Commentary on Proposition XII.18¶
Back to Proposition XII.18.
It is the method of this proposition which Legendre adopted for his proof of XII. 2 (see note on that proposition).
The argument can be put very shortly. We will suppose $S, S'$ to be the volumes of the spheres, and $d, d'$ to be their diameters; and we will for brevity express the triplicate ratio of $d$ to $d'$ by $d^3 : d'^3$.
If
$$d^3 : d'^3 \ne S : S',$$
then
$$d^3 : d'^3 = S : T,$$
where $T$ is the volume of some sphere either greater or less than $S'$.
Suppose, if possible, that $T < S'$.
Let $T$ be supposed concentric with $S'$.
As in XII. 17, inscribe a polyhedron in $S'$ such that its faces do not anywhere touch $T$; and inscribe in $S$ a polyhedron similar to that in $S'$.
Then $S : T = d^3 : d'^3$ $= (\text{polyhedron in } S) : (\text{polyhedron in } S');$ or, alternately, $S : (\text{polyhedron in } S) = T : (\text{polyhedron in } S').$
And $S > (\text{polyhedron in } S);$ therefore $T > (\text{polyhedron in } S').$
But, by construction, $T < (\text{polyhedron in } S'):$ which is impossible.
Therefore $T \not< S'$.
Suppose, if possible, that $T > S'$.
Now $d^3 : d'^3 = S : T$ $= X : S'$, where $X$ is the volume of some sphere less than $S$, [XII. 2, Lemma] or, inversely, $d'^3 : d^3 = S' : X$, where $X < S$.
This is proved impossible exactly as in Part I.
Therefore $T \not> S'$.
Hence $T$, not being greater or less than $S'$, is equal to it, and $d^3 : d'^3 = S : S'$.