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Volume of dodecahedron & icosahedron
By John Hewson on November 28, 1997:

What is the volume of a regular dodecahedron?

What is the volume of an icosahedron?

By Anonymous on Tuesday, September 12, 2000 - 02:57 pm:

Dear John,

Thanks for asking a question -- it's the first one we've had. However, you chose a tricky one!! I won't say that I don't know the answer (although I don't), but I can make some first guesses and go away and think about it (and ask some other people) -- I just wanted to reply reasonably quickly.

So, first of all, I think the best idea is to split the problem up. Consider the icosahedron first. It consists of 20 faces which are all equilateral triangles. Join the vertices of each triangle to the centre of the body. Then you have 20 congruent tetrahedra. It's not immediately obvious to me whether these are regular tetrahedra or not -- probably not in fact. The volume V of the icosahedron is then 20V1, where V1 is the volume of each tetrahedron.

Now, as far as I remember, the volume of ANY pyramid is 1/3 x (height) x (area of base). So we need to know the height and the area of the base.

Similarly for the dodecahedron -- V' = 12V2, where V2 is the volume of the pentagonal pyramid formed by joining the vertices to the centre. Again, we need V2 = 1/3 x (height) x (area of base).

So now the only problem is calculating the volume of these pyramids.

It seems to me that there is a question of how you specify what size the solids are. Do you give the length of each edge? Or the radius of an inscribed or circumscribed sphere? Probably one of these will be much easier to use, but I can't tell which one.

One final point : once you have the answer, clearly it has to be reasonably similar to that for a sphere!

You'll notice that I haven't actually solved the problem! But I'll go away and think about it and get back to you.

I hope this has helped, but maybe it was all obvious. If so, I'm sorry.

It's very hard from this end knowing what level to aim things at. Feel free to raise any points at all.

Yours,

David.

By Anonymous on Tuesday, September 12, 2000 - 03:02 pm:

Dear John,
Sorry it's taken me so long to get back to you on the icosahedron question. If it's any consolation, I've got a reasonable amount to tell you about it now, so here goes.

Firstly let me point you to the website http://www.astro.virginia.edu/~eww6n/math/math.html. It's an online dictionary of maths, with lots of interesting things in, including quite long, detailed articles on icosahedra and dodecahedra. For example, there is a (long and not particularly enlightening, perhaps) derivation of the volume. But it's quite interesting just to browse. Since it's in the States, it's probably better to try it in the morning. There's also an icosahedron site at http://neon.ci.lexington.ma.us/~jrosen/icosahedron/gallery.html. I found it using one of the search engines by typing in icosahedron - there are lots more where that came from!

[Unfortunately the two links referred to here are currently unavailable. See the end of this discussion for some others. - The Editor]

One of my friends came up with the following clever method for the icosahedron (I think she'd seen it before). The basic problem, as we have already decided, is finding the height of the tetrahedra. The tetrahedra are definitely not regular. (If you solved the problem of finding the distance between the opposite faces of a dodecahedron, then you have effectively done this already, since the height of the pyramids is just half of this length! If your method of doing this is applicable to the icosahedron, you've done that too!)

So, how do you find the required height? Her method is as follows. Put the icosahedron, say its side length is 1, on one of its vertices, with the opposite vertex directly above it. (This is all much easier to follow with a diagram. However, it's quite hard to draw an icosahedron. There is a diagram of one in the first website above.) Cut vertically down through both of the vertices. The plane of the cut should pass through one of the vertices adjacent to the top one. The intersection of the plane with the icosahedron is a hexagon. This hexagon has 4 sides of length a, and 2 of length b. These 2 are opposite each other. Let Q be one of the vertices between 2 of the same length sides. Let P be the point half way along from Q to one of its neighbouring vertices on the hexagon. (Draw all this!!) Then the triangle OPQ, where O is the centre of the icosahedron, is a right angle, since P is on the centre of one of the triangular faces and Q is half-way along one of the sides of the icosahedron. So now you know the distance OQ, call it s.

