Sweet Dreams Are Made Of This...

This month, Wilf finds that a jar of mint humbugs is the perfect way to get in shape...

In the seaside town of my birth (I am a Man of Kent) there was a shop that sold lettered rock and similar sweets. Not only did they sell it: they made it in plain view at specified hours of the day – a visual treat, especially for children eager to learn the answer to important questions such as “How do they get the letters so small yet perfect, down the whole length of the rock?” Another question silently but eloquently answered by demonstration was how humbugs got their peculiar shape....

To us a humbug was no mere striped sweet lozenge: it was a peculiar package almost spherical in the middle but tapering to an edge on top and bottom. Curiously these two edges were not in line: they were set at ninety degrees. The candymaker had rolled out long strips of molten sugars, eventually forming long cylindrical wands with stripes (of a different sugar) running the full length of each wand. When the wand had cooled and started to harden, she snipped off each separate humbug, going down the wand until it had all gone. Between each snip, the wand was rotated ninety degrees, so the edges at both ends of each humbug were out of line.

Effectively the shape produced was that of a tetrahedron (ancient Greek meaning “four facets”). You can construct a model of a tetrahedron by joining together four triangles. If all four triangles are equilateral, the tetrahedron is said to be “regular”.

Exercise your mind for a few moments by picking up one of these tetrahedron humbugs: resist the compulsion to stuff it into your mouth. Instead, let it rest in your (imaginary) flat hand, It resembles a pyramid with three sides – that is, three visible sides coming to a point at the top. The volume of this shape must be less than that of a cube with the same length edges: for one thing it isn’t quite as tall, and for another, the base (face) on which it sits is smaller, being a triangle rather than a square. (See figure i).

Figure i

When in my youth I mused at the creation of rock candy and humbugs, I realised that it gave a good exercise to portions of my brain – parts that prove useful in visualisation, co-ordination and logical thinking. I set myself a “quest” – a problem to think through, to solve without paper or pencil. Since this first “quest” I have devised many more (some more successfully than others) and use them to give my mind a couple of mental laps round the course every so often.

My first quest was to determine the volume of a tetrahedron where the four edges were one unit in length. Even though I have done it now so often that I have memorised the answer, it still presents a stimulating challenge. Perhaps you will go through this first quest with me now, and you can judge for yourself.

Calculating the volume of a regular tetrahedron needs very little in the way of formulae. The most complicated part of the process involves the Pythagorean Theorem – in a right-angled triangle in plane space, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

These handmade stripe-free mint humbugs (available from Mrs O'Malley's sweet shop) clearly show the rather peculiar shape...

Now where did we put that imaginary humbug: oh yes, there you have it in the palm of your imaginary hand. Generally the volume of a shape is related to the area of its base and its overall height, and the tetrahedron, you will be pleased to learn, is no exception. If you were to slice through a cube some distance upward from its base but parallel to the ground, the cross-section would be a square exactly the same size as the base. The formula for the volume of such shapes is B x H where B is the area of the base, and H is the overall height. The tetrahedron falls into a different category. When you slice into it above the base and parallel to the ground, you will get the same shape as the base (a triangle this time) but it will be smaller in size. As a matter of fact, the cross-section evenly drops in area until it dwindles to zero at the point. The volume of such objects is (base x height / 3): this is called the “cone” rule, because it works for cones, or for any shape tapering evenly to a point.

That volume calculation entails the area of the base: in the case of a tetrahedron, this base is an equilateral triangle: we will have to know how to calculate the area of that triangle at some stage, so let’s get it out of the way immediately. The area of a triangle is (baseline x height)/2 where “height” is the widest extent away from the baseline. Well the baseline is easy enough – it is one unit long. How can we know the height, except to know that it is something less than one unit? The trick is to see this equilateral triangle as two triangles – mirrored duplicates, divided by a vertical line from the distant point to the middle of the baseline. These two triangles are right-angled, so let’s look closer at one of them to see if Pythagoras will help. We have divided the big triangle half-way along its base, so the new base is half a unit in length. The hypotenuse is one of the sides of the original triangle, so is one unit long. The vertical line marking the splitting of the original triangle must obey Pythagoras, so its square plus one quarter (the square of the half-base) must add to one, the square on the hypotenuse.

By direct calculation, the height of the triangle is √3/2. This is the height of the equilateral triangle as well, so we can now jump to the answer: Baseline = 1, height = √3/2 so the area of the triangle is √3/4. If you don’t like to clutter your memory with surds, you can convert to decimal and say the area is 0.861 square units, but believe me when I say that this is premature. After a few repetitions you will get to know some squares, cubes, logs off by heart without the slightest inconvenience: 1.732 is a very close approximation of √3, and 0.861 is just half that.

Look again at the tetrahedron in your hand. Imagine it to be translucent, and observe a line starting at the uppermost point dropping to the middle of the base triangle. If you name the peak A, the three other corners B, C and D, this new midpoint can be E. In addition think of a line running from corner B to the middle of the edge CD. The line AE is the tetrahedron’s height: if we knew its length, we would have everything needed to solve the quest. (See figure ii)

Figure ii

There are a few ways to proceed from here, but I have found the easiest is to think of the triangle ACE: it is right-angled, and we know the length of AC (one unit, of course). If we knew the length EC we could work out AE via Pythagoras. We don’t know the length of EC, but it is part of the small right-angled triangle ECF. As a matter of fact, you should be able to see that the angle ECF is 30 degrees, so we have another 30-60-90 degree triangle: the ratio of its sides are one, two and the square root of three. We know the length of CF (one half unit) so we can now calculate the length of EC. It works out to be (1 / √3). Don’t bother calculating this out to decimal form, though; it isn’t necessary yet. But now we can use Pythagoras again, since EC² plus AE² is AC². Remember that AC is one of the original edges of the tetrahedron, so its square is one.

