Circumscribe a disc with a regular hexagon, and circumscribe the hexagon with a circle. This gives what I call the hexagonally circumscribed circle of the original disc. It is a concentric circle whose radius is times the original one. If P is any point outside the original disc, the two tangents from P to that disc bound what I call the cap corresponding to P and the disc. The key property we shall need later on is that the vertex angle at P is a decreasing function of the distance of P from the disc. This vertex angle will be exactly 120^o precisely when P lies on the hexagonally circumscribed circle, because then it is a vertex of a circumscribed hexagon. So when this angle is less than 120^o the point P must lie outside the hexagonally circumscribed circle. This vertex angle will be exactly 120^o precisely when P lies on the hexagonally circumscribed circle, and when this angle is less than 120^o the point lies outside the hexagonally circumscribed circle.

We shall need also another property of these circumscribed circles. Suppose two of them intersect, but that the discs themselves do not intersect. Then that intersection can only intersect the Voronoi cell of a third disc if the three discs are mutually touching in the configuration of discs in a hexagonal packing.

For suppose P to be a point in the intersectionof two circumcribed circles, and also in the Voronoicell of a third disc.The two discs are close enough that thepoint exactly half-way between them will be in the dead region where thetriple point cannot be located.The triple point must therefore lie between P and this half-way point,and also in both hexagonally circumscribed circles.But each of the vertex angles at the capped discsfrom this triple point must then be at least120^o. This can only happen ifthis angle is exactly 120^o and the three discs are mutually touching.

In the diagram to the right, this means that the pointsin the yellow rhombus are never in the Voronoi cellof the third disc.