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

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.
 A: Denote by $S_1, S_2, \ldots, S_n$ the $n$ disjoint spheres of radius $r_1, \ldots, r_n$, centered at $p_1, \ldots, p_n$ respectively, and let $D_i$ be the dark region of each one. It's clear that $D_i$ is an open and connected subset of $S_i$ since it's defined by
$$
D_i= \{x\in S_i: \langle x-p_i, p_j-p_i\rangle< -r_i(r_j-r_i), \text{ for all } j\neq i\}.
$$
Following your proof, define the projection maps
$$
\pi_i: D_i\to S^2, \\
x\mapsto \frac{x-p_i}{r_i}.
$$
It's clear that $\pi_i$ is injective. Following your proof, we want to show that
$$
S^2= A\cup\bigcup_{i=1}^n \pi_i(D_i),
$$
where $A$ is a set of measure zero, and
$$
\bigcap_{i=1}^n \pi_i(D_i)\setminus A= \emptyset. 
$$
This would show the claim since then
$$
4\pi= |S^2|= \int_{S^2}\text{d}S= \sum_i\int_{\pi_i(D_i)}\text{d}S= \sum_i\int_{D_i}\frac{\text{d}S}{r_i^2}= \frac{A_1}{r_1^2}+ \ldots+ \frac{A_n}{r_n^2},$$
using the change of variables formula.
For the first claim, observe $v\in\pi_i(D_i)$ if and only if
$$
\langle p_i+r_i v- p_i, p_j-p_i\rangle< -r_i(r_j-r_i),
$$
for all $j\neq i$, equivalently
$$
\langle v, p_j-p_i\rangle< -(r_j-r_i),
$$
for all $j\neq i$. Now, define
$$
A= \bigcup_{i,j=1}^n\{v\in S^2: \langle v, p_i-p_j\rangle=-(r_j-r_i)\}.
$$
This is clearly a set of measure zero. Fix $v\in S^2$. Without loss, we'll show that if $v\notin \pi_2(D_2)\cup\ldots\cup\pi_{n}(D_n)\cup A$, then $v\in \pi_1(D_1)$. But, if $v\notin \pi_2(D_2)\cup\ldots\cup \pi_n(D_n)\cup A$, and $i\in\{2, \ldots, n\}$, there is some $\phi(i)\neq i$ such that
$$
\langle v, p_{\phi(i)}-p_i\rangle>-(r_{\phi(i)}-r_i).
$$
Note that, for $k>l$
$$
\langle v, p_{\phi^k(i)}- p_{\phi^l(i)}\rangle= \sum_{s=l}^{k-1}\langle v, p_{\phi^{s+1}(i)}-p_{\phi^s(i)}\rangle> \sum_{s=l}^{k-1}-(r_{\phi^{s+1}(i)}-r_{\phi^s(i)})= -(r_{\phi^k(i)}-r_{\phi^l(i)}).
$$
Since $\phi:\{2, \ldots, n\}\to \{1, \ldots, n\}$, and $\phi(i)\neq i$, there must me some $k$ such that $\phi^k(i)=1$, otherwise there would be some $k>l$ such that $\phi^k(i)=\phi^l(i)$, but then
$$
0= \langle v, p_{\phi^k(i)}-p_{\phi^l(i)} \rangle> -(r_{\phi^k(i)}-r_{\phi^l(i)})=0,
$$
contradiction. As a result, $\langle v, p_1-p_i\rangle> -(r_1-r_i)$, for all $i\geq 2$ which proves the claim.
For the second claim, let $v\in \pi_i(D_i)\setminus A$, that is
$$
\langle v, p_j-p_i\rangle< -(r_j-r_i),
$$
for all $j\neq i$. Then, $\langle v, p_i-p_j\rangle> -(r_i-r_j)$, i.e., $v\neq \pi_j(D_j)$ for all $j\neq i$.

Let me describe why this is the right definition for $D_i$. Let us work on the plane. Suppose $S_1= \partial D(0,r)$, and $S_2= \partial D((-a,0), R)$, for some $a>0$ such that $a> |R-r|$. We want to find the largest $x\in (-r,r)$, such that $(x,\sqrt{r^2-x^2})$ is visible from $S_2$. This means that the tangent
$$
(x, \sqrt{r^2-x^2})+ t(-\sqrt{r^2-x^2}, x),
$$
intersects $S_2$ at a single point. Equivalently, the equation
$$
|(x- t\sqrt{r^2-x^2}, \sqrt{r^2-x^2} +tx)- (-a,0)|=R,
$$
has a unique solution. We may rewrite this equation as:
$$
r^2 t^2- 2a\sqrt{r^2-x^2}t+ (a^2+2xa+r^2-R^2)=0.
$$
This has a unique solution if and only if the discriminant
$$
\Delta= 4a^2(r^2-x^2)-4r^2(a^2+2xa+r^2-R^2)=0,
$$
i.e.,
$$
(ax+r^2)^2= r^2R^2. 
$$
If $r<R$, we want $x>0$ so we take $ax+r^2= rR$. This means that for any $x>\frac{rR-r^2}{a}$, $(x, \pm\sqrt{r^2-x^2})$ is in the dark region of $S_1$, or equivalently when $-ax+0y< -r(R-r)$.
If $r>R$ then we want $x<0$ so we still take $ax+r^2= rR$, resulting in the same equation.

I would also like to expand on @JyrkiLahtonen comment since I thought his suggestion was excellent. Another way to do this is note that the convex hull
$$
C= \text{conv}\{S_1, \ldots, S_n\}= D_1\cup\ldots\cup D_n\cup \{\text{cylinders}\}\simeq S^2,
$$
and hence all the curvature of $C$ is concentrated on $D_1, \ldots, D_n$. Now my Gauss-Bonnet
$$
4\pi= \int_{C}K\text{ d} S= \int_{D_1}\frac{1}{r_1^2}\text{ d}S+ \ldots+ \int_{D_n}\frac{1}{r_n^2}\text{ d}S= \frac{A_1}{r_1^2}+ \ldots+ \frac{A_n}{r_n^2}. 
$$

A: 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.

