Construction of Regular Tetrahedron within Given Sphere

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

From :
 * let the center of the circle $EFG$ be $H$.

Let $EH, HF, HG$ be joined.

From :
 * 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 :
 * $KH$ 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 :
 * $DA = KE$

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

Therefore:
 * $AD^2 = 3 \cdot DC^2$

But from :
 * $FE^2 = 3 \cdot EH^2$

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.

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 :
 * $AC : CD = CD : CB$

while:
 * $AC = KH$
 * $CD = HE$
 * $CB = HL$

Therefore:
 * $KH : HE = EH : HL$

Therefore from :
 * $KH \cdot HL = EH^2$

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.

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.