First slide

Cartesian plane

Question

If $α, β, γ$ are the real roots of the equation $x^{2} - 3 a x^{2} + 3 b x - 1 = 0$ then the centroid of the triangle with metrices $(α, \frac{1}{α}), (β, \frac{1}{β})$ and $(γ, \frac{1}{γ})$ is at the point

Easy

A

$(a, b)$
B

$(a / 3, b / 3)$
C

$(a + b, a - b)$
D

$(3 a, 3 b)$

Solution

We have
$α + β + γ = 3 a, α β + β γ + γ α = 3 b$ and $α β γ = 1$
The centroid of the given triangle is at the point
$\frac{a + β + γ}{3}, \frac{\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}}{3} = (\frac{α + β + γ}{3}, \frac{α β + β γ + γ α}{3 α β γ}) = (a, b)$

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