Volumes and areas of inspheres and incircles

By Stuart Ocock :

Why is it that those plane figures which have 'area = perimeter' have an incircle of radius 2, and why is it that in 3D the objects with 'volume = surface area' have an insphere of radius 3? Is the fact that the radius = dimension significant? Must the objects be convex?

By sk269 on July 29, 1998 :

It can be proved that, if an 'insphere' of a polyhedron exists which touches ALL the faces of the polyhedron, then the polyhedron must be convex. The corresponding results can be proved in the plane and in

n

dimensions.

Suppose $P$ is a plane polygon (not necessarily regular) with an incircle (assuming this means a circle that touches ALL sides of the polygon). Now suppose that $P$ satisfies ''area=perimeter''. Take a typical side of $P$ of length $s$ and let the incircle have radius $r$ .

diagram

For the single triangle as illustrated we have

area $= 1 / 2 s r$ , length contributing to perimeter $= s$ .

Adding over all triangles, we get

$\sum_{s} \frac{1}{2} s r = \sum_{s} s$ so $r = 2$ .

The radius of the incircle is indeed the dimension of the space.

In 3 dimensions, for example, we 'replace' the formula $1 / 2 s r$ by $1 / 3 r (area of face)$ [consider the formula for the volume of a pyramid] so the 'proof' becomes

$\sum_{A} 1 / 3 r A = \sum_{A} A$ (where $A$ is the (variable) area of a typical face) hence $r = 3$ .

You need $n$ -dimensional geometry to prove in general but it is true that in all dimensions the radius of the incircle is the dimension of the space.

The relationship between the radius of the insphere and the dimension applies more generally and not just to objects for which 'area = perimeter' or 'volume = surface area' etc. Stated more generally for an $n$ dimensional 'polyhedron' this is:

Radius of $n$ dimensional 'insphere' $= n \times$ 'volume' of polyhedron/ 'surface area' of polyhedron