Ratio of Volumes of Regular Dodecahedron and Regular Icosahedron in Same Sphere/Lemma

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Lemma to Ratio of Volumes of Regular Dodecahedron and Regular Icosahedron in Same Sphere

In the words of Hypsicles of Alexandria:

If two straight lines be cut in extreme and mean ratio, the segments of both are in one and the same ratio.

(The Elements: Book $\text{XIV}$: Proposition $8$ : Lemma)


Proof

Euclid-XIV-8-Lemma.png

Let $AB$ be cut at $C$ in extreme and mean ratio where $AC$ is the greater segment.

Let $DE$ be cut at $F$ in extreme and mean ratio where $DF$ is the greater segment.

It is to be demonstrated that:

$AB : AC = DE : DF$


Thus:

\(\ds AB \cdot BC\) \(=\) \(\ds AC^2\)
\(\ds DE \cdot EF\) \(=\) \(\ds DF^2\)
\(\ds \therefore \ \ \) \(\ds AB \cdot BC : AC^2\) \(=\) \(\ds DE \cdot EF : DF^2\)
\(\ds \therefore \ \ \) \(\ds 4 \cdot AB \cdot BC : AC^2\) \(=\) \(\ds 4 \cdot DE \cdot EF : DF^2\)
\(\ds \therefore \ \ \) \(\ds \left({4 \cdot AB \cdot BC + AC^2}\right) : AC^2\) \(=\) \(\ds \left({4 \cdot DE \cdot EF + DE^2}\right) : DF^2\) Proposition $18$ of Book $\text{V} $: Magnitudes Proportional Separated are Proportional Compounded
\(\ds \therefore \ \ \) \(\ds \left({AB + BC}\right)^2 : AC^2\) \(=\) \(\ds \left({DE + EF}\right)^2 : DF^2\) Proposition $5$ of Book $\text{II} $: Difference of Two Squares
\(\ds \therefore \ \ \) \(\ds \left({AB + BC}\right) : AC\) \(=\) \(\ds \left({DE + EF}\right) : DF\)
\(\ds \therefore \ \ \) \(\ds \left({AB + BC + AC}\right) : AC\) \(=\) \(\ds \left({DE + EF + DF}\right) : DF\) Proposition $18$ of Book $\text{V} $: Magnitudes Proportional Separated are Proportional Compounded
\(\ds \therefore \ \ \) \(\ds 2 \cdot AB : AC\) \(=\) \(\ds 2 \cdot DE : DF\)
\(\ds \therefore \ \ \) \(\ds AB : AC\) \(=\) \(\ds DE : DF\)

$\blacksquare$


Historical Note

This proof is Proposition $8$ of Book $\text{XIV}$ of Euclid's The Elements.


Sources