# Proposition 13

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.

Set out the diameter AB of the given sphere, cut it at the point C so that AC is double CB, describe the semicircle ADB on AB, draw CD from the point C at right angles to AB, and join DA.

Set out the circle EFG with radius equal to DC, inscribe the equilateral triangle EFG in the circle EFG, take the center H of the circle, and join EH, HF, and HG.

Set HK up from the point H at right angles to the plane of the circle EFG, cut off HK equal to the straight line AC from HK, and join KE, KF, and KG. XI.Def.3

Now, since KH is at right angles to the plane of the circle EFG, therefore it makes right angles with all the straight lines which meet it and are in the plane of the circle EFG. But each of the straight lines HE, HF, and HG meets it, therefore HK is at right angles to each of the straight lines HE, HF, and HG.

I.4

And, since AC equals HK, and CD equals HE, and they contain right angles, therefore the base DA equals the base KE. For the same reason each of the straight lines KF and KG also equals DA. Therefore the three straight lines KE, KF, and KG equal one another.

And, since AC is double CB, therefore AB is triple BC.

But that AB is to BC as the square on AD is to the square on DC will be proved afterwards.

XIII.12

Therefore the square on AD is triple the square on DC. But the square on FE is also triple the square on EH, and DC equals EH, therefore DA also equals EF.

But DA was proved equal to each of the straight lines KE, KF, and KG, therefore each of the straight lines EF, FG, and GE also equals each of the straight lines KE, KF, and KG. Therefore the four triangles EFG, KEF, KFG, and KEG are equilateral.

Therefore a pyramid has been constructed out of four equilateral triangles, the triangle EFG being its base and the point K its vertex.

It is next required to comprehend it in the 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.

I.3

Produce the straight line HL in a straight line with KH, and make HL equal to CB.

Now, since AC is to CD as CD is to CB, while AC equals KH, CD equals HE, and CB equals HL, therefore KH is to HE as EH is to HL. Therefore the rectangle KH by HL equals the square on EH.

cf. VI.8, III.31

And each of the angles KHE, EHL is right, therefore the semicircle described on KL passes through E also.

If then, KL remaining fixed, the semicircle is carried round and restored to the same position from which it began to be moved, then it also passes through the points F and G, since, if FL and LG are joined, then the angles at F and G similarly become right angles, and the pyramid is comprehended in the given sphere. For KL, the diameter of the sphere, equals the diameter AB of the given sphere, since KH was made equal to AC, and HL to CB.

I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.

Since AC is double CB, therefore AB is triple BC, and, in conversion, BA is one and a half times AC.

But BA is to AC as the square on BA is to the square on AD. Therefore the square on BA is also one and a half times the square on AD. And BA is the diameter of the given sphere, and AD equals the side of the pyramid. Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.

Q.E.F.

# Lemma

It is to be proved that AB is to BC as the square on AD is to the square on DC. I.46

Set out the figure of the semicircle, join DB, describe the square EC on AC, and complete the parallelogram FB.

Since the triangle DAB is equiangular with the triangle DAC, therefore BA is to AD as DA is to AC. Therefore the rectangle BA by AC equals the square on AD.

VI.1

And since AB is to BC as EB is to BF, and EB is the rectangle BA by AC, for EA equals AC, and BF is the rectangle AC by CB, therefore AB is to BC as the rectangle BA by AC is to the rectangle AC by CB.

VI.8,Cor.

And the rectangle BA by AC equals the square on AD, and the rectangle AC by CB equals the square on DC, for the perpendicular DC is a mean proportional between the segments AC and CB of the base, because the angle ADB is right. Therefore AB is to BC as the square on AD is to the square on DC.

Q.E.D.

## Guide

This figure is usually called a regular tetrahedron, that is, a solid figure contained by four equal and equilateral triangles. Euclid simply calls it “a pyramid” with the understanding that by that he means not just any pyramid, but a regular tetrahedron. A similar ambiguity occured in ancient Greek when the word “tetragon” was used. It meant either any four-angled figure or specifically a square, depending on the context.

#### Summary of the construction

 Standardize the radius of the sphere at 1 unit, so that AB = 2. Then cut AB at C so that AC = 4/3 and BC = 2/3. Let DC be their mean proportional (2/3)√2. Then AD = (2/3)√6. This line AD will end up being the length of the side of the tetrahedron. Note that it has the correct value so that “the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.” Set out the circle EFG of radius EH = (2/3)√2, and inscribe in that circle an equilateral triangle. Then each side of the triangle will be (2/3)√6, the same as AD (XIII.12). Make HK of length 4/3 and perpendicular to the plane of the triangle, and connect KE, KF, and KG. Then K lies on the surface of the sphere. And since the triangle HKE is a right triangle, therefore its hypotenuse KE = (2/3)√6, the same as AD. Likewise KF and KG have the same length. That constructs the tetrahedron in the sphere. #### Coordinates for the vertices of the tetrahedron

A cube can be easily constructed from a tetrahedron since the four vertices of a tetrahedron are four of the eight vertices of a cube. See proposition XIII.15. That being the case, an obvious coordinate system will make eight vertices of a cube have the coordinates

(1,1,1)   (1,1,–1)   (1,–1,1)   (1,–1,–1)   (–1,1,1)   (–1,1,–1)   (–1,–1,1)   (–1,–1,–1)

that is, all eight combinations of –1 and 1 in all three coordinates. The radius for such a cube is √3, so if a unit sphere is desired, then all the coordinates would have to be divided by √3.

The tetrahedron has only half of these eight vertices, and they can be chosen to be

(1,1,1)   (1,–1,–1)   (–1,1,–1)   (–1,–1,1)

that is, the points which have an odd number of positive coordinates. There is another tetrahedron which has as its vertices the remaining four points which have an even number of positive coordinates.

#### Use of this construction

Constructing this regular tetrahedron is an end in itself. In the last proposition of the Elements XIII.18, the five regular polyhedra are compared, and this construction is needed there as well as constructions of the other four regular polyhedra.