Triangle Centroid

Vectors prove to be convenient in this problem.

Let $\vec{u} = \vec{AB}$ and $\vec{v} = \vec{AC}$

and let $a = | \vec{BC} |, b = | \vec{AC} |=| \vec{v} | , c = | \vec{AB} | = | \vec{u} |$

and let $d = | \vec{OA} |, e = | \vec{OB} | , f = | \vec{OC} |$

Then

$\vec{AO} = \dfrac{1}{3} ( \vec{u}+\vec{v} ) \cdots (1)$

$\vec{OB} = \vec{u} - \vec{AO} = \dfrac{2}{3} \vec{u} - \dfrac{1}{3} \vec{v} \cdots (2)$

and

$\vec{OC} = \vec{v} - \vec{AO} = \dfrac{2}{3} \vec{v} - \dfrac{1}{3} \vec{u} \cdots (3)$

In addition,

$\vec{BC} = \vec{u} - \vec{v} \cdots (4)$

Now using the length formula for a vector, and equation (1), we have

$d^2 = \dfrac{1}{9} ( \vec{u} \cdot \vec{u} + \vec{v} \cdot \vec{v} + 2 \vec{u} \cdot \vec{v} )$

but $\vec{u} \cdot \vec{u} = c^2$ and $\vec{v} \cdot \vec{v} = b^2$

Hence,

$d^2 = \dfrac{1}{9} ( c^2 + b^2 + 2 \vec{u} \cdot \vec{v} ) \cdots (5)$

similarly, we obtain from equations (2) through (4), the following,

$e^2 = \dfrac{1}{9} ( 4 c^2 + b^2 - 4 \vec{u} \cdot \vec{v} ) \cdots (6)$

and

$f^2 = \dfrac{1}{9} ( c^2 + 4 b^2 - 4 \vec{u} \cdot \vec{v} ) \cdots (7)$

and finally, from equation (4) , we have

$a^2 = b^2 + c^2 - 2 \vec{u} \cdot \vec{v} \cdots (8)$

Adding equations (5) through (7), we get

$d^2 + e^2 + f^2 = \dfrac{1}{9} ( 6 b^2 + 6 c^2 - 6 \vec{u} \cdot \vec{v}) \cdots (9)$

Using equation (8) to replace the last term, we get

$d^2 + e^2 + f^2 =\dfrac{1}{9}( 6 b^2 + 6 c^2 + 3 (a^2 - b^2 - c^2 ) )$

which becomes,

$d^2 + e^2 + f^2 = \dfrac{1}{3} (a^2 + b^2 + c^2) \cdots (10)$

So that finally, we have

$\dfrac{a^2 + b^2 + c^2}{d^2 + e^2 + f^2} = 3$