- Dear Hyacynthians,

looking for variations of the Van Lamoen-Grinberg-Wolk-transform I

discovered the following properties:

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.

I found the following relationships using the ETC

P = Incenter X(1) ==> S = X(104)

P = Centroid X(2) ==> S = X(98)

P = Circumcenter X(3) ==> S = X(74)

P = Orthocenter X(4) ==> S = undetermined

P = Euler-center X(5) ==> S = X(1141)

P = Lemoine point X(6) ==> S = X(74)

P = Isogonal conjugate of Euler-center X(54) ==> S = X(74)

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,

I found the following using the ETC

P = Incenter X(1) ==> T = undetermined

P = Centroid X(2) ==> T = X(1296)

P = Circumcenter X(3) ==> T = X(110)

P = Orthocenter X(4) ==> T = X(110)

P = Euler-center X(5) ==> T = X(1291)

P = Lemoine point X(6) ==> T = X(1296)

P = Isogonal conjugate of Euler-center X(54) ==> T = X(1291)

It seems that isogonal conjugates map to the same point.

Are these properties known ?

Greetings from Belgium

Eric Danneels - Dear Eric Danneels,

You wrote:

>> Consider a point P and a triangle ABC.

Do you have a synthetic proof? The property was

>> 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.

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

This is very interesting.

>> 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.

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