Well, this is quite enjoyable. In $\mathbf R^3,$ as long as we are taking polyhedra or other figures that are strictly similar, with some parameter $w$ to which any length measurement of the figure keeps a constant ratio, it follows that the volume is some $V_1 w^3$ and the surface area is some $ S_1 w^2.$ Furthermore, some choice of length scale or units causes $3 V_1 = S_1$ and we get a pure derivative law such as the earlier examples I gave in comments. For several very symmetric figures, the "inradius" or radius of an inscribed sphere gives such a fortunate relationship.

I thought it might depend critically on symmetry, but perhaps not as much as I thought. For the square pyramid with the square base on the $xy$-plane and one of the triangular faces within the plane
$x + y + z = (1 + \sqrt 3 ) w,$ we get the center of the inscribed sphere at
$(0,0,w), $ the point of tangency on that triangle
$w(\frac{1}{\sqrt 3}, \; \frac{1}{\sqrt 3}, \; 1 + \frac{1}{\sqrt 3}), $
the length of any edge
$ w (1 + \sqrt 3 ) \sqrt 2,$ the volume $ V = \frac{2}{3} (1 + \sqrt 3 )^3 w^3, $ finally the surface area $S = 2 (1 + \sqrt 3 )^3 w^2.$ Notice that the inscribed sphere does touch the square at its center but the triangles elsewhere.

I will fiddle with this some more. As I mentioned, there is either a truncated cube or a truncated octahedron which possesses an inscribed sphere that is tangent to all the faces, the amount of truncation must be just right. Interesting to check.

EDIT: it always works, symmetry is not important, and it is utterly trivial but cute. Take an irregular polyhedron that is convex, and is such that every face is tangent to the same sphere of radius $r.$ For each face, find the similar face that would be tangent to a sphere of radius 1, call that of area $A_j,$ so that the actual area of the original face $j$ is $A_j r^2.$ Because we use the same $r$ for all faces, the surface area is
$$ S = \sum A_j r^2 = r^2 \sum A_j.$$ But the volume of a cone or pyramid is one third of the altitude times the area of the base, and the altitude is $r$ because of the tangency condition. So $$ V = \frac{1}{3} r^3 \sum A_j $$ and the derivative relationship holds.

Same for irregular polygons with all edges tangent to the same circle.

Who knew?

EDIT 2: To turn it around, the full set of polygons or polyhedra for which this recipe works comes from taking a finite set of distinct points on a circle or sphere, a sufficient condition for a closed bounded figure being that the set of points not be contained in any closed semicircle/hemisphere. For each point, take the tangent plane, and indeed the half space on the same side of that tangent as the sphere. The intersection of these half-spaces is the convex polygon/polyhedron. In the polyhedron case, it will not be obvious ahead of time how many sides each polygonal face will possess. Next, the example I called Archimedes' favorite cylinder can be thought of as a limit of regular $n$-sided prisms as $n$ increases.

Platonic and Archimedean Solidsby Daud Sutton. Willie's co-area formula trumps all other descriptions. But I am still curious about my inradius formulation for the rhombic dodecahedron, various polyhedra where the inradius is constant and each "radius" meets a regular polgon at its center. $\endgroup$ – Will Jagy Oct 16 '10 at 19:554more comments