Construction of Regular Tetrahedron within Given Sphere

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Theorem

In the words of Euclid:

To construct a pyramid, to comprehend it in a given sphere, and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.

(The Elements: Book $\text{XIII}$: Proposition $13$)


Lemma

In the words of Euclid:

It is to be proved that, as $AB$ is to $BC$, so is the square on $AD$ to the square on $DC$.

(The Elements: Book $\text{XIII}$: Proposition $13$ : Lemma)


Proof

Euclid-XIII-13.png

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 $DA$ be joined.

Let the circle $EFG$ be constructed whose radius equals $DC$.

From Proposition $2$ of Book $\text{IV} $: Inscribing in Circle Triangle Equiangular with Given:

let the equilateral triangle $EFG$ be inscribed within the circle $EFG$.

From Proposition $1$ of Book $\text{III} $: Finding Center of Circle:

let the center of the circle $EFG$ be $H$.

Let $EH, HF, HG$ be joined.

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

let $HK$ be drawn from $H$ perpendicular to the plane of the circle $EFG$.

Let $HK$ equal to $AC$ be cut off from $HK$.

Let $KE, KF, KG$ be joined.

We have that $KH$ is perpendicular to the plane of the circle $EFG$.

So from Book $\text{XI}$ Definition $3$: Line at Right Angles to Plane:

$KH$ is perpendicular to each of the straight lines which meet it and are in the plane of the circle $EFG$.

But each of $HE, HF, HG$ join $HK$ and are in the plane of the circle $EFG$.

Therefore $HK$ is perpendicular to each of $HE, HF, HG$.

We have that:

$AC = HK$

and:

$CD = HE$

Both $AC, CD$ and $HK, HE$ contain right angles.

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

$DA = KE$

For the same reason:

$KF = DA$
$KG = DA$

Therefore the three straight lines $KE, KF$ and $KG$ are equal to one another.


We have that:

$AC = 2 \cdot CB$

Therefore:

$AB = 3 \cdot CB$

But from Lemma to Proposition $13$ of Book $\text{XIII} $: Construction of Regular Tetrahedron within Given Sphere:

$AB : BC = AD^2 : DC^2$

Therefore:

$AD^2 = 3 \cdot DC^2$

But from Proposition $12$ of Book $\text{XIII} $: Square on Side of Equilateral Triangle inscribed in Circle is Triple Square on Radius of Circle:

$FE^2 = 3 \cdot EH^2$

We have that:

$DC = EH$

Therefore:

$DA = EF$

But we have that:

$DA = KE = KF = KG$

Therefore:

$EF = FG = GE = KE = KF = KG$

Therefore each of $\triangle EFG, \triangle KEF, \triangle KFG, \triangle KEG$ are equilateral.

Therefore a regular tetrahedron $EFGK$ has been constructed out of four equilateral triangles.

In this construction, $\triangle EFG$ is its base and $K$ is its apex.

$\Box$


It is next to be demonstrated that $EFGK$ is inscribed in the given sphere.

Let $KH$ be produced to $L$ such that $HL = CB$.

From Porism to Proposition $8$ of Book $\text{VI} $: Perpendicular in Right-Angled Triangle makes two Similar Triangles:

$AC : CD = CD : CB$

while:

$AC = KH$
$CD = HE$
$CB = HL$

Therefore:

$KH : HE = EH : HL$

Therefore from Proposition $17$ of Book $\text{VI} $: Rectangles Contained by Three Proportional Straight Lines:

$KH \cdot HL = EH^2$

We have that $\angle KHE$ and $\angle EHL$ are both right angles.

Therefore from:

Proposition $8$ of Book $\text{VI} $: Perpendicular in Right-Angled Triangle makes two Similar Triangles

and:

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

the semicircle described on $KL$ also passes through $E$.

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

Then it will pass through the points $F$ and $G$.

We have that:

$KH = AC$

and:

$HL = AB$

Thus:

$KL = AB$

Therefore $KL$ is the diameter of the given sphere.

Thus the regular tetrahedron $EFGK$ has been inscribed in the given sphere.

$\Box$


It remains to be shown that the square on the diameter of the given sphere is $1 \frac 1 2$ the square on the side of the regular tetrahedron $EFGK$.

We have that:

$AC = 2 \cdot CB$

Therefore:

$AB = 3 \cdot CB$

and so:

$2 \cdot AB = 3 \cdot AC$

But:

$BA : AC = BA^2 : AD^2$

Therefore:

$BA^2 = \dfrac 3 2 AD^2$

But we have that:

$BA$ equals the diameter of the given sphere

and:

$AD$ equals the side of the regular tetrahedron $EFGK$.

Hence the result.

$\blacksquare$


Historical Note

This theorem is Proposition $13$ of Book $\text{XIII}$ of Euclid's The Elements.


Sources