TITLE: Surface area of spheres not seen by any other.
QUESTION [8 upvotes]: I found this question recently looking around the internet, apparently it was on the IMO shortlist many years back. I haven't been able to solve it, and I am looking for hints and/or full solutions. Apparently it can be done very elegantly.
Consider N non overlapping spheres of equal radius placed in 3D space. Let $S$ be the set of points on the surface of these spheres which are not visible from any other sphere. Show that the total area of $S$ is equal to the surface area of one sphere.
Thoughts so far: the problem is trivial in one dimension, and likely equivalent in difficulty in 2D and 3D (it looks like it still holds in 2D, hence I imagine it works for any number of dimensions). It is also obviously true for 2 spheres, and can be verified with some effort for 3. I've had many ideas but none of them have yet led me anywhere I thought promising.

REPLY [3 votes]: As I'm seeing it: Let $c_i$ $(1\leq i\leq N)$ be the centers of the given spheres and $P$ be the convex hull of the $c_i$. Then $P$ is a compact convex polyhedron in ${\mathbb R}^3$, bounded by vertices, edges, and faces. Let $$Q:=P+B_r=\bigl\{x+y\bigm| x\in P, \>\|y\|\leq r\bigr\}$$ be the closed "$r$-neighborhood" of $P$. This $Q$ is bounded by a spherical piece for each vertex of $P$, a cylindrical piece for each edge of $P$, and a plane piece for each face of $P$. The spherical pieces are exactly the parts of the given spheres that cannot be seen from some other of the given spheres. The total curvature of these parts is $4\pi$, by Gauss'-Bonnet's theorem. Therefore the total unseen area on the given spheres is $4\pi r^2$, or the area of a single sphere.