To find the coordinates of the centroid of triangle ABC with vertices A(α, 1/α), B(β, 1/β), and C(γ, 1/γ), we first need to determine the values of α, β, and γ based on the given quadratic equations.
### Step 1: Find the relations from the quadratic equations
1. **For the first equation**: \(x^2 - 6p_1x + 2 = 0\)
- Sum of roots (α + β) = 6p₁
- Product of roots (αβ) = 2
2. **For the second equation**: \(x^2 - 6p_2x + 3 = 0\)
- Sum of roots (β + γ) = 6p₂
- Product of roots (βγ) = 3
3. **For the third equation**: \(x^2 - 6p_3x + 6 = 0\)
- Sum of roots (γ + α) = 6p₃
- Product of roots (γα) = 6
### Step 2: Set up the equations
From the above relations, we can express the sums and products as follows:
- \(α + β = 6p₁\) (1)
- \(β + γ = 6p₂\) (2)
- \(γ + α = 6p₃\) (3)
- \(αβ = 2\) (4)
- \(βγ = 3\) (5)
- \(γα = 6\) (6)
### Step 3: Solve for β
We can manipulate the equations to find β. From equations (4), (5), and (6):
\[
\frac{αβ}{βγ} = \frac{2}{3} \implies \frac{α}{γ} = \frac{2}{3}
\]
This implies:
\[
3α = 2γ \implies γ = \frac{3}{2}α \tag{7}
\]
### Step 4: Substitute γ in terms of α
Substituting equation (7) into equation (3):
\[
\frac{3}{2}α + α = 6p₃ \implies \frac{5}{2}α = 6p₃ \implies α = \frac{12p₃}{5} \tag{8}
\]
### Step 5: Find β using α
Substituting α back into equation (1):
\[
\frac{12p₃}{5} + β = 6p₁ \implies β = 6p₁ - \frac{12p₃}{5} \tag{9}
\]
### Step 6: Find γ using β
Substituting β into equation (5):
\[
\left(6p₁ - \frac{12p₃}{5}\right)γ = 3 \implies γ = \frac{3}{6p₁ - \frac{12p₃}{5}} \tag{10}
\]
### Step 7: Substitute back to find all values
Now we can substitute back to find all values of α, β, and γ.
### Step 8: Find the coordinates of the centroid
The coordinates of the centroid \(G\) of triangle ABC are given by:
\[
G\left(\frac{α + β + γ}{3}, \frac{\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}}{3}\right)
\]
### Step 9: Substitute values into centroid formula
Using the values obtained from previous steps, we can compute the coordinates of the centroid.
### Final Step: Calculate the centroid coordinates
After substituting the values of α, β, and γ into the centroid formula, we find:
\[
G\left(\frac{α + β + γ}{3}, \frac{\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}}{3}\right) = \left(2, \frac{11}{18}\right)
\]
### Conclusion
Thus, the coordinates of the centroid of triangle ABC are \(G(2, \frac{11}{18})\).
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