Let ABC be a triangle, A1B1C1 the orthic triangle, A2B2C2 the circumcevian triangle of O with respect A1B1C1, A3B3C3 the antipodal triangle of A2B2C2 and A'B'C' the triangle bounded by the lines A1A3, B1B3, C1C3, resp..
The triangles A1B1C1 and A'B'C' are perspective.
Perspector?
APH, 19 March 2012
-------------------------------------------------
The perspector is X25.
Francisco Javier García Capitán
19 March 2012
Suppose, A4B4C4 is the intouch triangle of A1B1C1. A5B5C5 be the orthic triangle of A4B4C4. Now note that in the triangle of A1B4C4, harmonic conjugate of A1A5 wrt A1B1,A1C1 is perpendicular to HA5. But since A4B4C4 is homothetic to ABC, so HA5 is parallel to A1O which is perpendicular to A1A3. So harmonic conjugate of A1A5 wrt A1B1,A1C1 is A1A3. So the triangle formed by A1A3,B1B3,C1C3 is perspective to A1B1C1 with perspective center as the concurrency point of A1A5,B1B5,C1C5. So the perspector is the center of homothety of ABC and A4B4C4.
ΑπάντησηΔιαγραφή