- Dear Alexei
> The polar transformation and the properties of Simson lines allow to

prove next interesting fact.

> Let I be the incenter of ABC. For an arbitrary point P we denote as

P_A the point on the line PA such that IA is the bisector (internal or

external) of angle PIP_A. The points P_B, P_C are defined similarly.> 1. If P_A, P_B, P_C are collinear then this line touches the incircle

of ABC.

> 2. The locus of points P with this property is the line.

Here is a better presentation of a generalization :

> Can anybody prove this without using the polar transformation?

Starting with a point J in the plane of ABC, the perpendicular lines

through J to JA, JB, JC intersect BC, CA, AB respectively at A', B', C'.

It is well known that A',B',C' are on a same line L(J).

For any point M,

Ma is the point of the line AM such as <MJA = <AJMa

Mb is the point of the line BM such as <MJB = <BJMb

Mc is the point of the line CM such as <MJC = <CJMc

Then Ma,Mb,Mc are collinear if and only if M lies on L(J)

In this case, the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the

inscribed conic with focus J

This can be proved with projective tools using the fact that M->Ma is

the harmonic homology with center A and axis the line JA'

* The harmonic homology (I don't know if this English name is correct)

with center O, axis L maps M to his harmonic conjugate wrt O and OM

inter L

I think that, if we consider three harmonic homologies with centers

A,B,C, the locus of M with three colinear images is generally a cubic,

even when the three axis concur. In the particular case of three axis

La,Lb,Lc going through a point J and such as there exists an involution

swapping (JA,La), (JB,Lb),(JC,Lc), then the cubic degenerates in three

lines :

the fixed lines of the involution and the line through La inter BC, Lb

inter CA, Lc inter AB

Friendly. Jean-Pierre - [APH]:Let ABC be a triangle P a point and A'B'C', A"B"C" the pedal, circumcevian triangles of P, resp.

Denote:

A*, B*, C* = the other (than A", B", C") intersections of A"A', B"B', C"C' and the circumcircle, resp.

S1, S2, S3 = the Simson lines of A*, B*, C*, resp.

AsBsCs = the triangle bounded by S1, S2, S3.

Which is the locus of P such that:

1. ABC, AsBsCs are parallelogic?

The entire plane?

2. S1, S2, S3 are concurrent?

[César Lozada]:2) Simson lines concur for P on: {sidelines} \/ { Line at infinity} \/ {circum-sixtic q6 through ETC’s: 4, 64, 84})

q6: Σ[ v*w*((SB-SC)*SB*SC*a*v^2*w^2+( ((SB-SA)*S^2+3*SA^2*SB)*b*v-(( SC-SA)*S^2+3*SA^2*SC)*c*w)*u^ 3)] = 0 (trilinears)

ETC-pairs (P,Z2(P)=point of concurrence): (4, 4), (64, 5562), (84,10)

__1)____Locus for parallelogic ABC and AsBsCs = the entire plane__The parallelogic center A->As is Za(P) = Isogonal(Reflection(Isogonal( P), O)),

i.e., the Isogonal-conjugate of the-reflection-in-O of the isogonal-conjugate-of-P.

The parallelogic center Zs(P)=As->A has a complicated expression. There are no ETC pairs (P,Zs(P)) for non-concurrent Simson lines.

Points Zs(P) seem not to be of interest: long coordinates and very few relations with existing 13600+ ETC centers.

César Lozada