Six Discs
Problem
Six circular discs, all of radius 1 unit, are packed in differentshaped boxes so that the discs touch their neighbours and the sides of the box.
 Without using any calculations, compare the areas of the bases of boxes A and B and say which is the larger.
 Use estimation to list the packings in order of increasing base area. Which are the hardest to put in order?
 Now calculate the area of the base of each box and compare the correct order with your estimate.
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Getting Started
What would happen if you superimposed B on A? What about D on E?
These diagrams show some rightangled triangles you can use to work out the lengths on shapes B and D using trigonometry. What angles do you know? What sides do you know?
Student Solutions
Hee Chan and Jeang Bin sent us their working with correct areas for shapes A and C:
The longer part of the rectangle is $4$ because each circle has radius $1$ unit.
The rightangled triangle with side length marked $x$ has angles $30^{\circ}, 60^{\circ}$ and $90^{\circ}$ so its hypotenuse must be $2$, and side $x$ is $\sqrt{3}$ according to Pythagoras's law.
So the total length of one side of the triangle is $4+2\sqrt{3}$.
The area of the equilateral triangle is
$$\frac{1}{2}(4+2\sqrt{3})^2 \times \sin 60^{\circ}$$ $$=\frac{1}{2}(4+2\sqrt{3})^2 \times\frac{\sqrt{3}}{2}$$ $$=7\sqrt{3}+12$$which is $24.12$ square units to two decimal places.
For C, for the bottom there are 3 circles and this means that the length of the bottom is 6 units. Also for the height there are 2 circles so the length is 4 units. The area is $4 \times 6 = 24$ square units.
In a similar problem Nicola Spittal, S4, Madras College, St Andrew's sent in this excellent solution, demonstrating that A must have a larger than B and also calculating the correct areas for shapes A, B and C.
By superimposing the triangular packing (A) on the parallelogram packing (B) as shown below we can see that $area(\Delta QRS) = area(\Delta STU)$ so that (A) has larger area than (B) and the difference in areas is equal to the area of the small triangle $\Delta UVW$.
We use many 306090 triangles with sides in the ratio $1:2:\sqrt {3}$. The areas of the packings are as follows.
Packing (A)
The triangle has side length $4+2\sqrt{3}$ so that area is ${1\over 2}(4+2\sqrt{3})^2 \sin 60^\circ$, or (equivalently, using half base times height) ${1\over 2}(4+2\sqrt{3})(3+2\sqrt{3})$, which is 24.12 square units to 2 decimal places.
Packing (B)
The base length is given by $PW=\tan 60^\circ +4+\tan 30^\circ = \sqrt{3}+ 4 + {1\over \sqrt{3}}= 6.3094.$ The height is given by $h = 2+2\cos 30^\circ = 2 + \sqrt{3}= 3.7321$, so that the area is 23.55 square units to 2 decimal places.
Packing (C)
As the rectangle is $6\times 4$, the area is 24 square units.
Here is one method of finding the area of the remaining shapes:
Packing (D)
$h = \sqrt{3} + 1$ (from diagram)
$\cos 30^\circ = \frac{h}{a} \Rightarrow a = \frac{h}{\cos 30^\circ} = 2(\frac{\sqrt{3}}{3} + 1)$
Splitting the hexagon in to two trapezia, the area is therefore:
$A_D = 2\{\frac{1}{2}(a + 2a)h\} = \frac{3}{2}ah = \frac{3}{2}(\frac{\sqrt{3}}{3}+1)(\sqrt{3}+1) = 8\sqrt{3}+12 = 25.86 \textrm{ (4 s.f.)}$
Packing (E)
Area composed of triangle, width $2h$ height $\frac{a}{2}$, and rectangle, width $2h$ height $2+\frac{a}{2}$
$\therefore \ A_E = \frac{1}{2}(2h)(\frac{a}{2}) + (2h)(2+\frac{a}{2}) = \frac{3}{2}ah + 4h = 8\sqrt{3} + 10 = 23.86 \textrm{ (4 s.f.)}$
Packing (F)
$A_F = 5(2h) = 10h = 10(\sqrt{3}+1) = 27.32 \textrm{ (4 s.f.)}$