Dear Eric Danneels,

You wrote:

>> Consider a point P and a triangle ABC.

>> Let A'B'C' be the circumcevian triangle of P

>> with respect to ABC.

>> Let A°, B° and C° be the circumcenters of

>> the triangles PB'C', PC'A' and PA'B'.

>>

>> ==> The lines AA°, BB° and CC° pass through

>> a point S on the circumcircle of ABC.

Do you have a synthetic proof? The property was

established by Paul Yiu with barycentrics in

Hyacinthos message #3957, but I think that there

must be some easy proof.

>> If we consider ABC as the circumcevian

>> triangle of P with respect to A'B'C' we have

>>

>> Let A*, B* and C* be the circumcenters of the

>> triangles PBC, PCA and PAB.

>>

>> ==> The lines A'A*, B'B* and C'C* pass

>> through a point T on the circumcircle of ABC,

>> It seems that isogonal conjugates map to the

>> same point.

This is very interesting.

Probably, an idea to prove that the lines A'A*,

B'B* and C'C* concur at a point on the

circumcircle of ABC analytically but without

much calculation would be using the angles

A' = angle BPC, B' = angle CPA, C' = angle APB.

If I don't mistake, the point A* has trilinears

A* ( -cos A : cos(C+A') : cos(B+A') ).

But what are the trilinears of A' ?

Sincerely,

Darij Grinberg