7223Re: Variation on the Van Lamoen-Grinberg-Wolk-transform
- Jun 1, 2003Dear Eric Danneels,
>> 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 circumcevianThis 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' ?
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