Dear Georg and Jean-Pierre,
> > > 1)In the triangle ABC we consider A',B' and C' the simetric points
> > of
> > > A,B,C regarding the opposite sides. Prove that the triangle ABC
> > and
> > > A'B'C' have the same G point(weight point) if and only if the
> > triangle
> > > ABC is equilateral!
> > Roughly, I would say that the fixed point of the affine mapping
> > ABC->A'B'C' is the symedian point and that symedian point = centroid
> > <=> the triangle is equilateral
> > Surely, there exists something more elementary and, may be, shorter.
> > Friendly. Jean-Pierre
The solution I wrote this morning assumed that A'AG shouls be collinear.
That is a wrong assumption, so basicly all I wrote was useless.
Let us assume that ABC and A'B'C' share centroid G. Then, since the
centroid of A'B'C' is the reflection of the centroid of ABC through the
centroid of the orthic triangel A*B*C*, we have that A*B*C* must have
the same centroid G.
This means that the Euler lines of A*B*C* and ABC must coincide: apart
from the shared G also the ninepointcenter N lies on both Euler lines.
This means that the orthocenter H of ABC lies on the Euler line of
A*B*C*. The orthocenter is an incenter or an excenter of A*B*C*. On the
other hand GN:NH = 1:3, or G*O*:O*Ix* = 1:3. This latter relation should
easily lead analyticly to A*B*C* is equilateral. So G*=O*, hence G=N,
hence G=H and ABC is equilateral as well.