Let `A(alpha, 1/(alpha)),B(beta,1/(beta)),C(gamma,1/(gamma))` be the vertices of a `DeltaABC` where `alpha, beta` are the roots of the equation
`x^(2)-6p_(1)x+2=0, beta, gamma` are the roots of the equation
`x^(2)-6p_(2)x+3=0` and `gamma, alpha` are the roots of the equation
`x^(2)-6p_(3)x+6=0, p_(1),p_(2),p_(3)` being positive. Then the coordinates of the cenroid of `DeltaABC` is

Options

A. `(1,11/18)`
B. `(0,11/8)`
C. `(2,11/18)`
D. None of these

Answer

To find the coordinates of the centroid of triangle ABC with vertices at points A(α, 1/α), B(β, 1/β), and C(γ, 1/γ), we will follow these steps:

### Step 1: Find the roots α and β from the first equation
The first equation is given as:
\[ x^2 - 6p_1x + 2 = 0 \]
Using Vieta's formulas, we know:
- The sum of the roots (α + β) = 6p_1
- The product of the roots (αβ) = 2

### Step 2: Find the roots β and γ from the second equation
The second equation is given as:
\[ x^2 - 6p_2x + 3 = 0 \]
Using Vieta's formulas again:
- The sum of the roots (β + γ) = 6p_2
- The product of the roots (βγ) = 3

### Step 3: Find the roots γ and α from the third equation
The third equation is given as:
\[ x^2 - 6p_3x + 6 = 0 \]
Using Vieta's formulas:
- The sum of the roots (γ + α) = 6p_3
- The product of the roots (γα) = 6

### Step 4: Multiply the three equations
From the three sets of equations, we can multiply the products of the roots:
\[ (αβ)(βγ)(γα) = 2 \cdot 3 \cdot 6 \]
This gives us:
\[ (αβγ)^2 = 36 \implies αβγ = 6 \]

### Step 5: Express α, β, and γ in terms of p_1, p_2, and p_3
Now we can express α, β, and γ using the products:
- From αβ = 2, we can express β as \( β = \frac{2}{α} \).
- From βγ = 3, we can express γ as \( γ = \frac{3}{β} = \frac{3α}{2} \).
- From γα = 6, we can express α as \( α = \frac{6}{γ} = \frac{6 \cdot 2}{3} = 4 \).

### Step 6: Substitute to find β and γ
Using α = 4:
- \( β = \frac{2}{4} = \frac{1}{2} \)
- \( γ = \frac{3}{\frac{1}{2}} = 6 \)

### Step 7: Find the coordinates of the centroid
The coordinates of the centroid (G) of triangle ABC can be calculated using the formula:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the coordinates:
- \( x_1 = α = 4 \)
- \( x_2 = β = \frac{1}{2} \)
- \( x_3 = γ = 6 \)

Calculating the x-coordinate of the centroid:
\[ G_x = \frac{4 + \frac{1}{2} + 6}{3} = \frac{10.5}{3} = 3.5 \]

Calculating the y-coordinates:
- \( y_1 = \frac{1}{α} = \frac{1}{4} \)
- \( y_2 = \frac{1}{β} = 2 \)
- \( y_3 = \frac{1}{γ} = \frac{1}{6} \)

Calculating the y-coordinate of the centroid:
\[ G_y = \frac{\frac{1}{4} + 2 + \frac{1}{6}}{3} = \frac{\frac{3}{12} + \frac{24}{12} + \frac{2}{12}}{3} = \frac{\frac{29}{12}}{3} = \frac{29}{36} \]

### Final Coordinates of the Centroid
Thus, the coordinates of the centroid G are:
\[ G\left(3.5, \frac{29}{36}\right) \]