NO COMPUTATIONS, PLEASE! -- 1

Problem:

Let ABC be a triangle, P a point, A'B'C' the pedal triangle of P and Ab, Ac the orthogonal projections of A' on PB',PC', resp. The line AbAc intersects BC at A". Similarly B' and C'. Which is the locus of P such that the points A',B',C' are collinear?

Lemma:

Let BC be a line segment, A' a fixed point on BC and A a variable point on the perpendicular to BC at A'. Let Ab,Ac be the orthogonal projections of A' on AB,AC, resp. The line AbAc passes through a fixed point.

Let A" be the intersection of the lines BC and AbAc.

We have that:

A"B/A"C = (A'B/A'C)^2 (the proof is left to the reader).

Therefore AbAc passes through the fixed point A".

Now, in the Problem we have:

A"B/A"C = (A'B/A'C)^2

and similarly:

B"C/C"A = (B'C/B'A)^2

C"A/A"B = (C'A/C'B)^2

A",B",C" are collinear ==>

(A"B/A"C).(B"C/C"A).(C"A/A"B) = [(A'B/A'C).(B'C/B'A).(C'A/C'B)]^2 = 1

==>

1. (A'B/A'C).(B'C/B'A).(C'A/C'B) = -1 ==> ABC, A'B'C' are perspective
(ie AA',BB',CC' are concurrent, by Ceva Theorem)

Or

2. (A'B/A'C).(B'C/B'A).(C'A/C'B) = +1 ==> A',B',C' are collinear (by Menelaus Theorem).

1: The locus is the Darboux cubic
2: The locus is the Circumcircle + the Line at Infinity.

Therefore the locus of P is the union of the Darboux cubic, the Circumcircle, and the Line at Infinity.