Congratulations to Nicola Spittal, S4, Madras College, St Andrew's for your excellent solution to this problem.

We have four packings of six circles, each of radius 1 unit that touch their neighbours and the sides of the box namely: (A) triangular; (B) parallelogram; (C) rectangle (unusual packing) and (D) rectangle (usual packing).

By superimposing the triangular packing (A) on the parallelogram packing (B) as shown below we can see that area(DQRS) = area(DSTU) so that (A) has larger area than (B) and the difference in areas is equal to the area of the small triangle DUVW.

We use many 30-60-90 triangles with sides in the ratio 1:2:Ö3. The areas of the packings are as follows.

Packing (A)

Triangle and parallelogram overlaid.

The triangle has side length 4+2Ö3 so that area is 1/2(4+2Ö3)2 sin60°, or (equivalently, using half base times height) 1/2(4+2Ö3)(3+2Ö3), which is 24.12 square units to 2 decimal places.

Packing (B)

The base length is given by
PW=tan60° +4+tan30° = Ö3+ 4 + 1
Ö3
 = 6.3094.

The height is given by h = 2+2cos30° = 2 + Ö3 = 3.7321, so that the area is 23.55 square units to 2 decimal places.

Packing (C)

Skewed rectangle.

The line joining AE is parallel to the bottom of the rectangle. The base of the rectangle has length AA ' + AD cosq+ DD ' which is 2+6 cosq. Similarly, the height of the rectangle is 2+6sinq. It remains to find q. As DCDE is equilateral, ÐA C E = 120 °. Using the Cosine Rule
AE2
= 16 + 4 - 16cos120° = 28
so that AE = 2Ö7. Using the Sine Rule for DACE,
sinq
CE
= sin120°
AE
 ,
hence sinq = Ö3/2Ö7. Thus cosq = 5/2Ö7, and the area is 30.40 square units to 2 decimal places.

Packing (D) : as the rectangle is 6×4, the area is 24 square units.