Construction of Regular Tetrahedron within Given Sphere
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:
(The Elements: Book $\text{XIII}$: Proposition $13$ : Lemma)
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 $DA$ 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 $EH, HF, HG$ be joined.
- 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 Congruence:
- $DA = KE$
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:
- $AB = 3 \cdot CB$
- $AB : BC = AD^2 : DC^2$
Therefore:
- $AD^2 = 3 \cdot DC^2$
- $FE^2 = 3 \cdot EH^2$
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 $HL = CB$.
- $AC : CD = CD : CB$
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:
and:
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:
and:
- $AD$ equals the side of the regular tetrahedron $EFGK$.
Hence the result.
$\blacksquare$
Historical Note
This proof is Proposition $13$ 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