Circumcircle of triangle $ABC$

Question

In an equilateral triangle $ABC$ if $P$ be any point of the circumcircle of $\triangle ABC$ and $R$ represent radius of that circle.

Then find $\displaystyle \frac{|\vec{PA}|^2+ |\vec{PB}|^2+|\vec{PC}|^2}{R^2}$

Try: assuming position vector of $C$ is $\vec{0}$ and $A,B,P$ respectively $\vec{a},\vec{b},\vec{p}$.

So $$|\vec{PA}|^2+ |\vec{PB}|^2+|\vec{PC}|^2=|\vec{p}-\vec{a}|^2+|\vec{p}-\vec{b}|^2+|\vec{p}-\vec{0}|^2=3|\vec{p}|^2+|\vec{a}|^2+|\vec{b}|^2-2\bigg(\vec{p}\cdot\vec{a}+\vec{p}\cdot\vec{b}\bigg)$$

Could some help me to solve it, thanks

Answer 1

Take the centre of the triangle as the origin $O$.

\begin{align*} &\;\frac{|\vec{PA}|^2+|\vec{PB}|^2+|\vec{PC}|^2}{R^2}\\ =&\;\frac{|\vec{OA}|^2+|\vec{OB}|^2+|\vec{OC}|^2+3|\vec{OP}|^2-2\vec{OP}\cdot(\vec{OA}+\vec{OB}+\vec{OC})}{R^2}\\ =&\;\frac{3R^2+3R^2-2\vec{OP}\cdot(\vec{0})}{R^2}\\ =&\;6 \end{align*}

Answer 2

$PA^2+PB^2+PC^2$ is the moment of inertia of $\{A,B,C\}$ with respect to $P$. In an equilateral triangle the centroid $G$ and the circumcenter $O$ are the same point, hence by the parallel axis theorem, for any point $P$ on the circumcircle of an equilateral triangle $ABC$ we have

$$ PA^2+PB^2+PC^2 = 3 OP^2 + OA^2+OB^2+OC^2 = 6R^2.$$