Take an arbitrary packing of the plane by nonoverlapping disks of radius 1. We will partition the plane into regions, and will show that each region has density at most $\pi /\sqrt{12}$. Center a larger circle of radius $2/\sqrt{3}$ around each disk (Figure i).

There cannot be a point interior to three large circles. Indeed, in the extreme case, three large circles meet at a point, the circumcenter, if the centers are the vertices of an equilateral triangle of edge $2$ (Figure iii).

This gives our partition of space: regions outside all of the larger circles (white), the isosceles triangles (cyan), and the part inside the larger circles but outside all triangles (purple) - see Figure iv.

The regions outside all of the larger circles have density $0$, which is certainly less than $\pi /\sqrt{12}$. The density of the interior of the larger circles is the square $3/4$ of the ratio $(1:2/\sqrt{3})$ of the smaller to larger radius, again less than $\pi /\sqrt{12}$. This inequality can be seen geometrically by drawing a hexagon that the small circle inscribes and the large one circumscribes (Figure v).

To calculate the density of an isosceles triangle, we apply a linear transformation to the triangle (preserving ratios of areas and hence densities) to transform it into an equilateral triangle with edge $2/\sqrt{3}$. The transformation scales along the orthogonal axes through the base and altitude of the triangle and fixes the vertex $v$ opposite the base of the isosceles triangle. The unit circle is transformed into an ellipse. (See Figure vi.)

The linear transformation preserves the lengths of the two equal edges of the isosceles triangle. Hence the ellipse cuts those edges at distance 1 from its center $v$. This identifies all four points of intersection of two conic sections, the ellipse and the unit circle centered at $v$.

Knowing these points of intersections, we deduce that the intersection of the ellipse with the interior of the equilateral triangle is contained in a disk of radius 1 centered at $v$. In particular, the density is of the equilateral triangle is increased by replacing the ellipse with a circle of radius 1. The equilateral triangles fit together to form hexagons with inscribed disks of radius 1. The density of these pieces is therefore $\pi /\sqrt{12}$. This completes the proof of Thue's theorem.