# Mock AIME 2 Pre 2005 Problems/Problem 11

$\alpha$, $\beta$, and $\gamma$ are the roots of $x(x-200)(4x+1) = 1$. Let $$\omega = \tan^{-1}(\alpha) + \tan^{-1}(\beta) + \tan^{-1} (\gamma).$$ The value of $\tan(\omega)$ can be written as $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Determine the value of $m+n$.

## Solution

We know that $\alpha, \beta, \gamma$ are the roots of $x(x-200)(x+1/4)-1/4 = x^3 - \frac{799}{4}x^2 - 50x - \frac{1}{4}$. By Vieta's formulas, we have:

$\alpha + \beta + \gamma = \frac{799}{4}$

$\alpha\beta + \beta\gamma + \gamma\alpha = -50$

$\alpha\beta\gamma = \frac{1}{4}$

Now, by tangent addition formulas, we have $\tan(\omega) = \frac{\alpha + \beta + \gamma - \alpha\beta\gamma}{1 - \alpha\beta - \beta\gamma - \gamma\alpha}$. Substituting Vieta's formulas, we obtain $\tan(\omega) = \frac{\frac{799}{4} - \frac{1}{4}}{1 - (-50)} = \frac{\frac{798}{4}}{51} = \frac{133}{34}$. Therefore, our answer is $133 + 34 = \boxed{167}$ and we are done.