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If $G$ be the centroid of a triangle ABC and $P$ be any other point in the plane, prove that $PA^2 + PB^2 + PC^2 = GA^2 + GB^2 + GC^2 + 3GP^2$.
Given:
$G$ is the centroid of a triangle ABC and $P$ is any other point in the plane.
To do:
We have to prove that $PA^2 + PB^2 + PC^2 = GA^2 + GB^2 + GC^2 + 3GP^2$.
Solution:
Let $G(u,v)$ be the centroid of a $\triangle ABC$ and the coordinates of $A$ are $(x_1, y_1)$, $B$ are $(x_2, y_2)$ and $C$ are $(x_3, y_3)$.
Let $P (h, k)$ be any point in the plane.
The coordinates of centroid \( \mathrm{G} \) are \( (\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}) \)
This implies,
\( u=\frac{x_{1}+x_{2}+x_{3}}{3} \) and \( v=\frac{y_{1}+y_{2}+y_{3}}{3} \)
Let us consider LHS,
\( \mathrm{PA}^{2}+\mathrm{PB}^{2}+\mathrm{PC}^{2}=\left(h-x_{1}\right)^{2}+\left(k-y_{1}\right)^{2}+\left(h-x_{2}\right)^{2}+\left(k-y_{2}\right)^{2}+\left(h-x_{3}\right)^{2}+\left(k-y_{3}\right)^{2} \)
\( = 3\left(h^{2}+k^{2}\right)+\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)+\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)-2 h\left(x_{1}+x_{2}+x_{3}\right)-2 k\left(y_{1}+y_{2}+y_{3}\right) \)
\( =\left(3 h^{2}+k^{2}\right)+\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)+\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)-2 h(3 u)-2 k(3 v) \)
\( =3\left(h^{2}+k^{2}\right)-6 h u-6 k v+\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)+\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right) \)
Now, let us consider RHS,
\( \mathrm{G} \mathrm{A}^{2}+\mathrm{GB}^{2}+\mathrm{GC}^{2}+3 \mathrm{GD}^{2} \)
\( =\left(u-x_{1}\right)^{2}+\left(v-y_{1}\right)^{2}+\left(u-x_{2}\right)^{2}+\left(v-y_{2}\right)^{2}+\left(u-x_{3}\right)^{2}+\left(v-y_{3}\right)^{2}+3\left[(u-h)^{2}+(v-k)^{2}\right] \)
\( =3\left(u^{2}+v^{2}\right)+\left(x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+x_{3}^{2}+y_{3}^{2}\right)-2 u\left(x_{1}+x_{2}+x_{3}\right)-2 v\left(y_{1}+y_{2}+y_{3}\right) \)
\( +3\left(u^{2}+h^{2}-2 u h+v^{2}+k^{2}-2 v k\right) \)
\( =6\left(u^{2}+v^{2}\right)+\left(x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+x_{3}^{2}+y_{3}^{2}\right)-2 u(3 u)-2 v(3 v)+3\left(h^{2}+k^{2}\right)-6 u h-6 v k \)
\( =\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)+\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)+3\left(h^{2}+k^{2}\right)-6 u h-6 v k \)
Therefore,
L.H.S. $=$ R.H.S.
Hence proved.
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