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

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## Contents

## 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

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:

\(\displaystyle AB \cdot BC\) | \(=\) | \(\displaystyle AC^2\) | |||||||||||

\(\displaystyle DE \cdot EF\) | \(=\) | \(\displaystyle DF^2\) | |||||||||||

\(\displaystyle \therefore \ \ \) | \(\displaystyle AB \cdot BC : AC^2\) | \(=\) | \(\displaystyle DE \cdot EF : DF^2\) | ||||||||||

\(\displaystyle \therefore \ \ \) | \(\displaystyle 4 \cdot AB \cdot BC : AC^2\) | \(=\) | \(\displaystyle 4 \cdot DE \cdot EF : DF^2\) | ||||||||||

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

\(\displaystyle \therefore \ \ \) | \(\displaystyle \left({AB + BC}\right)^2 : AC^2\) | \(=\) | \(\displaystyle \left({DE + EF}\right)^2 : DF^2\) | Proposition $5$ of Book $\text{II} $: Difference of Two Squares | |||||||||

\(\displaystyle \therefore \ \ \) | \(\displaystyle \left({AB + BC}\right) : AC\) | \(=\) | \(\displaystyle \left({DE + EF}\right) : DF\) | ||||||||||

\(\displaystyle \therefore \ \ \) | \(\displaystyle \left({AB + BC + AC}\right) : AC\) | \(=\) | \(\displaystyle \left({DE + EF + DF}\right) : DF\) | Proposition $18$ of Book $\text{V} $: Magnitudes Proportional Separated are Proportional Compounded | |||||||||

\(\displaystyle \therefore \ \ \) | \(\displaystyle 2 \cdot AB : AC\) | \(=\) | \(\displaystyle 2 \cdot DE : DF\) | ||||||||||

\(\displaystyle \therefore \ \ \) | \(\displaystyle AB : AC\) | \(=\) | \(\displaystyle DE : DF\) |

$\blacksquare$

## Historical Note

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

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): The So-Called Book $\text{XIV}$, by Hypsicles