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Problem 18

# If $$\tan \alpha, \tan \beta$$, tan $$\gamma$$ are the roots of the equation $$x^{3}-3 a x^{2}+3 b x-1=0$$. Find the centroid of the triangle whose vertices are $(\tan \alpha, \cot \alpha),(\tan \beta, \cos \beta),(\tan \gamma, \cot \gamma)$.

Expert verified
The centroid of the triangle is at point $$(a, b)$$.
See the step by step solution

## Step 1: Find the roots of the cubic equation

The roots of the cubic equation are given as $$\tan{\alpha}, \tan{\beta}, \tan{\gamma}$$. According to Vieta's formulas, the sum of the roots of a cubic equation $$x^{3}-3 a x^{2}+3 b x-1=0$$ is equal to $$3a$$. By the addition formula for tangent function, $$\tan{\alpha} + \tan{\beta} + \tan{\gamma} = \tan{\alpha} + \tan{\beta} - \tan{(\pi - \gamma)} = \tan{(\alpha +\beta)} = 3a$$.

## Step 2: Calculate the $$x$$-coordinate of centroid

Substituting the values into the formula for the $$x$$-coordinate of the centroid, we get $$\frac{\tan \alpha + \tan \beta + \tan \gamma}{3} = a$$.

## Step 3: Calculate the $$y$$-coordinate of centroid

Similarly, we can follow the same steps to find the $$y$$-coordinates of the triangle vertices. We know that $$\cot(\alpha) = \frac{1}{\tan(\alpha)}$$, so we can compute the $$y$$-coordinate of the centroid as $$\frac{\frac{1}{\tan \alpha} + \frac{1}{\tan \beta} + \frac{1}{\tan \gamma}}{3}$$. But knowing that $$\frac{1}{\tan \alpha} + \frac{1}{\tan \beta} + \frac{1}{\tan \gamma} = 3b$$, the $$y$$-coordinate of the centroid simplifies to $$b$$.

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