If you inscribe a regular polygon into a given circle, the larger the number of sides, the larger the area of the polygon. I guess I always thought that the same would apply to Platonic solids inscribed in a sphere..... It doesn't. I noticed this as I was looking though "The Penny cyclopædia of the Society for the Diffusion of Useful Knowledge

By Society for the Diffusion of Useful Knowledge" (1841). As I browsed the book, I came across the table below:

The table gives features of the Platonic solids when inscribed in a one-unit sphere. At first I thought they must have made a mistake, but not so. The Dodecahedron fills almost 10% more of a sphere (about 66%) than the icosahedron (about 60%). So the Dodecahedron is closer to the sphere than the others.

Interestingly, if you look at the radii of the inscribing spheres, it is clear that solids which are duals are tangent to the same internal sphere.

But if you look at the table of volumes when the solids are inscribed with a sphere inside tangent to each face:

When you put the Platonic solids around a sphere, the one smallest, and thus closest to the sphere is the icosahedron.

This leads to the paradox that when platonic solids are inscribed with a sphere, the icosahedron is closest to the circle in volume (thus most spherical?) but when they are circumscribed by a sphere, the dodecahedron is the closest to the volume of a sphere (and thus most spherical?)... hmmm

Here is a table of the same values when the surface area (superficies) is one square unit.Notice that for a given surface area, the icosahedron has the largest volume, so it is the most efficient "packaging" of the solids (thus more spherical?).

I guess that makes it 2-1 for the icosahedron, so I wasn't completely wrong all along.

POSTSCRIPT::: Allen Knutson's comments on the likely cause of this reversal of "closeness" to the sphere:

- I think it's about points of contact. On the inside, the dodecahedron touches the sphere at the most points (20), and on the outside, the icosahedron touches the sphere at the most points (again 20).
- Indeed: my recipe would suggest that inside, the 8-vertex cube is bigger than the 6-vertex octahedron, and outside, the 8-face octahedron is smaller than the 6-face cube. Both are borne out by your tables. Thank you, Allen
- Some other pertinent comments too good to ignore:
I think "most-spherical" would need to be better defined.

Not that I want to do it (and not that I am certain that I could), but perhaps summing the squares of the distances from each point on the surface of the solid to the closest point on the sphere would be the way to go (like a sort of physical variance).

But as I say, I'm not sure I could do this (maybe I can kill some time today, I'm visiting a puzzle-friend), and I'm not sure that it makes a lot of sense.

Oh, oh, and inscribe or circumscribe or find the best match in-between?

(it's play with this, read on-line, or grade. Choices.)

Jonathan(Love When anonymous signs his comment)

What a fascinating post! Here is a nice way to think about your first result in terms of the empty space.

The icosahedron leaves ~20% of the space in its circumscribing sphere empty. The dodecahedron leaves ~12% of the space in its circumscribing sphere empty.

That means that if you start with a solid sphere (let's imagine it's something easy to carve, like soap!) and carefully cut away the 20 portions needed to turn it into an icosahedron, each of the 20 pieces you trimmed off will have volume of about 1% of the sphere.

Similarly, if you do the same thing to carve a dodecahedron out of a sphere, each of the 12 pieces you trimmed off will also have volume of about 1% of the sphere.

From now on, whenever I look at either polyhedron (icosa or dodeca) I will always think of it a little differently--because I will think of those ~1% extensions on each face needed to round it out to a sphere.I really liked this idea, and want to take time to find the amount of volume reduced by cutting each face into a sphere.

It seems that the cube is third best, just ahead of the octahedron. The cube volume in a unit sphere is 1.5396.. and the volume of a unit sphere is 4/3 Pi, or 4.1888, cutting off 2.64919 cubic units (more than half the sphere, about 63%) . That means to sculpt a cube

from a unit sphere we cut off an average of just over 10% for each face. (I did that quickly,

do check)

It might be a wonderful challenge to imagine a slicing approach to take out just the right fraction of the volume

to be removed to cut the same amount when revealing each face. I think I see some of them, but some are.... "difficult"?

