Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 74 / 3 / a / Solution

A square circle packing of equal radii has one circle per cell of side $2r_1$. Its packing fraction is thereforeFor the alternating-radius square circle packing shown in the source figure, let $d$ be the nearest unlike-centre separation. A two-circle repeat cell has area $2d^2$ and solid area $\pi (r_1^2+r_2^2)$. Unlike circles require $d\ge r_1+r_2$; like circles on the two sublattices require $\sqrt{2}d\ge 2r_1$ and $\sqrt{2}d\ge 2r_2$. ThusAll three nonoverlap constraints matter when varying the radius ratio.

Take $r_1\ge r_2$ as in the illustrated arrangement and set $s=r_2/r_1$. If $0\le s\le \sqrt{2}-1$, the large circles determine the spacing and $\eta =(\pi /4)(1+s^2)$ increases with $s$. If $\sqrt{2}-1\le s\le 1$, unlike contacts determine it and $\eta =(\pi /2)(1+s^2)/(1+s{)}^2$, whose derivative is proportional to $s-1$ and is nonpositive. The maximum is therefore at the joining point:The large circles then touch their large neighbours, while each small circle exactly fills a square interstice. Interchanging the species gives the reciprocal radius ratio. This is the highest packing fraction within the specified alternating square arrangement; it is not a claim about every possible unequal-circle packing.