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TETRAHEDRON {3, 3}
At right is the Schlegel diagram, the existence of which guarantees that Euler’s formula will hold. Check it out: does it have the right number of vertices, edges, triangular faces? We can also measure the angle between two faces; this is called the dihedral angle.
Point M is the midpoint of edge BC, so BM = Here is a summary of some of the metric properties of the regular tetrahedron of edge length s.
Centres of gravity are usually calculated using the integral calculus, but for triangles and tetrahedra there is a simpler approach. A uniform triangle ABC of weight W behaves in the same way as a congruent but weightless triangle having a weight
Arguing in the same way as for the triangle, the weights at B, C, D are equivalent to a weight of For the regular tetrahedron, each median is an altitude perpendicular to the face, (use a little calculation with Pythagoras’ theorem), so the centroid is the common intersection of the four altitudes.
If you are feeling energetic, you might like to calculate the inradius, the midradius and the circumradius of a regular tetrahedron of sidelength s!
With each of the polyhedra we consider, it is instructive to construct a cardboard model. You can quickly obtain a fine collection! It is worth buying some good quality cardboard from an art shop. For extensive advice and hints on model-making with cardboard look at Card Modelling FAQ. There are essentially three basic ways of constructing polyhedral models out of card. (1) Construct (or copy) a net of joined faces. The outline is cut out, with extra tabs allowed for on each alternate outside edge. The edges between the faces are scored by running a blunt knife over them (I use a compass point), and the faces carefully folded backwards (away from the scoring), forming a good clean fold. The model is then constructed by glueing the tabs behind the appropriate faces one by one, using a good quality craft glue. Be careful to make each glueing as accurate as possible. The glueing of the final face can be a little tricky as you have to work from the outside only. Have a damp cloth handy to wipe off any excess glue. The final model can be painted. (2) From a a constructed (or copied) face template make the required number of separate faces. A good way of doing this is to start with the original and prick through the vertices using a needle or compass point. Join the vertices with straight lines, and carefully cut out the faces with sharp scissors or a guillotine. Next glue the faces edge to edge (keeping in mind the final model!) using a good craft glue. Wipe off any excess glue. Different coloured card can be used in this construction, or the final model can be painted. (3) Sometimes a combination of methods (1) and (2) works well, with small components being constructed using method (1) and then assembled using method (2). Generally, I have found that method (2) gives better results than method (1). The grade of cardboard used will depend on the size of your models. Larger models require a thicker grade of card. In any case, a thicker, more rigid grade of cardboard is advisable for the tetrahedron and cube than for the other polyhedra, as the faces are relatively larger. We will give face templates which are sized to give a pleasing set of models. The templates can be scaled if you wish to modify the size of your set. Small nets are also provided as a help for assembly. So, finally, for the tetrahedron ... ! Neat, small, peel and stick models of the Platonic solids are available from Ozzigami (see references).
The tetrahedron has further interesting properties. We shall discover some of these using some hands-on exploration! It will also enable you to get a ‘feel’ of this fascinating solid.
A regular tetrahedron can be placed so that pairs of opposite edges form the diagonals of opposite faces of a cube. This surprising property can be useful for placing the regular tetrahedron in a coordinate context. For example, we might take the eight vertices of the cube to be (
Volume = 1/3 x area of base x height. This formula is usually obtained using calculus, but we can demonstrate that six (non-regular) tetrahedra of equal volume pack into a cube. This illustrates the above formula. We begin by dividing the cube into two right triangular prisms:
If we look at the right triangular prism formed from half the cube, the volume of the prism is area of base x height. Since three tetrahedra having the same volume occupy this space, and noting that the tetrahedron base has the same area as that of the prism, each tetrahedron has volume 1/3 x area of base x height. 1/3 x area of base x height =
The tetrahedron has little obvious symmetry relative to a set of coordinate axes. We probably do best to start with the cube. Assuming for the moment that we can take the vertices of a cube to be ( (1, 1, 1), (–1, –1, 1), (–1, 1, –1), (1, –1, –1). In this choice, all the entries are
Here are some ideas.
Properties MathWorld : http://mathworld.wolfram.com/Tetrahedron.html Centre of gravity Yaglom, I. M., Boltyanskii, V. G., Convex Figures, Holt, Rinehart and Winston 1961, pp34 – 36 Model making Card Modelling FAQ : http://www.cardfaq.org/faq/tips.html#s3.4
Ozzigami has peel and stick models: www.ozzigami.com.au |
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Simple tetrahedron properties
We notice that the conditions for this to be a regular polyhedron are all satisfied: congruent equilateral triangles for faces, and vertices all alike. Each vertex figure is again an equilateral triangle. The face angles are clearly all 60°. 
• Use the diagram at right and the following steps to calculate the dihedral 
and use your calculator to find the 
, and we can find length AM using Pythagoras’ theorem.
= 
. A calculator now gives
at each vertex. The weights 
at the segment midpoint M. Now the weight 
We can argue the same way with the tetrahedron (regular or not).
at each vertex.
at the triangle centroid M. We next deduce that the centre of gravity G of the tetrahedron lies on AM and with AG : GM = 3 : 1. If we define AM as a median of the tetrahedron, we can say that the four medians intersect at the centre of gravity (centroid). We also notice that the line segments joining opposite edges of the tetrahedron have G as their common midpoint.
• A cube is given at right. In the applet, can you see how to place the tetrahedron so that its four vertices coincide with four of the vertices of the cube? If the cube has edge length 1, what is the edge length of the tetrahedron?
The first slicing problem obviously gives rise to a sequence of equilateral triangles. In the second case, we obtain a family of rectangles, beginning with long thin horizontal rectangles, and progressing through to long thin vertical rectangles. In the middle, at the midpoints of the four sides which the rectangle vertices are transversing, there is a transition rectangle which is a square. There is a puzzle based on this property, giving a dissection of a regular tetrahedron into two identical pieces, which is surprisingly hard for some people to do.
1, 
You may be into drawing polyhedra on a computer or graphics calculator by calculating coordinates, or perhaps producing animated models using a programming language such as Java. In such cases, it is useful to find the simplest and most symmetric set of coordinates for the vertices of your solid. There are no fixed rules for this, apart from trying to build in any obvious symmetries of the polyhedron.