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- Later I tried working on the Archimedian Solids.
### Measuring Sphereocity of The Archimedian Solids

mathworld.wolfram.com

A few years ago I wrote a blog about which of the Platonic solids (above) was most spherical. I compared which ones had the most volume inscribed into a unit sphere, and the which had the smallest volume when circumscribed about a unit sphere. Surprisingly I got different answers to the two methods, and lots of good mathematical comments about why this might be so. Then a while ago (July 2015) I posted it again. As part of a tongue-in-cheek exchange with Adam Spencer @adambspencer I challenged him to find the roundest of the thirteen Archimedean Solids. For those who are not familiar with the distinction, both the Platonic and Archimedean solids are made of up faces that are regular polygons, and both have the property that the view at each vertex is identical to every other, but where the Platonic solids consist of only a single type of regular polygon, (for example the tetrahedron is made up of four equilateral triangles), the Archimedean solids may have more than one (in the cubeoctahedron each vertex is surrounded by two squares and two equilateral triangles).

Then only a day or so later, I got to wondering about the actual answer and started doing some research. Along the way I found a couple of papers on the topic, one of which was published earlier in the same month I began my search for the Platonic Solids Sphere-ness. What was great was that they re-exposed me to a formula for comparing according to George Polya from his Mathematics and Plausible Reasoning: Patterns of plausible inference. I sheepishly admit that I had read this book ( it's on my bookcase nos) several times years ago, but somehow this didn't pop up when I was thinking of the Platonic solids.

The problem of comparing the ratio of the numerical values of the surface area to the volume is that the answer changes over size. In a sphere, for instance, if the radius is one, then the volume is 4 pi/3, and the surface area is 4pi, so the V= 1/3 SA. Now increase the radius to three units and the volume is 36 pi, and the surface area is also 36 pi, Now V = SA, and if we keep making the radius bigger, the volume becomes larger than the surface area, in fact the ratio of Volume to surface area can be reduced to \( \frac{V}{SA} = \frac{r}{3} \). Now that kind of thing happens with all the solids, the larger the volume gets, the larger the ratio of V to SA gets. It is one of those things that amazes students (and well it should) that for any solid, there is some scalar multiplication which will transform it into a solid with Volume = Surface Area. In this sense, every solid is isoperimetric (same measure)

So Polya found a way to neutralize this growth. He created an "Isoperimetric Quotient" that served to null out this scalar alteration. By setting the IQ = \( \frac{36 \pi V^2}{S^3} \) With this weapon he was able to compare, for instance, the "roundness" of the Platonic Solids. Try this with any sphere and you always get one. Try it with anything else, you always get less than one.

The IQ of the Platonic Solids follows the number of faces, with the tetrahedron at the bottom with an IQ of about .3 and the icosahedron at the top with an IQ of about .8288.

SO what about the Archimedean Solids? Well here they are

TruncatedTetrahedron...... 0.4534

TruncatedOctahedron ....... 0.749

TruncatedCube ............. 0.6056

TruncatedDodecahedron ..... 0.7893

TruncatedIcosahedron ...... 0.9027

Cuboctahedron ............. 0.7412

Icosidodecahedron ......... 0.8601

SnubCube .................. 0.8955**SnubDodecahedron .......... 0.94066**

Rhombicuboctahedron ....... 0.8669

TruncatedCuboctahedron .... 0.8186

Rhombicosidodecahedron .... 0.9357

TruncatedIcosidodecahedron. 0.9053

So the roundness winner is the snubdodecahedron, with a pentagon and four equilateral triangles around each vertex. That is four 60^{o}angles and one of 108^{o}for a total of 348^{o}. Students might check if any other of the Archimedean solids can top that. Is that "flatness" at vertices somehow related to "roundness"? I have to admit I first thought it might be the truncated icosahedron. Students may have heard of this one more than others. A molecule of C-60, or a “Buckyball”, consists of 60 carbon atoms arranged at the vertices of a truncated icosahedron. It's roundness is a feature of many of its applications, but it only comes in fourth.

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