Now the icosahedron consists of 2 pentagonal pyramids opposite each other, with a ten-sided ring of triangles in between. Cut straight through the middle of this ten-sided ring, giving a regular decagon, with side length half (by similar triangles) and distance from centre to each vertex given by s. From this, using cos(2×p/5) = (sqrt(5) -1) /4, it follows that the height of one of the tetrahedra is (3+sqrt(5)) / (4.sqrt(3)). Hence the volume is 5/12 × (3+sqrt(5)). Multiply this by a3 if the side length is a.

So there you are. Another way, that another of my friends suggested, is to project from the centre of the icosahedron onto a circumscribing sphere. This gives triangles on the sphere which tesselate it. From this you can find the angle between e.g. the centre of one of the faces and a vertex, and from there, using trigonometry, the height etc.

One of the things I read made the interesting point that if you inscribe a dodecahedron and an icosahedron into the same sphere, the dodecahedron will fill MORE space in the sphere! This is because it's more nearly spherical.

I hope you've been able to follow all that. If anything isn't clear, please ask! I might think of a sensible way of adding the diagrams to this somewhat difficult-to-follow description eventually.

Yours, David.

By Gareth McCaughan on December 12, 1997:

I suspect there's a frighteningly elegant way to do this, but if so I don't know it. Here's a rather nasty way to do it for the dodecahedron.

You can make a dodecahedron by taking a cube and putting a cap on each face. (I suggest getting pencil and paper if you want to follow this.) Each cap is like a house roof: if we pretend that the face in question is the square with corners at (0,0,0), (1,0,0), (0,1,0), (1,1,0) then the cap
has two other vertices, at (t,1/2,h) and (1-t,1/2,h). There are two ways to fit a cap on a face of the cube, because there are two possible orientations for the "ridge", differing by a 90-degree rotation. We fit the caps so that a triangular face of one cap meets a quadrilateral face of the adjacent cap. (I meant it about the pencil and paper.) And if we choose t and h correctly, then those two faces will merge nicely to produce a regular pentagon: one face of the dodecahedron.


So, what values of t and h do we want? The "corner to ridge" length and the "along the ridge" length are both supposed to be edges of a regular pentagon, so they must be equal. That is,
sqrt(t2 + ½2 + h2) = 1-2t,

or in other words
3t2 - 4t - h2 + 3/4 = 0. . . . . . . . (1)


And those two faces have to merge properly, so their slopes have to be related correctly. A bit of scribbling indicates that what we want is for the product of the slopes to equal 1:
h/(1/2) . h/t = 1
or in other words
t = 2h{2} . . . . . . . . . . . . . . . . . . (2)


If we substitute (2) into (1) we get a quadratic in t; one solution is bigger than 1 (and therefore impossible) and the other is [3-sqrt(5)]/4. It's then not hard to find that h = [sqrt(5)-1]/4.


Now, what's the volume of one cap? Slice it into three bits by the planes x=t and x=-t. The bit in the middle clearly has volume 1/2 . (1-2t) . h, and each of the other bits is a pyramid (albeit a rather odd one) with base area t and height h, so with volume th/6. So the whole cap has volume
[(1-2t)h]/2 + 2.th/6 = (3h-6th+2th)/6; so the whole dodecahedron (consisting of the cube and 6 caps) has volume 1+3h-4th, which turns out to equal (9-sqrt(5))/4.

That's in units where one side of the cube is 1 unit long.

We probably actually want units where an edge of the dodecahedron is 1 unit long. Well, an edge of the dodecahedron is (1-2t) units long; that is, [sqrt(5)-1]/2 units long. So we need to divide our answer by the cube of [sqrt(5)-1]/2, and it turns out that we then get [13+7sqrt(5)]/4.

I've probably made at least one mistake in the foregoing, so you'd be well advised to check it if you actually need the answer for anything.

By The Editor:

Some links to look at:

Volume of dodecahedron with diagrams

Nice method of constructing the platonic solids giving a neat method for the volumes of both an icosahedron and a dodecahedron

Table of radius and volume for all the platonic solids