Let ABC be a triangle and A1B1C1, A2B2C2 the pedal triangles of two isogonal points P,P*, resp.
Let Q be a point on the line PP* and P1P2P3 the circumcevian triangle of Q with respect the triangle A1B1C1.
Conjecture:
The triangles A2B2C2, P1P2P3 are perspective. The perspector Q' lies on PP*.
APH, 22 February 2012
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The locus of Q such that P1P2P3 and A2B2C2 are perspective includes the line PP*
I find that the locus includes also the points on the circle with the midpoint
of PP* as center and and radius (AP BP CP)/(2 |OP^2 - R^2|).
Francisco Javier, Hyacinthos #20856.
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Let R1R2R3 be the reflection of the triangle P1P2P3 in the line PP*.
The triangles A2B2C2, R1R2R3 are perspective.
APH, 24 February 2012
It is true. Here is a proof:-
ΑπάντησηΔιαγραφήSuppose, A3B3C3 is the circumcevian triangle of P* wrt A2B2C2. R be the reflection of Q on O. A4B4C4 be the circumcevian triangle of R wrt A3B3C3. Clearly, P1P2P3 is the antipodal triangle of A4B4C4. Suppose, A2P1 intersects B2P2 at Q" and A2B4 intersects B2A4 at S. Applying Pascal's theorem on A2P1A4B2P2B4 we get that Q",S,R are collinear and applying it on A2A3A4B2B3B4 we get that P*,S,R are collinear. So Q",S',R,P* are collinear. So A2P1,B2P2 intersects on PP*, similarly A2P1,C2P3 intersects on PP* and so. So A2P1,B2P2,C2P3 are concurrent on PP*.
Suppose, PP* intersects the circumcircle of A1B1C1 at U,V. So (A2P1,A2R1;A2U,A2V)=-1. So A2R1 passes through the harmonic conjugate of Q' wrt UV. Similarly B2R2,C2R3 also passes through that point. Hence proved.
ΑπάντησηΔιαγραφή