An equilateral triangle is inscribed in a circle of radius $r$. If $P$ is any point on the circumference, find the value of $PA^2 + PB^2 + PC^2$. 
An equilateral triangle is inscribed in a circle of radius $r$.  If $P$ is any point on the circumference find the value of $PA^2 + PB^2 + PC^2$.  

I have managed to solve this problem  using co-ordinate geometry by taking a triangle with its centroid at (0,0) and assuming a point at $(r\cos\theta , r\sin \theta)$.  
A simpler proof using geometry however eludes me. Is it possible to prove this using Euclidean geometry or without a calculation intensive approach?  

The answer is   

$6r^2$

 A: $PA^2+PB^2+PC^2$ is the moment of inertia of $\{A,B,C\}$ with respect to $P$, which by the parallel axis theorem only depends on the distance of $P$ from the centroid of $ABC$. In particular, if $ABC$ is equilateral and $P$ belongs to the circumcircle of $ABC$, $PA^2+PB^2+PC^2 = AA^2+AB^2+AC^2=6r^2. $
A: Let $\omega:=e^{2\pi i/3}$. Then $\omega\bar\omega=1$, and $1+\omega+\bar\omega=0$. For any $z\in{\mathbb C}$ one therefore has
$$(z-1)(\bar z-1)+(z-\omega)(\bar z-\bar \omega)+(z-\bar\omega)(\bar z-\omega)=3(|z|^2+1)\ .$$
Translated to your problem this means that $|PA|^2+|PB|^2+|PC|^2=6r^2$.
A: Let $\Gamma$ be the circumcircle of the equilateral triangle $ABC$.  Without loss of generality, assume that $P$ is on the arc $BC$ of $\Gamma$ that does not contain $A$.  By Ptolemy's Theorem,
$$PA\cdot BC=PB\cdot CA+PC\cdot AB\,.$$
Since $AB=BC=CA$, we get $PA=PB+PC$.  Thus, the desired expression is
$$PA^2+PB^2+PC^2=2(PB^2+PC^2+PB\cdot PC)\,.$$
Now, observe that $\angle BPC=\dfrac{2\pi}{3}$, so using the Law of Cosines, we obtain
$$PB^2+PC^2+PB\cdot PC=PB^2+PC^2-2\,PB\cdot PC\,\cos(\angle BPC)=BC^2\,.$$
Obviously, $BC^2=3r^2$ (you can use Pythagoras or trigonometry here), whence $$PA^2+PB^2+PC^2=2(3r^2)=6r^2\,.$$
A: Without the loss of generality, we can assume that $P$ is between $A$ and $B$. Thus $m\angle{AOP}+m\angle{BOP}=120°$. Next, either $m\angle{AOP} \ge m\angle{BOP}$ or $m\angle{AOP} < m\angle{BOP}$. Let's assume $m\angle{AOP} \ge m\angle{BOP}$ (this we can do due to the symmetry of the following proof). Then $m\angle{COP}=120°+m\angle{BOP}$. 
Let $m\angle{BOP}/2=x$, then
$PA=2r\sin (60°-x)$, $PB=2r\sin x$, $PC=2r\sin (60°+x)$ and we have $$PA^2+PB^2+PC^2=4r^2 \bigl( \sin^2 (60°-x)+\sin^2 x+ \sin^2 (60°+x) \bigr)=4r^2 \bigl( \sin^2 60° \cos^2 x -2\sin 60° \cos 60° \sin x \cos x + \cos^2 60° \sin^2 x+\sin^2 x+ \sin^2 60° \cos^2 x +2\sin 60° \cos 60° \sin x \cos x + \cos^2 60° \sin^2 x \bigr)=4r^2 \bigl( \frac{3}{2} \cos^2 x +\sin^2 x+ \frac{1}{2} \sin^2 x \bigr)=6r^2$$
A: I will show that
if there are $n$ equally spaced points
on the unit circle
and point $P$ is distance $r$
from the origin,
then the sum of the squares
of the distances is
$n(r^2+1)$.
The OP is
$n=3, r=1$
which gives the answer $6$.
This is all analytic geometry.
I'll start general 
and work towards
the original problem.
Let the angles
of the points
on the unit circle be
$(t_k)_{k=1}^n$
so the points are
$(\cos(t_k), \sin(t_k))_{k=1}^n$.
Let $P$ be at
$(r\cos(p), r\sin(p))$.
The sum
$\begin{array}\\
S
&=\sum_{k=1}^n ((r\cos(p)-\cos(t_k))^2 +(r\sin(p)-\sin(t_k))^2)\\
&=\sum_{k=1}^n (r^2\cos^2(p)+\cos^2(t_k))-2r\cos(p)\cos(t_k)\\
&\quad +r^2\sin^2(p)+\sin^2(t_k)-2r\sin(p)\sin(t_k))\\
&=\sum_{k=1}^n (r^2(\cos^2(p)+\sin^2(p))+\cos^2(t_k)+\sin^2(t_k)\\
&\quad -2r(\cos(p)\cos(t_k)+\sin(p)\sin(t_k))\\
&=\sum_{k=1}^n (r^2+1
-2r\cos(p-t_k))\\
&=n(r^2+1)-2r\sum_{k=1}^n \cos(p-t_k)\\
&=n(r^2+1)-2r(\cos(p)\sum_{k=1}^n \cos(t_k)+\sin(p)\sum_{k=1}^n \sin(t_k))\\
\end{array}
$
In the OP,
where $n=3$
and $r=1$,
the first term
gives the proposer's
answer of $6$.
Let's see if
spacing the points
equally around the circle
makes the sums zero.
Let
$t_k
=2\pi(k-1)/n
$.
Then
$\begin{array}\\
\sum_{k=1}^n e^{t_ki}
&=\sum_{k=1}^n e^{2\pi i(k-1)/n}\\
&=\sum_{k=0}^{n-1} e^{2\pi ik/n}\\
&=\sum_{k=0}^{n-1} (e^{2\pi i /n})^k\\
&=\dfrac{1-(e^{2\pi i /n})^n}{1-e^{2\pi i/n}}\\
&=\dfrac{1-e^{2\pi i}}{1-e^{2\pi /n}}\\
&=0\\
\end{array}
$
Since
$\sum_{k=1}^n e^{t_ki}
=\sum_{k=1}^n \cos(t_k)+i\sum_{k=1}^n \sin(t_k)
$,
this implies that
$0
=\sum_{k=1}^n \cos(t_k)
=\sum_{k=1}^n \sin(t_k)
$
so the sums
are zero
and the overall sum is
$n(r^2+1)$.
A: Given $n$ points $A_1, \ldots, A_n$, and a constant $K$, what is the locus of a point $P$ such that
$\sum {PA_i}^2 = K$? Move the axes so that the origin is at the centroid of the $A_i$, and let $A_i = (a_i, b_i)$. Let the distance of $A_i$ from the centroid be $r_i$, so that ${r_i}^2 = {a_i}^2 + {b_i}^2$. Let $P = (x,y)$. We require
$$\sum[(x - a_i)^2 + (y - b_i)^2] = K.$$
Since $\sum a_i = \sum b_i = 0$, this is equivalent to
$$x^2 + y^2 = (K - \sum {r_i}^2)/n$$
so provided $K > \sum{r_i}^2$ the locus is a circle with centre at the centroid of the $A_i$. If the $A_i$ form a regular polygon on a circle of radius $r$, then the locus of $P$ is the same circle when $K = 2nr^2$. (In fact the polygon needn't be regular, it only needs to have its centroid at the centre of the circle.)
