TITLE: Normalized area of the dark region is $4\pi$.
QUESTION [6 upvotes]: $\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\ab}[1]{\langle #1\rangle}$ $\newcommand{\vp}{\varphi}$ $\newcommand{\R}{\mathbf R}$

IMO 1982, Longlisted Problem 10.
Let $r_1, \ldots, r_n$ be the radii of $n$ pairwise disjoint spheres.
Let $A_1, \ldots, A_n$ be the areas of the set of points of each sphere from which one cannot see any point of any other sphere.
Prove that
$$
\frac{A_1}{r_1^2} + \cdots + \frac{A_n}{r_n^2} = 4\pi
$$

I could solve the problem assuming $r_1= \cdots = r_n = 1$. Here is the solution for this special case.
Let $p_1, \ldots, p_n$ be the centers of the spheres, and let us denote the sphere with center $p_i$ as $S_i$.
Let $S^2$ denote the unit sphere in $\R^3$.
For $v\in S^2$, we say that the point $p_i+v$ on $S_i$ is visible from $S_j$, where $j\neq i$, if $\ab{v, p_i-p_j}> 0$.
The dark region $D_i$ of $S_i$ is defined as the part of $S_i$ not visible from any other sphere.
Thus $A_i$ is the area of $D_i$.
Note that if $v\in S^2$ and $p_i+v$ is not visible from $S_j$, for some $j\neq i$, then $p_j+v$ is visible from $S_i$ (except for a measure zero set of points $v$ in $S^2$).
Thus for any $v\in S^2$, we have at most one $i\in \set{1, \ldots, n}$ such that $p_i+v$ is in the dark region of $S_i$ (again, modulo a set of measure zero in $S^2$).
Define
$$
D_i' = \set{v\in S^2:\ p_i+v\in D_i}
$$
From the above discussion it is clear that $D_1', \ldots, D_n'$ are pairwise (measure theoretically) disjoint.
We show that their union is all of $S^2$, up to a set of measure $0$.
Let $v$ in $S^2$ be arbitrary and assume that for each $i\in \set{1, \ldots, n}$ there is $\vp(i)\in \set{1, \ldots, n}\setminus{\set{i}}$ such that $p_i+v$ is visible from $S_{\vp(i)}$.
Thus $\ab{v, p_i-p_{\vp(i)}} > 0$.
Since there are finitely many spheres, there must be $i\in \set{1, \ldots, n}$ and a positive integer $k\geq 2$ such that $\vp^k(i) = i$.
Thus we have
$$
\sum_{r=0}^{k-1} \ab{v, p_{\vp^r(i)} - p_{\vp^{r+1}(i)}} > 0
$$
which is a contradiction since the left hand side of the above is zero.
So we have that the area of $D_i'$ is same as the area of $D_i$ and the $D_1', \ldots, D_n'$ partition $S^2$ (module a set of measure zero).
Thus we have the desired result.

REPLY [2 votes]: This is how I would approach it.

Show that each area $A_i$ is the intersection of several solid caps (obtained through intersections with the other spheres $S_j, j\neq i$). This avoids measure-theoretic concerns.
Associate the area to the corresponding solid angle measure. Note that the value of the solid angle measure is $ \frac{A_n}{ r_n^2}$.
Show that this association is a bijection with the angle measure of a unit sphere, except possibly at the boundaries of $A_i$.


Given any point on the unit sphere, rotate the sphere+confiuration so that the point is at the bottom. Then, we map this point to all the lowest point in the configuration.
It is possible that  there are multiple lowest points in the configuration.  Show that this occurs only on the boundary of $A_i$.


Hence, the sum of $ \frac{ A_n}{ r_n^2}$ is the solid angle measure of the unit sphere which is $ 4 \pi$.

Note: There are similarities to IMO 2002/6 which involves unit circles. A particular solution uses the fact that the convex hull of the circles has curved perimeter of $2\pi$, which is similar to the above via the associated central angle measure.
Speaking as a contestant of the IMO 2002, I believe that (but am not fully confident)

Contestants with a complete proof apart from merely stating this fact without proof were not docked points.
Merely stating this fact wasn't sufficient to yield partial credit. Those that also tied it to the corresponding angle measure and related it to the question, were given partial credit.