perspective

Let ABC be a triangle, HaHbHc the cevian triangle of H (orthic triangle), GaGbGc the cevian triangle of G (medial triangle), H1H2H3, G1G2G3 the circumcevian triangles of H and G, resp.

Denote:

A1 = the 2nd intersection of H1Ga and the circumcircle (other than H1)
A2 = the 2nd intersection of G1Ha and the circumcircle (other than G1)

B1 = the 2nd intersection of H2Gb and the circumcircle (other than H2)
B2 = the 2nd intersection of G2Hb and the circumcircle (other than G2)

C1 = the 2nd intersection of H3Gc and the circumcircle (other than H3)
C2 = the 2nd intersection of G3Hc and the circumcircle (other than G3)

The lines A1A2, B1B2, C1C2 bound a triangle A*B*C*
(A* = B1B2 /\ C1C2, B* = C1C2 /\ A1A2, C* = A1A2 /\ B1B2)

The triangles ABC, A*B*C* are perspective.

Perspector?

Generalization:
Locus of P instead of 1.H, 2.G

APH 29 January 2012

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The coordinates are:

{a^2 (a^2 + b^2 - c^2) (a^2 - b^2 + c^2) (3 a^4 - 6 a^2 b^2 + 3 b^4 +
c^4) (3 a^4 + b^4 - 6 a^2 c^2 + 3 c^4),
b^2 (a^2 + b^2 - c^2) (-a^2 + b^2 + c^2) (3 a^4 - 6 a^2 b^2 + 3 b^4 +
c^4) (a^4 + 3 b^4 - 6 b^2 c^2 + 3 c^4),
c^2 (a^2 - b^2 + c^2) (-a^2 + b^2 + c^2) (3 a^4 + b^4 - 6 a^2 c^2 +
3 c^4) (a^4 + 3 b^4 - 6 b^2 c^2 + 3 c^4)}

Not in ETC

The result is true for any P, Q.
The Isogonal Conjugate of the perspector for P=(u:v:w) and Q(x:y:z) is:.

{u x (w^2 y^2 - v w y z + v^2 z^2), v y (w^2 x^2 - u w x z + u^2 z^2), wz (v^2 x^2 - u v x y + u^2 y^2) }

Francisco Javier García Capitán
30 January 2012