Question

If $G$ is the centroid of any $△ABC$ then find the value of $\frac{A{B}^2+B{C}^2+C{A}^2}{A{G}^2+B{G}^2+C{G}^2}$.

Answer 1

Let $A(x_1,y_1)B(x_2,y_2)$ and $C(x_3,y_3)$ be vertices of $\Delta ABC$

Assume that centroid of $\Delta ABC$ to be at the origin

i.e; G = (0,0)

centroid of $\Delta ABC=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

$\therefore \frac{x_1+x_2+x_3}{3}=0$ and $\frac{y_1+y_2+y_3}{3}=0$

so $x_1+x_2+x_3=0$ & $y_1+y_2+y_3=0$

squaring on both sides we get

$x_1^2+x_2^2+x_3^2+2x_1{x}_2+2x_2{x}_3+2x_3{x}_1=0$ and

$y_1^2+y_2^2+y_3^2+2y_1{y}_2+2y_2{y}_3+2y_3{y}_1=0\cdots \left(1\right)$

$A{B}^2+B{C}^2+C{A}^2=\left[\right(x_2-x_1{)}^2+(y_2-y_1{)}^2]+\left[\right(x_3-x_2{)}^2+(y_3-y_2{)}^2]+\left[\right(x_1-x_3{)}^2+(y_1-y_3{)}^2]$

$=[x_1^2+x_2^2-2x_1{x}_2+y_1^2+y_2^2-2y_1{y}_2]+[x_3^2+x_2^2-2x_3{x}_2+y_2^2+y_3^2-2y_2{y}_3]+[x_1^2+x_3^2-2x_1{x}_3+y_1^2+y_3^2-2y_1{y}_3]$

$=[2x_1^2+2x_2^2+2x_3^2-2x_1{x}_2-2x_2{x}_3-2x_1{x}_3]+[2y_1^2+2y_2^2+2y_3^2-2y_1{y}_2-2y_2{y}_3-2y_1{y}_3]$

$=(3x_1^2+3x_2^2+3x_3^2)+(3y_1^2+3y_2^2+3y_3^2)$

$=3(x_1^2+x_2^2+x_3^2)+3(y_1^2+y_2^2+y_3^2)\cdots \left(2\right)$

$3(G{A}^2+G{B}^2+G{C}^2)=3\left[\right(x_1-0{)}^2+(y_1-0{)}^2+(x_2-0{)}^2+(y_2-0{)}^2+(x_3-0{)}^2+(y_3-0{)}^2]$

$=3[x_1^2+y_1^2+x_2^2+y_2^2+x_3^2+y_3^2]$

$=3(x_1^2+x_2^2+x_3^2)+3(y_1^2+y_2^2+y_3^2)\cdots \left(3\right)$

from (2) and (3) we get

$A{B}^2+B{C}^2+C{A}^2=3(G{A}^2+G{B}^2+G{C}^2)$