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.

Another try.

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.

Kind regards,

Sincerely,

Floor.