A unit square lamina contains $n$ independent uniformly random points. Each point is the centre of a disc whose perimeter touches the nearest neighboring point. Here is an example with $n=20$.
In this example, the total area of the discs (i.e. the sum of areas of the discs, not the area of the union of the discs) is approximately $1.0105$.
What does the total area of the discs approach, as $n\to\infty$ ?
The answer is $1$, i.e. the area of the square itself. I give a non-intuitive proof below. But since this result is quite elegant, I am looking for an intuitive explanation.
My non-intuitive proof
As $n\to\infty$, the points approach a 2D Poisson process with intensity $n$. For a chosen point, let $x$ be the distance to its nearest neighbor, which has density function
$$f(x)=2n\pi xe^{-n\pi x^2}$$
So the total area of the discs approaches
$$n\int_0^\infty \pi x^2 f(x)\mathrm dx=1$$
by integration by parts twice.
Remarks
The shape of the lamina (e.g. square) does not matter, since we are taking $n\to\infty$.
The answer is the same in one, two or three dimensions. (In one dimension, the nearest neighbor distance density function is $f(x)=2ne^{-2nx}$. In three dimensions, it is $f(x)=4n\pi x^2e^{-\frac{4n\pi}{3}x^3}$.)
This question was inspired by a related question, "A square contains many random points. From each point, a disc grows until it hits another disc. What proportion of the square is covered by the discs?".