OCTAHEDRON {3, 4} Simple octahedron properties The third in our listing of the regular polyhedra is theoctahedron.
The name octahedron comes from the Greek
Each face is an equilateral triangle, each vertex figure a square. The face angles are clearly all 60°. The octahedron has an obvious centre where the three diagonals meet. This point is the centre of gravity, incentre, midcentre and circumcentre. At right is the Schlegel diagram for the octahedron. Try to visualize how you would obtain this from the figure above. What does this representation tell us? Let the octahedron have edge length
Here is a summary of some of the metric attributes of a regular octahedron of side length
The octahedron is usually pictured balancing on one vertex, giving us the picture of it as a double pyramid. There is a whole family of these bipyramids, but the octahedron is the only member which is regular.
But the octahedron can also be pictured lying on a face.
Check your answers to the above three exercises ... The octahedron is usually pictured balancing on one vertex, giving us the picture of it as a double pyramid. There is a whole family of these bipyramids, but the octahedron is the only member which is regular.
But the octahedron can also be pictured lying on a face. In this position it can be classed as an antiprism.
Let’s tackle these questions in order. First the bipyramids. Each of the bipramids with (internal) bases {3}, {4}, {5} can have all their faces equilateral triangles. However, for the bipyramid with base {6}, and beyond, we run into trouble. Why is this? The octahedron is the only member of this class that can have all vertices alike, as it is the only member with the same number of faces about each vertex. All members of the prism family can have all their faces regular, taking the vertical faces as squares. Here, the cube is the only member having all vertices alike, as this is the only case where all the faces are congruent. With the antiprisms, again all members can have all their faces regular, taking all the linking triangles to be equilateral. Again taking all the triangles to be equilateral, each members can have all its vertices alike. The octahedron is now the only member having all vertex figures regular, as it is the only case where all the faces are congruent. It can be fun exploring the different relationships between the various solids. Model making You can make your own model octahedron, using the face template of suitable size given here.
We could play the net game as before, but here is the solution: We notice that there are eleven possibilities. Now a really interesting question arises. ‘Eleven’ is a reasonably uncommon number. So is there some explanation as to why the cube and the octahedron have the same number of nets? Typically, we would like to set up some one-to-one correspondence between the nets or between the solids, but it is not clear how this might be done, as the number of faces is different for the cube and the octahedron. Some answers can be found in the references below. Space tessellations We have seen that cubes tessellate space, but that none other of the regular solids have this property. However, there is an interesting space tessellation involving the regular octahedron.
Our results suggest that we might be able to pack tetrahedra around an octahedron with the solids sitting on a plane surface. This encourages us to think about space filling possibilities. The adjacent figure shows an octahedron resting on one of its faces, and four tetrahedra. See how the solids pack together. However, there is a much easier way.
Check your answer ... . Cube inscribed in an octahedron It is interesting to see how the various solids fit inside one another, as the tetrahedron in the cube above. Here is a figure of the cube inside an octahedron – a very simple combination of solids, with the eight vertices of the cube placed at the midpoints of eight of the twelve edges of the octahedron.
We next explore some rather different properties.
Finally, and just for fun
In the case of the sections, in (a) we obtain a sequence of squares (probably placed on the diagonal!) which grow from a point and then diminish back to a point. In (b) we begin with (say) a horizontal line segment which turns into a rectangle, thickening and shortening to become a square, and then converting to a vertical segment. The sequence then reverses as we move to the far edge of the octahedron. Case (c) is the most interesting, beginning with an equilateral triangle, passing through a regular hexagon, and then changing to an inverted equilateral triangle. The behaviour is illustrated at right.
Since there are 3 axes and 6 vertices, we place the vertices on the axes, symmetrically about the origin (1, 0, 0), (0, 1, 0), (0, 0, 1). The numbers 3 and 6 are obviously related by 6 = 2 x 3. Real life occurrences
Here are a few examples ... . References Properties Octahedron properties : http://mathworld.wolfram.com/Octahedron.html Bipyramids and antiprisms Antiprisms : http://mathworld.wolfram.com/Antiprism.html Model making The excellent model book: Wenninger, M. J., Number of nets:
Space tessellations Space filling: http://whistleralley.com/polyhedra/octahedron.htm Further properties Sections and projections: http://whistleralley.com/polyhedra/octahedron.htm |

(a) Suppose the octahedron has unit side length. Then looking at some simple right-angled triangles, we find that the dihedral angle for the octahedron is 2a, where a = 2.(b) We discovered earlier that the dihedral angle of the regular tetrahedron is 70.53°. This means that the two dihedral angles are actually supplementary – they add to 180°. |

(a) Each diagonal is the diagonal of a square of side length s, and so has length 2s. Thinking of the octahedron as a double pyramid, its volume is now ‘1/3 area of base times height’, which gives ^{1}/_{3}.s^{2}. 2s = 2s^{3}/3.(b) The area of each face is 3 s^{2}/4. There are eight faces so the required surface area is 23s^{2}.(c) The circumradius is clearly half the length of a diagonal : 2s/ _{2}. Similarly, the midradius is the distance from the centre to an edge : /^{s}_{2}.Finding the inradius is a little harder. Let _{2}, and AB has length 3s/_{2}. We need to find the length r of the perpendicular from C to AB. Calculating twice the area of triangle ABC in two ways gives:
_{2} , whence r = s/6. |