Multiple choice

If $\alpha , \beta , \gamma $ are the real roots of the equation $x^{3}-3px^{2}+3qx-1=0$, then the centroid of the triangle with vertices $\displaystyle \left ( \alpha , \frac{1}{\alpha } \right )$, $\left ( \beta , \dfrac{1}{\beta } \right )$ and $\displaystyle \left ( \gamma , \frac{1}{\gamma } \right )$ is at the point

  1. $(p, q)$
  2. $\left(\dfrac p3, \dfrac q3\right)$
  3. $(p+q, p-q)$
  4. $(3p, 3q)$
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A Correct answer
Explanation

The centroid of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is ((x1+x2+x3)/3, (y1+y2+y3)/3). For the given cubic equation, the sum of roots alpha + beta + gamma = 3p and the sum of reciprocals (alpha*beta + beta*gamma + gamma*alpha) / (alpha*beta*gamma) = 3q/1 = 3q. Dividing the sums by 3 gives the centroid coordinates (p, q).

AI explanation

By Vieta's formulas for the cubic equation x^3 - 3px^2 + 3qx - 1 = 0, the sum of the roots is alpha + beta + gamma = 3p. Factoring the equation as (x^3 - 1) - 3px^2 + 3qx = 0 and dividing by x gives 1/alpha + 1/beta + 1/gamma = 3q. The centroid is the average of the vertices, so its x-coordinate is (alpha+beta+gamma)/3 = 3p/3 = p. The y-coordinate is (1/alpha + 1/beta + 1/gamma)/3 = 3q/3 = q, giving the point (p, q).

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