Yet another constant associated to regular polygons! Consider any regular polygon and an arbitrary circle with center at the centroid of the polygon. Let $L$ be an arbitrary line tangent to the circle. Then, the sum of the distances from the vertices of the polygon to $L$ is constant.
Any reference for this problem?
EDIT: I have found an extension of this result for a tetrahedron. Consider a tetrahedron ABCD and its circumsphere. Let $P$ be a plane tangent to the circumsphere. Then, the sum of the distances from the vertices of the tetrahedron to the plane $P$ is constant. I suspect this is true for any sphere with center at the centroid of the tetrahedron ABCD. Does it generalize to other regular polyhedra?
 A: Here is the proof of this property in a certain category of cases.
Let us take as unity the radius of the circumscribed circle to the polygon.
Let $R$ be the radius of the "arbitrary" circle.
We assume that 
$$\tag{0} R \geq 1.$$
Let $O$ be the common center to the 2 circles, $N$ the number of sides of the regular polygon.
Let $V_k$ be the name of the $k$th vertex ($k=0,1 \cdots N-1$).
A classical result that we will use later on is : 
$$\tag{1}\sum_{k=0}^{N-1}\vec{OV_K}=0.$$
Let $\theta$ be the polar angle of a normal vector to the tangent line. Thus the rotation by $-\omega$ 


*

*turns this tangent line into the tangent line at $A(R,0)$. 

*brings points $V_k$ onto points $V'_k$. 
Let $W_k$ be the projection of $V'_k$ on the $x$ axis.
We have to compute $\sum_{k=0}^{N-1}length([AW_k])$
These lengths can be computed by taking abscissa $R$ of $A$ minus abscissa of $W_k$ without need for taking signs because the abscissa $R$ of $A$ is bigger that the abscissas of all $W_k$s that are $\leq 1$ due to assumption $(0)$.
Using a complex number representation where 


*

*$V_k$ is associated with $e^{2i \pi k/N}$,

*rotation by $-\omega$ with multiplication by $e^{-i \omega}$,

*projection onto the $x$ axis being realized by taking the real part $\Re$,
we have to show that the following sum is a constant:
$$\tag{2}\sum_{k=0}^{N-1}\Re(R-e^{2i \pi k/N}e^{-i\omega})=constant.$$
Expanding and grouping:
$$\sum_{k=0}^{N-1}(R)-\Re\left(e^{-i\omega}\sum_{k=0}^{N-1}e^{2i \pi k/N}\right)=constant$$
But the last sum is zero because of $(1)$.
Thus, it is true that the LHS is equal to a constant, which is NR, being understood that this proof is under condition (0).
