TITLE: Does an Icosidodecahedron have an equation?
QUESTION [5 upvotes]: 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.

REPLY [5 votes]: 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.