You may also like

Coke Machine

The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design...

At a Glance

The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?

Cosines Rule

Three points A, B and C lie in this order on a line, and P is any point in the plane. Use the Cosine Rule to prove the following statement.

Six Discs

Age 14 to 16
Challenge Level

Hee Chan and Jeang Bin sent us their working with correct areas for shapes A and C:

Shape A

The longer part of the rectangle is $4$ because each circle has radius $1$ unit.

The right-angled 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 30-60-90 triangles with sides in the ratio $1:2:\sqrt {3}$. The areas of the packings are as follows.

Packing (A)

Triangle and parallelogram overlaid.

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)

Circles packed into hexagon with labelling

$h = \sqrt{3} + 1$ (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)

circles packed into pentagonal shape with labelling

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)

Circles packed into rectangle with labelling

$A_F = 5(2h) = 10h = 10(\sqrt{3}+1) = 27.32 \textrm{ (4 s.f.)}$