Ratio of Lengths of Sides of Cube and Regular Icosahedron in Same Sphere

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Theorem

In the words of Hypsicles of Alexandria:

If any straight line whatever be cut in extreme and mean ratio, then, as is $(1)$ the straight line the square on which is equal to the sum of the squares on the whole line and on the greater segment to $(2)$ the straight line the square on which is equal to the sum of the squares on the whole and on the lesser segment, so is $(3)$ the side of the cube to $(4)$ the side of the icosahedron.

(The Elements: Book $\text{XIV}$: Proposition $7$)


Proof

Euclid-XIV-7.png

Let a regular dodecahedron, a regular icosahedron and a cube be inscribed in a given sphere.

From Proposition $3$ of Book $\text{XIV} $: Circle Circumscribing Pentagon of Dodecahedron and Triangle of Icosahedron in Same Sphere:

the circle which circumscribes the regular pentagon which is the face of the regular dodecahedron is the same size as the circle which circumscribes the equilateral triangle which is the face of the regular icosahedron.

Let $AHB$ be the circle which circumscribes both that regular pentagon and that equilateral triangle.

Let $C$ be the center of the circumscribing circle.

Let $CB$ be a radius of $AHB$.

Let $CB$ be divided at $D$ in extreme and mean ratio where $CD$ is the greater segment.

Then from:

Proposition $9$ of Book $\text{XIII} $: Sides Appended of Hexagon and Decagon inscribed in same Circle are cut in Extreme and Mean Ratio

and the converse of:

Proposition $5$ of Book $\text{XIII} $: Straight Line cut in Extreme and Mean Ratio plus its Greater Segment

it follows that:

$CD$ is the side of a regular decagon which has been inscribed in $AHB$.

Let $E$ be the side of the equilateral triangle that is the face of the regular icosahedron.

Let $F$ be the side of the regular pentagon which is the face of the regular dodecahedron.

Let $G$ be the side of the square which is the face of the cube.

From Porism to Proposition $17$ of Book $\text{XIII} $: Construction of Regular Dodecahedron within Given Sphere:

if $G$ is divided in extreme and mean ratio, the greater segment is equal to $F$.


Thus:

\(\ds E^2\) \(=\) \(\ds 3 \cdot BC^2\) Proposition $12$ of Book $\text{XIII} $: Square on Side of Equilateral Triangle inscribed in Circle is Triple Square on Radius of Circle
\(\ds CB^2 + BD^2\) \(=\) \(\ds 3 \cdot CD^2\) Proposition $4$ of Book $\text{XIII} $: Area of Squares on Whole and Lesser Segment of Straight Line cut in Extreme and Mean Ratio
\(\ds \therefore \ \ \) \(\ds E^2 : CB^2\) \(=\) \(\ds \left({CB^2 + BD^2}\right) : CD^2\)
\(\ds \therefore \ \ \) \(\ds E^2 : \left({CB^2 + BD^2}\right)\) \(=\) \(\ds CB^2 : CD^2\)
\(\ds \) \(=\) \(\ds G^2 : F^2\)
\(\ds \therefore \ \ \) \(\ds G^2 : E^2\) \(=\) \(\ds F^2 : \left({CB^2 + BD^2}\right)\)
\(\ds \therefore \ \ \) \(\ds G^2 : E^2\) \(=\) \(\ds \left({BC^2 + CD^2}\right) : \left({CB^2 + BD^2}\right)\) Proposition $10$ of Book $\text{XIII} $: Square on Side of Regular Pentagon inscribed in Circle equals Squares on Sides of Hexagon and Decagon inscribed in same Circle

Hence the result.

$\blacksquare$


Historical Note

This proof is Proposition $7$ of Book $\text{XIV}$ of Euclid's The Elements.


Sources