Surface area of spheres not seen by any other.

Question

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.

Answer 1

The set S consists of several pieces corresponding to spheres.

Use translations to move all the pieces to one sphere.

We will prove that 1) the translated pieces do not overlap 2) any point of that sphere (except some set of area 0) is covered by some piece.

For the sake of brevity I will call points A and B equivalent if they lie on different spheres and the translation which takes one sphere to the other takes A to B.

1) If translated pieces have a point in common, that means that there are two equivalent points in S. But this cannot be true (just draw a picture): if A and B are equivalent, the line AB intersects the two spheres in points P, Q. In any possible arrangement of A,B,P,Q on a line either A or B sees some point on the other sphere. (the arrangement cannot be APQB, since then A and B are not equivalent)

2) draw all possible lines connecting centers of the spheres and for each such line delete the big circle on our sphere perpendicular to the line. Let's prove that for any remaining point there is an equivalent point in S. Let that point be X and the center of our sphere be O. Look in the direction XO. What you see is a bunch of circles. Some of them are close to us and some are remote. Take the closest point. It is equivalent to X and it belongs to S. (We deleted the circles to be sure that there is only one closest point: if there are two closest points, then they lie on a big circle perpendicular to center line of two spheres)

Answer 2

The following are some preliminary considerations that might help to better focus the problem.

a) For two spheres in direct sight of each other, the visible portions are those contained into the cylinder connecting the two.
b) The visible and dark portions are separated by a major circle, on the plane normal to the line connecting the centers.
c) The problem is not affected by value of the common radius of the spheres: it can in the limit be reduced to zero.
d) For any number of aligned spheres, the hypothesis is true.
e) For three spheres ( a triangle) the dark area of each corresponds to the (dihedral) angle comprised brtween the normals to the sides thus equal to $\pi $ minus the internal angle: the tesis is true.

f) Any sphere inside the triangle is totally visible: the thesis is true for any triangle, containing whatever number of internal spheres.
g) Same for any convex polygon, the dark dihedral angle is $\pi$ minus internal angle, and the total is $(n-(n-2)) \pi = 2\pi$ .
h) Passing to a polyhedron, the dark area will be that included in the solid angle individuated by the planes normal to the edges.
i) The spheres inside a convex polyhedron are again totally visible.
Consider in fact a sphere $\to$ point inside the polyhedron. By definition of convexity, any plane through the point will leave at least one vertex of the polygon on either side of the plane. Therefore any point on the internal sphere will see at least one vertex.
j) The same consideration above is valid for any convex polytope.

k) Given a convex polytope, keeping one vertex (O) firm and moving the other spheres (A) along the prolongation of the edges (B) won't alter the total amount of dark area, because of the consideration in a):
i.e., $OA$ and $OB$ are equivalent.
The choice of the vertex O is arbitrary, so the projection can be repeated from any vertex in A or B.
l) In addition to the above, any angle-preserving transformation won't alter the situation.

Answer 3

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.