Finding a geometric interpretation I recently solved a question of complex numbers which was this:
$A\left( \frac{2}{\sqrt{3}} e^\frac{i\pi}{2}\right)$, $B\left( \frac{2}{\sqrt{3}} e^\frac{-i\pi}{6}\right)$, $C\left( \frac{2}{\sqrt{3}} e^\frac{-5i\pi}{6}\right)$ are the vertices of an equilateral triangle. If $P$ be a point on the incircle of the triangle, prove that $AP^2 + BP^2 + CP^2 = 5$.
My approach:
The point $P$ is given by $z = \frac{1}{\sqrt{3}}e^{i\theta}$, since the radius of the incircle of an equilateral triangle is half its circumradius. Then,
$$
AP² = |A - z|^2 \\
= (A - z)(A^* - z^*) \\
= AA^* - Az^* - A^*z + zz^* \\
= |A|² + |z|² - Az^* - A^*z \\
= \frac43 + \frac13 - Az^* - A^*z \\
= \frac53 - Az^* - A^*z.
$$
Now,
$$
AP^2 + BP^2 + CP^2 \\
= 3\times\frac53 - z^*(A+B+C) - z(A^*+B^*+C^*) \\
= 5
$$
since $A+B+C = A^*+B^*+C^* = 0$ because the position vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$ are coplanar and are mutually separated by $120^\circ$.
This, I guess, is a pretty neat solution. But what I'm looking for is a more intuitive solution, rather a geometric interpretation. Does anyone know such an approach?
 A: Hint: for any triangle $\triangle ABC$ and any point $P$ the following holds, where $G$ is the centroid of the triangle (see here for example):
$$
PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3 PG^2
$$
For an equilateral triangle all the centers coincide, so $G$ is also the center of the incircle, thus $PG$ is constant for any point $P$ on the incircle. Therefore the RHS is constant, and you can choose for example $P$ to be the midpoint of $BC$ in order to calculate the value of the constant.
A: The intuition lies in the fact that the three points in question $A$, $B$, $C$ has a bary-center which is at the center $O$ of the inscribed circle.
If $P$ is on the inscribed circle then using scalar product and letting $a=\vec{OA}$, $b=\vec{OB}$, $c=\vec{OC}$ and $m=\vec{OP}$: 
$$ |AP|^2+|BP|^2+|CP|^2= \sum_{p\in \{a,b,c\}} (p-m,p-m) = 
3 |p|^2 + 3 |m|^2 - 2(m,\sum_{p\in \{a,b,c\}} p) =  3(4/3+1/3) -0=5$$
A: The sides of your triangle are $2$. E.g. the distance of $A$ and $B$ is the length of $$\frac{2}{\sqrt{3}}(e^{-\frac{i\pi}{6}}-e^{\frac{i\pi}{2}})=\frac{2}{\sqrt{3}}(\frac{\sqrt{3}}{2}-\frac{3}{2}i)=1-\sqrt{3}i.$$
We can calculate $PA^2+PB^2+PC^2$ using elementary methods, if $P$ is a point on the incircle of an equilateral triangle $ABC$ of side $a$.
In this case the radius of the incircle is $r=\frac{a\cdot\sqrt{3}}{6}$. Let $O$ be the incenter. Then $OA=OB=OC=\frac{a\sqrt{3}}{3}$. Let $\angle POC$ be $\varphi$. Then using the cosine theorem for the triangle $POC$ $$PC^2=r^2+OC^2-2r\cdot OC\cdot\cos\varphi = \frac{a^2}{12}+\frac{a^2}{3}-\frac{a^2}{3}\cos\varphi.$$
Since $\angle POA=120^{\circ}-\varphi$ and $\angle POB=120^{\circ}+\varphi$, similarly we get
$$PA^2 = \frac{a^2}{12}+\frac{a^2}{3}-\frac{a^2}{3}\cos(120^{\circ}-\varphi)$$
and
$$PB^2 = \frac{a^2}{12}+\frac{a^2}{3}-\frac{a^2}{3}\cos(120^{\circ}+\varphi).$$
Summarizing and using the addition theorem
$$PA^2+PB^2+PC^2=a^2(\frac{1}{4}+1-\frac{1}{3}\cos\varphi-\frac{1}{3}(-\frac{1}{2}\cos\varphi+\frac{\sqrt{3}}{2}\sin\varphi)-\frac{1}{3}(-\frac{1}{2}\cos\varphi-\frac{\sqrt{3}}{2}\sin\varphi))=\frac{5}{4}a^2.$$
Since in our case $a=2$, this gives $PA^2+PB^2+PC^2=5.$