A little more juggling of symbols and you should come to the conclusion that the length of AE is (√2 / √3). Now you can plug this into the volume formula: (base x height / 3). The area of the base is (√3 / 4) and the height is (√2 / √3). This comes out to a volume of (√2 / 12): a little calculation from memory can change this to 0.118 cubic units.

I must confess, the first time I followed this quest to its conclusion, I feared I had failed somewhere. The result of 0.118 cubic units is disturbingly small: about an eighth the volume of a cube with one unit sides. So I tried a rough approximation method. The area of the base of a tetrahedron is somewhat less than half the area of the base of the corresponding cube. (You can confirm this by noting that the height of the triangle is less than one). If instead of a tetrahedron we were dealing with a wedge one unit high, it would be less than half the volume of the cube. The fact that it tapers to a point reduces its volume by a further factor of three. This confirms that by rough estimation, the tetrahedron has a volume less than one-sixth of the cube. It seems our mental calculations aren’t so far from what we might have expected.

This quest may seem a little daunting the first time though, but try it on your own, visualising the shapes and it will become easier – and you will have acquired a new skill. You needed very little more than a working knowledge of the Pythagorean theorem.

Recently the son of a friend asked me if I would explain why there are an infinite number of “-gons” but a finite number of “-hedra” in geometry. Words ending in “-gon” are usually two-dimensional shapes with a certain number of sides: a pentagon has five sides, and the prefix “penta-“ is Greek for five; the “-gon” suffix means edge. The names for three-dimensional shapes are similar: “hexa-“ means six, and “-hedron” means “facet” or “face”. So a hexahedron is a 3-D shape with six faces.

My friend’s son was correct, though he left out a few details: he was talking about “regular” polyhedra – ones where every facet is the same size and shape. The 2-D shapes can have any number of equal-length edges, as long as there are at least three. A trigon is almost always called a triangle, and a tetragon is likewise called a square. Beyond that you can find pentagons, hexagons, heptagons, octagons, nonagons and decagons (ten-sided). You can continue as far as you like: a polygon can have any number of sides – no limit. From ancient times, though, mathematicians had found only five polyhedra – no more. These six are very interesting: you will have seen all of them at one time or another.

The simplest is the tetrahedron – the shape that is the focus of my quest. It has four sides, all triangles. The next simplest is the hexahedron, made up of six sides, all squares. We usually call this a cube. The next is the dodecahedron – a lovely name, so why should we change it? The “dodeca-“ part is from the Greek for twelve: it is a twelve-faceted shape, each facet a pentagon. When it rests on one of its facets, another is flat on top. There are two other regular polyhedra: the octahedron has eight facets, each a triangle, and the icosahedron (“eye-COH-sa-dee-dron”) has twenty facets. It is likewise made up of triangles. That’s all five regular polyhedra – there are no more. But why not? How can we be sure?

The simplest of these five shapes, the tetrahedron, is produced by putting together equilateral triangles with a corner from three meeting at one place to become a point. A fourth triangle fits in nicely so that four points are produced, each being the juncture of three of the triangles. The next simpler shape is the cube. Again, each point is the juncture of three corners – this time each of the facets is a square. Next is the icosahedron, and again each point is the juncture of three shapes – this time pentagons. What about trying to make a polyhedron with each facet made from six sides? The regular six sided polygon is the hexagon. When you come to join three hexagons together you will not form a point. Hexagons fit together to form a flat surface.

A triangle’s corner takes up 60 degrees; a square’s corner takes up 90 degrees; a pentagon’s corner takes up 108 degrees; a hexagon’s corner takes up 120 degrees. There are a couple of things to note here: (1) you need at least three corners to form a 3-D point; (2) you cannot form a point if the combined corners take up more than 360 degrees. There is no use looking for polyhedra made up from polygons any more complex than five sides. There are two more regular polyhedra, and you can think of these as special cases. Combining triangles with four corners to a point makes an eight-sided shape, the octahedron. It looks very unlike the other polyhedra, most resembling two pyramids glued back-to-back. It does not seem to be ball-shaped to any extent. It can be used as a die, but it looks a bit peculiar in that role. You can combine five triangles at a point as well (since five times 60 degrees is still less than 360 degrees), and you will end up with a very handsome shape with twenty facets. Its name is “icosahedron” – I wonder what “icosa-“ means in ancient Greek? We hit the 360-wall when we try to combine more than five triangles, and we get nowhere when we try four squares to a point.

There we have it – five polyhedra and no more possible. You can go somewhat further by combining a few different shapes: for example a judicious combination of pentagons (5-sided) and hexagons (six-sided) shapes will produce a nice shape familiar to us as a soccer football. But there exactly are just five regular polyhedra, no more – no less.

These are called “the Platonic solids” in honour of Plato (427 – 347 BC) who explained this peculiar ceiling to their number. Let’s take a final summary in table form:

Many thanks to Mrs O'Malley's (Purveyors of Fine Confectionary) for supplying the photographs of traditional mint humbugs.

January 2006