## Re: [EMHL] Re: a geometry problem regarding equilater triangle!

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• Dear Georg and Jean-Pierre, ... Let A*B*C* be the the orthic triangle. The midpoints of BC, B C and B*C* are collinear. On the other hand A, the midpoint of
Message 1 of 2 , Sep 30, 2003
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
>

Let A*B*C* be the the orthic triangle. The midpoints of BC, B'C' and
B*C* are collinear. On the other hand A, the midpoint of B*C* and the
symmedian point are collinear. So the symmedian point and G must be both
on the same Cevian from A. Since these Cevians are normally symmetric
through the angle bisector, this means that ABC must be isosceles with
top angle A. In the same way ABC isosceles with top angle B. Hence ABC
is equilateral.

Kind regards,
Sincerely,
Floor.
• Dear Georg and 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
Message 2 of 2 , Oct 1, 2003
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.
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