Construction of Cube within Given Sphere
Theorem
In the words of Euclid:
- To construct a cube and comprehend it in a sphere, like the pyramid; and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube.
(The Elements: Book $\text{XIII}$: Proposition $15$)
Proof
Let $AB$ be the diameter of the given sphere.
Let $AB$ be cut at $C$ where $AC = 2 \cdot CB$.
Let $ADB$ be a semicircle on the diameter $AB$.
Let $CD$ be drawn from $C$ perpendicular to $AB$.
Let $DB$ be joined.
Let the square $EFGH$ be set out whose sides are all equal to $DB$.
- let $EK, FL, GM, HN$ be drawn from $E, F, G, H$ respectively perpendicular to the plane of the square $EFGH$.
Let $EK, FL, GM, HN$ be made equal to one of the straight lines $EF, FG, GH, HE$.
Let $KL, LM, MN, NK$ be joined.
The cube $FN$ can be seen to have been constructed which is contained by six squares.
$\Box$
Next it is to be demonstrated that the cube $FN$ can be inscribed in the given sphere.
Let $KG, EG$ be joined.
We have that $KE$ is perpendicular to the plane of the square $EFGH$.
So from Book $\text{XI}$ Definition $3$: Line at Right Angles to Plane:
- $KE$ is perpendicular to the straight line $EG$.
Therefore $\angle KEG$ is a right angle.
Therefore also if $FK$ is joined, then $GF$ is perpendicular to $FK$.
For this reason, the semicircle described on $GK$ will also pass through $F$.
Similarly it will also pass through the remaining vertices of the cube.
Let $GK$ remain fixed, and let that semicircle be carried around and restored to the same position from which it began.
Thus the cube $FN$ has been inscribed in a sphere.
We have that:
- $GF = FE$
and that $\angle F$ is a right angle.
Therefore from Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem:
- $EG^2 = 2 EF^2$
But:
- $EF = EK$
Therefore:
- $EG^2 = 2 \cdot EK^2$
Therefore from Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem:
- $GK^2 = EG^2 + EK^2 = 3 EK^2$
We have that:
- $AB = 3 BC$
while:
- $AB : BC = AB^2 : BD^2$
Therefore:
- $AB^2 = 3 \cdot BD^2$
But it has been proved that:
- $GK^2 = 3 EK^2$
while:
- $KE = DB$
Therefore:
- $KG = AB$
But $AB$ is the diameter of the given sphere.
Therefore:
and:
$\blacksquare$
Historical Note
This proof is Proposition $15$ of Book $\text{XIII}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XIII}$. Propositions