RADICAL AXES

Let ABC be a triangle, P a point, A'B'C' the pedal triangle of P.

Denote:

(Oa) = the circumcircle of (PBC)

(O1) = the reflection of (Oa) in BC

(O'1) = the reflection of (O1) in PA'

Ra = the radical axis of the pedal circle of P and (Oa)

R1 = the radical axis of the pedal circle of P and (O1)

R'1 = the radical axis of the pedal circle of P and (O'1)

Similarly Rb,Rc, R2,R3, R'2,R'3

Sa = the radical axis of the antipedal circle of P and (Oa)

S1 = the radical axis of the antipedal circle of P and (O1)

S'1 = the radical axis of the antipedal circle of P and (O'1)

Similarly Sb,Sc, S2,S3, S'2,S'3

Ab, Ac =

1. the orthogonal projections of Oa on Rb,Rc, resp.

APH, Hyacinthos #22150

2. the orthogonal projections of Oa on R2,R3, resp.

APH, Hyacinthos #22153

3. the orthogonal projections of Oa on R'2,R'3, resp.

APH, Hyacinthos #22153

4. the orthogonal projections of Oa on Sb,Sc, resp.

5. the orthogonal projections of Oa on S2,S3, resp.

6. the orthogonal projections of Oa on S'2,S'3, resp.

Similarly Bc,Ba and Ca, Cb

Which is locus of P such that the perpendicular bisectors of the line segments BaCa, CbAb,AcBc are concurrent?

1.

The perpendicular bisectors of the line segments BaCa, CbAb, AcBc are concurrent for all P.

If P=(x:y:z), the perpendicular bisectors concurrent in Q with first barycentric coordinate:

a^6y^2z^2(4x^2+4y*z+3x(y+z)) + a^4x*y*z(c^2y(5x^2+2x*y+2(y-z)z) + b^2z(5x^2+2x z+2y(-y+z))) - a^2x(c^4y^2(z^2(y+z)+2x*z(3y+2z)+2x^2(y+3z)) + b^4z^2(y^2(y+z)+2x^2(3y+z)+2x*y(2y+3z)) + 2b^2c^2y z(4x^2(y+z)-y*z(y+z) + x(y^2+4y*z+z^2))) + (b^2-c^2)x^3(c^4y^2(2y-z)+3b^2c^2y(y-z)z + b^4(y-2z)z^2)

The unique pairs of points {P, Q}, both being in ETC are, {X(1),X(3)} and {X(4),X(546)}.

If P=(x:y:z) lies on the circumcircle or line at infinity, the construction does not make sense.

However, as the coordinates sum of Q is 4(a-b-c)(a+b-c)(a-b+c)(a+b+c)xyz(x+y+z)(c^2xy+b^2xz+a^2yz), then Q lies on the line at infinity.

In fact, if P is in the circumcircle then Q is the isogonal conjugate of P.

Angel Montesdeoca, Hyacinthos #22152