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Question
what is the value of \[\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}\] .
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Solution
The co-ordinates of the vertices are (a, b); (b, c) and (c, a)
The co-ordinate of the centroid is (0, 0)
We know that the co-ordinates of the centroid of a triangle whose vertices are `(x_1 , y_1 ) ,(x_2 ,y_2 ) , (x_3 , y_3)` is-
`((x_1 + x_2 + x_3 ) / 3 , (y_1 + y_2 +y_3 ) / 3)`
So,
`( 0 , 0) = ((a + b +c ) / 3 , ( b +c +a ) / 3 )`
Compare individual terms on both the sides-
`(a + b+ c ) / 3 = 0`
Therefore,
a + b + c = 0
We have to find the value of -
=`(a^2)/(bc) + (b^2)/(ca) + (c^2)/(ab) `
Multiply and divide it by (abc) to get,
` = (1/(abc)) ( a^3 + b^3 + c^3 )`
Now as we know that if,
a + b + c = 0
Then,
`a^3 + b^3 + c^3 = 3abc`
So,
`(a^2 ) /( bc) +(b^2)/(ca) + (c^2)/(ab) = (1/(abc)) (a^3 + b^3 +c^3)`
`= (1/(abc))(3abc)`
= 3
