Let A_1,...,A_n be the vertices of a regular n -gon…
Let $A_1,\ldots,A_n$ be the vertices of a regular $n$-gon inscribed in a circle of radius $R$ centered at $O$. $\ 1^\circ$ Prove that for $M$ on the circle, $\sum_k|MA_k|^2=2nR^2$, independent of $M$'s position. $\ 2^\circ$ Prove that for $M$ anywhere in the plane, $\sum_k|MA_k|^2=n(R^2+\ell^2)$ where $\ell=|OM|$. $\ 3^\circ$ Prove $2^\circ$ still holds for $M$ outside the plane.
Explanation of Method
Expand $|M-A_k|^2$ using $|u-v|^2=|u|^2+|v|^2-2\text{Re}(u\bar v)$ and sum over $k$. The cross term collapses to zero because the vertices, as roots of unity scaled by $R$, always sum to zero — this single fact drives all three parts.
Step 1 · Set up coordinates
Center the polygon at the origin: $\ A_k=Re^{2\pi ik/n}$, $k=0,\ldots,n-1$. Let $M$ be an arbitrary point, $\ell=|M|$.
Step 2 · Expand the sum of squared distances
$$ \sum_{k=0}^{n-1}|M-A_k|^2=\sum_{k=0}^{n-1}\left(|M|^2+|A_k|^2-2\,\text{Re}(M\bar A_k)\right)=n\ell^2+nR^2-2\,\text{Re}\left(M\sum_{k=0}^{n-1}\bar A_k\right) $$
Step 3 · The key identity: vertices of a regular polygon sum to zero
$$ \sum_{k=0}^{n-1}A_k=R\sum_{k=0}^{n-1}e^{2\pi ik/n}=R\cdot0=0 $$
(a geometric series of the $n$-th roots of unity), so also $\sum_k\bar A_k=\overline{\sum_kA_k}=0$.
Step 4 · The cross term vanishes — for any $M$
Since $\sum_k\bar A_k=0$, the cross term in Step 2 is $0$ regardless of where $M$ is:
$$ \sum_{k=0}^{n-1}|M-A_k|^2=n\ell^2+nR^2=n(R^2+\ell^2) $$
Part $1^\circ$ · $M$ on the circle
If $M$ is on the circle, $\ell=R$, so:
$$ \boxed{\sum_k|MA_k|^2=n(R^2+R^2)=2nR^2} $$
independent of $M$'s position on the circle, as claimed.
Part $2^\circ$ · $M$ anywhere in the plane
Step 4 already proves this for arbitrary $M=x+iy$ in the plane:
$$ \boxed{\sum_k|MA_k|^2=n(R^2+\ell^2)}, \qquad \ell=|OM| $$
Part $3^\circ$ · $M$ outside the plane (in 3D)
Place the polygon in the $xy$-plane, $A_k=(R\cos\frac{2\pi k}n,R\sin\frac{2\pi k}n,0)$, and let $M=(x,y,z)$ with $\ell=\sqrt{x^2+y^2+z^2}$. Expanding each squared distance:
$$ |MA_k|^2=x^2+y^2+z^2+R^2-2xR\cos\frac{2\pi k}n-2yR\sin\frac{2\pi k}n $$
Summing over $k$, and using $\sum_k\cos\frac{2\pi k}n=\sum_k\sin\frac{2\pi k}n=0$ (real/imaginary parts of Step 3's identity):
$$ \boxed{\sum_k|MA_k|^2=n\ell^2+nR^2=n(R^2+\ell^2)} $$
— exactly the same formula, now with $\ell$ measured as the full 3D distance from $M$ to the polygon's center.
Verification
Consistency check: setting $\ell=R$ in the general formula from $2^\circ$/$3^\circ$ recovers $S=n(R^2+R^2)=2nR^2$, matching $1^\circ$ exactly.
Result
The sum depends only on the distance $\ell$ from $M$ to the center — never on direction — because the vertices of a regular polygon always sum to zero, killing every cross term regardless of where $M$ sits.