# Densest Packing of Identical Circles

## Theorem

The densest packing of identical circles in the plane obtains a density of $\dfrac \pi {2 \sqrt 3} = \dfrac \pi {\sqrt {12} }$:

$\dfrac \pi {2 \sqrt 3} = 0 \cdotp 90689 \, 96821 \ldots$

This happens when they are packed together in a hexagonal array, with each circle touching $6$ others.

## Proof

Consider the rectangular area $ABCD$ of the densest packing of circles. Let the radius of one circle be $1$.

The length $AB$ is $2$.

The length $AC$ is $2 \sqrt 3$.

Thus, from Area of Rectangle, the area of $\Box ABCD$ is $4 \sqrt 3$.

Within $ABCD$ there is one complete circle and one quarter of each of $4$ other circles.

That makes a total of $2$ circles.

Thus, from Area of Circle, the area of $ABCD$ which is inside a circle is $2 \pi$.

So the density is:

 $\ds \dfrac {\text {Area of Circles} } {\text {Area of Rectangle} }$ $=$ $\ds \dfrac {2 \pi} {4 \sqrt 3}$ $\ds$ $=$ $\ds \dfrac {\pi} {2 \sqrt 3}$ $\ds$ $=$ $\ds \dfrac {\pi} {12}$

as required.