<s>[INST] A trigonometry question involving medians and a sum of cotangents The medians of a triangle $ABC$ make angles $\alpha$, $\beta$ and $\gamma$ with each other. I have to prove that:

$$\cot\alpha+ \cot\beta+ \cot\gamma +\cot(A)+\cot(B)+\cot(C)= 0.$$

Any ideas?
 [/INST] 
By the sine and cosine theorems,
$$\cot(\widehat{BGC}) = \frac{\cos\widehat{BGC}}{\sin\widehat{BGC}} = \frac{\frac{BG^2+CG^2-BC^2}{2\, BG\cdot CG}}{\frac{2[BGC]}{BG\cdot CG}}=\frac{1}{4}\cdot\frac{BG^2+CG^2-BC^2}{[BGC]}\tag{1}$$
but $BG=\frac{2}{3}m_a$ and Stewart's theorem gives $m_a^2 = \frac{2b^2+2c^2-a^2}{4}$, hence:
$$\cot(\widehat{BGC}) = \frac{3}{4}\cdot\frac{BG^2+CG^2-BC^2}{[ABC]} = \frac{b^2+c^2-5a^2}{12[ABC]}\tag{2}.$$
On the other hand,
$$ \cot(\widehat{BAC}) = \frac{\cos\widehat{BAC}}{\sin\widehat{BAC}}=\frac{\frac{AB^2+AC^2-BC^2}{2\,AB\cdot AC}}{\frac{2[ABC]}{AB\cdot AC}}=\frac{b^2+c^2-a^2}{4[ABC]}\tag{3}$$
hence:
$$ \cot A+\cot B+\cot C = \frac{a^2+b^2+c^2}{4[ABC]} = -\left(\cot\alpha+\cot\beta+\cot\gamma\right)\tag{4} $$
as wanted.
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