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## Variation on the Van Lamoen-Grinberg-Wolk-transform

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• Dear Hyacynthians, looking for variations of the Van Lamoen-Grinberg-Wolk-transform I discovered the following properties: Consider a point P and a triangle
Message 1 of 2 , Jun 1, 2003
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, ... Do you have a synthetic proof? The property was established by Paul Yiu with barycentrics in Hyacinthos message #3957, but I think that
Message 2 of 2 , Jun 1, 2003
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
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