# Generalizing the apothem characterization of area for regular polygons, to convex regular polyhedra (and more?).

Recall: Given a convex, regular $$n$$-gon, of side length $$s$$, and apothem $$a$$, we may calculate its area $$A$$ as

$$A = \frac 1 2a n s$$

or, if we let $$P$$ be the perimeter (i.e. $$P=ns$$), then we have the formula we (or at least I) was typically given in school:

$$A = \frac 12 aP$$

To the uninitiated, the apothem is the line segment which is the radius of the inscribed circle of the regular polygon. It goes from the center of the polygon (in terms of this circle) to the midpoint of any side.

Question/Goal: Is there a way to generalize this formula to three dimensions, somehow relating the volume and surface area, to the length of some segment (from the center of some sphere to the middle of a face)? Up to some constant, of course. And presumably as a result, we can only consider convex, regular polyhedra and polytopes.

So, what I seek is some formula like

$$V = kaS$$

where $$S$$ denotes the surface area, and $$k$$ is some constant.

How would we prove this formula?

What about four dimensions and higher, if possible? Can it be done there as well?

Attempts & Research: (three dimensions)

So my first guess was to check the regular three-dimensional solids -- namely the convex ones, the Platonic solids.

Looking into things a bit more, the term I want (in 3D at least) to generalize the apothem is the "radius of the insphere" or "inradius" -- the insphere is a sphere whose faces are tangent to every face of the polyhedron. You can thus think of a 2D insphere giving you the length of the apothem for polygons (if we take the usual convention that a $$2$$-sphere is a circle).

Anyhow. Suppose we have a cube of edge length $$s$$. There should be no difficulty in seeing that the radius of the insphere is $$s/2$$, the volume is $$s^3$$, and the surface area is $$6s^2$$. Thus, we see that

$$V = kaS \implies s^3 = k \frac s 2 6s^2 \implies k = \frac 1 3$$

This hints at a possible formula for 3D, $$V = aS/3$$, but it needs to be checked.

Luckily, the platonic solids page on Wikipedia has a list of inradii among other things. (We'll still call it $$a$$ to match our convention.) For the tetrahedron, we have $$S = s^2 \sqrt 3$$, $$a = s/2\sqrt 6$$, and $$V = s^3/6\sqrt 2$$. Then we try our formula:

$$kaS = \frac{1}{3} \cdot \frac{s}{2\sqrt 6} \cdot s^2 \sqrt 3 = \frac{s^3}{6 \sqrt 2} =V$$

So we might be on the right track. We can check the others too for the necessary constant $$k$$. We use the Wikipedia notations of

$$\varphi := \frac{1+\sqrt{5}}{2} = 2 \cos \frac \pi 5 \qquad \xi := 2 \sin \frac \pi 5 = \sqrt{ \frac{5 - \sqrt 5}{2}}$$

as necessary.

• Octahedron: $$\displaystyle k = \frac{V}{aS} = \frac{s^3 \sqrt{2}}{3} \cdot \frac{\sqrt 6}{s} \cdot \frac{1}{2\sqrt 3 s^2} = \frac{\sqrt{12}}{6 \sqrt{3}} = \frac 1 3$$
• Dodecahedron: $$\displaystyle k = \frac{V}{aS} = \frac{5}{2} \frac{\varphi^3}{\xi^2} s^3 \cdot \frac{2 \xi}{\varphi^2 s} \cdot \frac{\sqrt{3-\varphi}}{15 \varphi s^2} = \frac{\sqrt{3-\varphi}}{3 \xi} = \frac 1 3$$ (final equality per WA)
• Icosahedron: $$\displaystyle k = \frac{V}{aS} = \frac{5 \varphi^2 s^3}{6} \cdot \frac{2 \sqrt 3}{\varphi^2 s} \cdot \frac{1}{5 \sqrt 3 s^2} = \frac 1 3$$

Thus, in each case of the platonic solids, the formula

$$V = \frac 1 3 a S$$

holds up.

