Construction of Regular Octahedron within Given Sphere

Theorem

In the words of Euclid:

To construct an octahedron and comprehend it in a sphere, as in the preceding case; and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron.

(The Elements: Book $\text{XIII}$: Proposition $14$)

Proof

Let $AB$ be the diameter of the given sphere.

Let $AB$ be bisected at $C$.

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 $HF$ and $EG$ be joined.

Let $K$ be the point where $HF$ and $EG$ meet.

From Proposition $12$ of Book $\text{XI} $: Construction of Straight Line Perpendicular to Plane from point on Plane:

let $KL$ be drawn from $K$ perpendicular to the plane of the square $EFGH$.

Let $KM$ be produced through to the other side of the plane of the square $EFGH$.

Let $KL$ and $KM$ be equal to any one of the straight lines $EK, FK, GK, HK$.

Let $LE, LF, LG, LH, ME, MF, MG, MH$ be joined.

We have that:

$KE = KH$

and:

$\angle EKH$ is a right angle.

Therefore from Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem:

$HE^2 = 2 \cdot EK^2$

We have that:

$LK = KE$

and:

$\angle LKE$ is a right angle.

Therefore from Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem:

$EL^2 = 2 \cdot EK^2$

But it has already been proved that:

$HE^2 = 2 \cdot EK^2$

Therefore:

$LE^2 = EH^2$

and so:

$LE = EH$

For the same reason:

$LH = HE$

Therefore $\triangle LEH$ is an equilateral triangle.

Similarly we can prove that each of the remaining triangles of which the sides of $EFGH$ are the bases are equilateral.

Therefore an octahedron has been constructed which is contained by eight equilateral triangles.

$\Box$

Next it is to be demonstrated that the octahedron $EFGHLM$ can be inscribed in the given sphere.

We have that:

$LM = KM = KE$

Therefore the semicircle described on $LM$ also passes through $E$.

Let $LM$ remain fixed, and let that semicircle be carried around and restored to the same position from which it began.

For the same reason as for $E$ it follows that the semicircle passes through the points $F, G, H$.

Thus the octahedron $EFGHLM$ has been inscribed in a sphere.

We have that:

$LK = KM$

while $KE$ is common.

These also contain right angles.

Therefore from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

$LE = EM$

From Proposition $31$ of Book $\text{III} $: Relative Sizes of Angles in Segments:

$\angle LEM$ is a right angle.

Therefore from Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem:

$LM^2 = 2 \cdot LE^2$

We also have that:

$AC = CB$

and so:

$AB = 2 \cdot BC$

But:

$AB : BC = AB^2 : BD^2$

Therefore:

$AB^2 = 2 \cdot BD^2$

But it has already been proved that:

$LM^2 = 2 \cdot LE^2$

and because $EH = DB$:

$DB^2 = LE^2$

Therefore:

$AB^2 = LM^2$

and so:

$AB = LM$

But $AB$ is the diameter of the given sphere.

Therefore $LM$ equals the diameter of the given sphere.

Therefore the (regular) octahedron $EFGHLM$ has been inscribed in the given sphere.

At the same time it has been demonstrated that the square on the diameter of the given sphere is double the square on the edge of the (regular) octahedron $EFGHLM$ inscribed within it.

$\blacksquare$

Historical Note

This proof is Proposition $14$ 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