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Let's say I have a triangle $ABC$, the middle of the sides are called $A'$, $B'$ and $C'$.

I have proved that $\Omega$, the orthocenter of $ABC$, is the barycentre of $A'B'C'$ with masses $\tan \alpha$, $\tan \beta$ and $\tan \gamma$ on $A'$, $B'$ and $C'$. Now I have to deduce from this that $\Omega$ is also the barycentre of $A$, $B$ and $C$ with masses $a$, $b$ and $c$. I want to find $a, b, c$ so that $\Omega$ is the barycentre of $ABC$.

Thank you in advance!

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I think this question is rather unclear as to what is being asked. Is my comment on the answer below essentially what is intended? If so the question needs to be modified accordingly. – Mark Bennet Oct 25 '11 at 18:21
I edited it, is that clearer like this ? – Skydreamer Oct 25 '11 at 18:43
up vote 3 down vote accepted

It's not true that the orthocentre of $ABC$ is the barycentre of $A'B'C'$ with masses $\tan\alpha$, $\tan\beta$ and $\tan\gamma$. In a right triangle, the orthocentre is at the corner $C$ with the right angle, whereas that barycentre would be at $C'$, since $\tan\gamma$ goes to infinity. More generally, as long as all angles are acute, that barycentre would be within $A'B'C'$, wheras the orthocentre of $ABC$ doesn't have to be.

If your result were correct, it would be straightforward to obtain a corresponding mass distribution on $A$, $B$ and $C$: Move half the mass on the midpoints to each of the adjacent corners; that moves all the mass to the corners without moving the barycentre. So the result would be masses of $(\tan\alpha+\tan\beta)/2$ on $C$ etc.

For the correct barycentric coordinates of the orthocentre, see this Wikipedia section.

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By what I recall, if $\omega$ is the orthocenter of a triangle and its barycenter, then, by $lal$, that the triangle is equilateral.

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Now that's an "If and only if". – Henrique Tyrrell Oct 25 '11 at 18:14
I think OP means that if there are masses $\tan \alpha$ etc at the vertices $A', B', C'$ then the barycentre of these is the orthocentre of triangle ABC, and wants to know what masses should be placed at A, B, C (up to a factor) so that the barycentre of those masses is also the orthocentre, using what he has already established. I think ... but the question is rather unclear, and I was going to post much the same as you. – Mark Bennet Oct 25 '11 at 18:20
You're right Mark, that's what I'm looking for. Sorry for my bad explanations... – Skydreamer Oct 25 '11 at 18:32

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