Problem: Given is a finite set of spherical planets, all of the same radius and no two intersecting. On the surface of each planet consider the set of points not visible from any other planet. Prove that the total area of these sets is equal to the surface area of one planet.

I found this problem this in the book ‘Putnam and Beyond’. This problem is was shortlisted for the 22^nd IMO proposed by the Soviet Union. I present a solution to this problem (different and simpler from the one given in the book) and say a little about the problem-solving process along the way. If you are intrigued by the problem you are recommended to stop reading and give it a try yourself. You can always come back for the solution later.

A first attempt

It is usually a good idea to reduce your problem to a simpler case that is easy to solve and then generalise. For instance, our problem, it may be easier to prove it for the case when all the centres of the spheres lie in one plane. As an extremely simple case consider just two spheres.

It is clear that the shaded region is on each sphere is the part that is not visible from the other sphere. It is also obvious that the total area of this region equals the area of one sphere. Now let us add one more sphere to the mix. The points that are not visible from the other spheres lie in the indicated angles.

The indicated angles are representative of a ‘spherical lune’ ( the ‘lemon wedge’ shaped region between two great circles).

Thus if the total area is to be equal to that of one sphere then the angles must add up to 360 degrees. This is easily shown to be true using some elementary geometry.

We are interested in the value of angle BCD. Line AD is parallel to the line passing through the centre of the lower sphere and C. The line AB is parallel to the line passing through C and the centre of the upper sphere. Thus the two angles marked X are equal by the property of the transversal. The two angles ADC and ABC are right angles by construction. As the sum of all angles in a quadrilateral is 360 degrees , . Similarly for the other two planets if we assume the angle between the line segments joining the centre to be Y and Z then the required angles are and respectively. Since X,Y and Z are the three angles of a triangle Thus the sum of the marked angles is

Can this be generalised to the case where arbitrary number of planets are situated on a plane in some random fashion? Let us see. Here is one such configuration.

The spheres that lie inside the polygon are exposed from all sides and do not contribute anything to the total ‘unseen’ area. Thus we only need to worry about the spheres that lie on the polygon. By essentially repeating the above argument we seen that if at a particular planet the interior angle of the polygon is then the angle corresponding the unseen region is . Let us assume that the outer polygon has sides. Thus when we sum up all these areas we get . Furthermore this can always be done. Meaning given an arbitrary set of finite points we can pick a subset which give a convex polygon such that all other points lie either on or inside the polygon. Such a subset is called a convex hull. With this we are done with the planar case altogether!

A dead end

The trouble with this approach is that it is very difficult to generalize to the 3D case. In 3D the convex hull of the centres of all spheres will form a convex polytope. Like in the 2D case the spheres inside this polytope contribute nothing.

The neighbouring vertices will subtend a solid angle at every vertex. To proceed in an analogous fashion to the 2D case we need to establish a relationship between this solid angle and the unseen area. I’m not saying it is impossible but merely complicated.

This happens often while solving problems. An approach that looks promising at first turns out to be problematic later. It is usually a good idea to try another approach when one is faced with such a situation.

Another approach

Although our earlier approach has not lead to the solution it has given us important insights into the nature of the problem. Observe that only angles were involved in both the calculations that we have done. Unseen area seem only to depend on the angles and not the lengths. So if we shrink a triangle to a smaller triangle keeping all the angles constant then the unseen area stays constant. If we continue doing this then eventually all the spheres collapse on the sphere at the vertex at which we are shrinking.

We can see that the three pieces fit together like pieces of a jigsaw puzzle. For the pieces to form one whole sphere the pieces must not intersect when they collapse onto the sphere. We show that they in fact don’t. To see this let us focus on a pair of spheres. The unseen areas of the spheres are some subsets of the hemisphere that does not face the other sphere. As we are slide the two spheres along parallel lines the two hemispheres stay apart and the unseen area don’t intersect.

We have yet to say what we really mean by squeezing a triangle. Choose a vertex (say A) and keep it fixed. Suppose B is some other vertex move B along the line AB so that its distance from A is a ‘c’ times its original distance from A with c<1.

Luckily this strategy can be generalized to 3D.

The final solution

We have already seen that in an arbitrary configuration of planets in 3D we can choose a subset of them that form a convex polytope. All other planets lie inside this polytope and contribute nothing to the unseen area. Now we select some point of the polytope and squeeze the polytope so that it eventually collapses on that point. Along this process the total unseen area stays fixed. Hence the unseen area at the beginning is equal to the unseen area at the end. But at the end we only have one sphere. Thus the total unseen area must be equal to that of one sphere!