On a constant associated to equilateral triangle and its generalization. I guess many of you are familiar with the result described as follows:
If ABC is an equilateral triangle, and P is any point on the incircle of ΔABC, then AP² + BP² + CP² is constant. See link below.
http://www.cut-the-knot.org/pythagoras/EquiIn3D.shtml
I was wondering whether this can be true for any regular polygon and found it to be true too. In the website it is mentioned that the result holds for any circle with center at the centroid of the triangle, but it does not mention whether the result holds for any other polygon. 
I will propose the following generalization:
Consider any regular polygon (n-gon) and a point $P$ on its circumcircle (or on any other circle with center at the centroid of the n-gon). Denote the vertices of the n-gon as $A_i$. Then, it follows
$$K_n=\sum_{1}^{n}{A_iP^2}$$
Where $K_n$ is a constant. 
On the other hand, in the case of a triangle, it admits a 3D-generalization(?) as you can see in this link (although they do not mention it in the link)
http://www.cut-the-knot.org/pythagoras/3DExercise.shtml
I have noticed that this is also true for the insphere of the tetrahedron, so I suspect that for any sphere centred at the centroid the result holds, just as the 2D case. I checking whether the result holds for a cube, but failed. Although it is true just for some segments of the cube since we can inscribed a tetrahedron in a cube.
Does anyone know whether this generalization is indeed new?
Thanks in advance.
EDITED 1: I have removed the golden ratio part for being irrelevant.
EDITED 2: Here is what I think is a more general result: 
http://geometriadominicana.blogspot.com/2016/07/sum-of-squares-of-distances-to-vertices.html

 A: In the complex plane the vertices of a regular polygon of $n$ sides and radius $1$ can be conveniently written as $A_i=e^{i2\pi/n}$. A generic point on the same unit circle is $P=e^{i\theta}$. Then:
$$
\sum_{i=1}^{n}\overline{A_iP}^2=\sum_{i=1}^{n}|P-A_i|^2=
\sum_{i=1}^{n}(2-2\,\hbox{Re}\ e^{i\theta-i2\pi/n})=
2n-2e^{i\theta}\,\hbox{Re}\sum_{i=1}^{n}e^{-i2\pi/n}=2n.
$$
For a regular polygon with $n$ sides and radius $r_n$ we find then $K_n=2nr_n^2$.
Regular polygons with $n$ sides of unit length have a radius $r_n=(1/2)/\sin(\pi/n)$, so in that case:
$$
K_5=10\cdot{2\over5-\sqrt5}=5+\sqrt5=6+2\phi,
$$
where $\phi=(\sqrt5-1)/2$.
EDIT 1.
An analogous reasoning can be made for $n$ points in $d$-dimensional space. Let $A_1,\dots, A_n$ be the points, $C$ their centroid and $P$ another point at distance $R$ from $C$. Define vectors $\vec{a}_i=A_i-C$ and $\vec{p}=P-C$; by definition of centroid we have then $\sum_{i=1}^{n}\vec{a}_i=0$ and:
$$
K_n=\sum_{i=1}^{n}\overline{A_iP}^2=
\sum_{i=1}^{n}(\vec{a}_i-\vec{p})^2=
\sum_{i=1}^{n}{a}_i^2+nR^2+2\vec{p}\cdot\sum_{i=1}^{n}\vec{a}_i=
\sum_{i=1}^{n}{a}_i^2+nR^2,
$$
where ${a}_i^2=\overline{A_iC}^2$.
This result does not change as long as $P$ stays on a sphere of center $C$ and radius $R$. In particular, if $\overline{A_iC}=R$ for every $A_i$, then we recover the result given for a regular polygon: $K_n=2nR^2$.
EDIT 2.
In particular, for a regular polyhedron with $n$ vertices, you have $a_i=r$ (polyhedron radius) for every $i$, so that
$$
K_n=nr^2+nR^2.
$$ 
The "more general result" cited in the question is nothing but this equality for $n=4$ (tetrahedron). The same equality holds if the $n$ points are all at the same distance $r$ from their centroid, even if they don't form a regular polyhedron: that's the case, for instance, of a semiregular polyhedron.