Attempts & Research: (four dimensions)

So far, we've seen two formulas:

$$A = \frac 1 2 a P \qquad V = \frac 1 3 a S$$

I conjecture, then: the $$n$$-dimensional volume $$V_n$$ is related to the $$n$$-dimensional surface area $$S_n$$ and the polytope's inradius $$a$$, by the formula

$$V_n = \frac 1 n a S_n$$

So it's natural to look at the four dimensional case as well. Indeed, we see that it works, as follows, going through the list of regular polytopes on Wikipedia:

• $$5$$-cell (post about volume): $$\displaystyle V_4 = \frac{1}{4} \cdot 1 \cdot V_3$$
• $$8$$-cell (tesseract, some info from here): $$\displaystyle V_4 = \frac 1 4 \cdot \frac{1}{2} s \cdot 8s^4 = s^4$$, as desired
• $$16$$-cell (some info from here): $$\displaystyle V_4 = \frac1 4 \cdot \frac{\sqrt{2}}{4} s \cdot 16 \frac{\sqrt{2}}{12} s^3 = \frac{1}{6} s^4$$, as desired
• $$24$$-cell (some info from here): $$\displaystyle V_4 = \frac 1 4 \cdot \frac{1}{\sqrt 2} s \cdot 24 \frac{\sqrt 2}{3} s^3 = 2s^4$$, as desired
• $$120$$-cell (some info from here): $$\displaystyle V_4 = \frac 1 4 \cdot \frac{7 + 3 \sqrt 5}{4} s \cdot 120 \frac{15 + 7 \sqrt 5}{4} s^3 = \frac{15}{4} (105 + 47 \sqrt 5)s^4$$, as desired (WA verified)
• $$600$$-cell (some info from here): $$\displaystyle V_4 = \frac 1 4 \cdot \sqrt{ \frac{9 + 4 \sqrt 5}{8} } s \cdot 600 \frac{\sqrt 2}{12} s^3 = \frac{25}{4} (2 + \sqrt 5) s^4$$ (WA verified)

Thus the conjecture holds for the regular four-dimensional convex polytopes.

Question (albeit reiterated somewhat):

The question remains, though: does this hold true in all dimensions? That is, for a convex, regular, $$n$$-dimensional polytope, with

• $$n$$-dimensional hypervolume $$V_n$$
• inradius $$a$$
• $$n$$-dimensional hyper-surface-area $$V_{n-1}$$ (or "surface volume" I guess? Not sure what the term is)

do we have

$$V_n = \frac 1 n a V_{n-1}?$$

If so, how do we prove this? Moreover, what (if any) is the underlying reason as to why this holds? (As in, how should one be able to "expect" this answer, from a motivational or intuitive aspect, if that makes any sense.)

This relation between volume and surface area holds true for all circumscribed polytopes, that is, polytopes for which there exists a sphere (the in-sphere) that touches every facet (e.g. the regular polytopes). If the polytope $$P\subset\Bbb R^n$$ is $$n$$-dimensional, its in-sphere has radius $$r$$, the volume is $$V_n$$ and the surface area is $$V_{n-1}$$, then indeed

$$V_n = \frac1n r V_{n-1}.$$

Proof.

Suppose that $$P$$ has facets $$f_1,...,f_m$$ and $$f_i$$ has $$(n-1)$$-dimensional volume $$v_i^{n-1}$$. Let $$p_i$$ be the pyramid with base face $$f_i$$ and apex at the center of the in-sphere. Since $$P$$ is circumscribed, each pyramid has then height $$r$$. The $$n$$-dimensional volume of this pyramid is known to be

$$v_i^n = \frac 1n r v_i^{n-1}.$$ The pyramids define a decomposition of the volume of $$P$$, as well as of its surface, and thus the volume $$V_n$$ is the sum of the $$v_i^n$$, and the surface area $$V_{n-1}$$ is the sum of the $$v_i^{n-1}$$. The desired relation follows.