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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 edited it, is that clearer like this ?2011-10-25

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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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    You're right Mark, that's what I'm looking for. Sorry for my bad explanations...2011-10-25