Area of triangle under tan conditions. For an acute triangle $ABC$, the following conditions hold.
$$\frac{1}{\tan A} + \frac{1}{\tan B} + \frac{1}{\tan C } =2$$
$$ a^2 + b^2 + c^2 =50 $$
Compute the area of such a triangle.
I used a tan addition law and get,
$$\tan A + \tan B + \tan C = \tan A \tan B \tan C $$.
And let $ \tan A \tan B \tan C = k $, so $\tan A , 
\tan B , \tan C $ are roots of the following cubic equation.
$x^3 - k x^2 +2k x - k $.
I am stuck right here.

We have $$2=\sum_{cyc}\frac{ab\cos\gamma}{ab\sin\gamma}=\frac{\sum\limits_{cyc}(a^2+b^2-c^2)}{4S}=\frac{50}{4S},$$ which gives $$S=\frac{25}{4}.$$
