# Does an Icosidodecahedron have an equation? It appears that using the absolute value function this is possible. Let $q = 1$ and $p = \frac{1 + \sqrt 5}{2}$ then , $$\left|\frac{z}{q} + \frac{y}{p} \right| + \left|\frac{z}{q} - \frac{y}{p} \right| + \left|\frac{x}{p} + \frac{y}{q} \right| + \left|\frac{x}{p} - \frac{y}{q} \right| + \left|\frac{z}{p} + \frac{x}{q} \right| + \left|\frac{z}{p} - \frac{x}{q} \right|= 64$$ describes an Icosidodecahedron.

The Circumsphere has radius $16(\sqrt 5 -1)$ . I was very surprised to find this! The general question is, what are equations for some familiar polyhedra? ( I'd include Platonic, Archimedean , and Catalan Solids since examples of each class have come up, along with many weird looking blobs! )

It appears these polyhedra are duals of Zonohedra. Quite a large collection, although as has been pointed out, the generic situation is fairly straightforward.

• A very beautiful sculpture! Thanks for the reference.
– Alan
Mar 14 '14 at 19:41
• Appeal may be due to GoldenRatio..formation is composed entirely of great circles, each divided into 10 parts, another with 60 degree rotation around a polar axis,the pair drawn for all vertices. Oct 3 '14 at 10:44

A convex polyhedron is the intersection of a finite number of half spaces, and a half space is described by a linear equation of the form

$$ax + by + cz + d \ge 0$$

Using the absolute value function, you can turn this into

$$(ax + by + cz + d) - \lvert ax + by + cz + d\rvert = 0$$

You can combine multiple inequalities by using the fact that a sum of squares of real numbers will only be zero if every input number is zero. So you'd get

$$\sum_i \bigl((a_ix+b_iy+c_iz+d_i)-\lvert a_ix+b_iy+c_iz+d_i\rvert\bigr)^2 = 0$$

So you now have a generic recipe to turn any convex polyhedron, described as a set of inequalities, into a single equation using the absolute value function. Not an elegant equation, to be sure, but very generic.