## Monday, 3 August 2015

### Measuring Sphereocity of The Archimedian Solids mathworld.wolfram.com

A few years ago I wrote a blog about which of the Platonic solids 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 short 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 Archimedian 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 60o angles and one of 108o for a total of 348o.  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.