Commentary on Proposition XII.5

Back to Proposition XII.5.

In the two preceding propositions it has been shown how we can divide a pyramid with a triangular base into (1) two equal prisms which are together greater than half the pyramid and (2) two equal pyramids similar to the original one, and that, if this process be continued with the two pyramids, then with the four resulting pyramids, and so on, and if, further, another pyramid of the same height as the original one be similarly divided, the subdivision being made the same number of times, the sum of all the prisms in one pyramid is to the sum of all the prisms in the other as the base of the first is to the base of the second.

We can now prove in the manner of XII. 2 that the volumes of the pyramids themselves are as the bases.

Let us call the pyramids $P$, $P'$ and their respective bases $B$, $B'$. If

$$P : P' \neq B : B'\text{,}$$

suppose that

$$B : B' = P : W.$$

  1. Let $W$ be $< P'$. Divide $P'$ into two prisms and two pyramids, subdivide the latter similarly, and so on, until the sum of the pyramids remaining is less than the difference between $P'$ and $W$ [X. 1], so that

$$P' > (\text{prisms in } P') > W.$$

Then divide $P$ similarly, the same number of times. Now

$$(\text{prisms in } P) : (\text{prisms in } P') = B : B' \quad \text{[XII. 4]}$$

$$= P : W\text{, by hypothesis,}$$

and, alternately,

$$(\text{prisms in } P) : P = (\text{prisms in } P') : W.$$

But

$$(\text{prisms in } P) < P\text{;}$$

therefore

$$(\text{prisms in } P') < W.$$

But, by construction,

$$(\text{prisms in } P') > W.$$

Hence $W$ cannot be less than $P'$.

  1. Suppose, if possible, that $W > P'$. Then, inversely,

$$B' : B = W : P$$

$$= P' : V\text{,}$$

where $V$ is some solid less than $P$. [Cf. XII. 2, Lemma, and note.] But this can be proved impossible exactly as in Part I.

Therefore $W$ is neither less nor greater than $P'$, so that __RSTBLOCKPLACEHOLDER_0003__

Legendre, followed by the American editors already mentioned, and by others, approaches the subject by a different route, proving the following propositions.

  1. If a pyramid be cut by a plane parallel to the base, (a) the lateral edges and the height will be cut in the same proportion, (b) the section by the plane will be a polygon similar to the base.

V A B C D E O a b c d e o X Y Z x y z

  1. Since a lateral face VAB of the pyramid V(ABCDE) is cut by two parallel planes in AB, ab,

AB $\parallel$ ab;

Similarly BC $\parallel$ bc, and so on. Therefore

VA $:$ Va $=$ VB $:$ Vb $=$ VC $:$ Vc $= \dots.$

And, if VO the height be cut in O, o,

BO $\parallel$ bo; and each of the above ratios is equal to VO $:$ Vo.

  1. Since BA $\parallel$ ba, and BC $\parallel$ bc,

[XI. 10]$\angle$ ABC $= \angle$ abc.

Similarly for all the other angles of the polygons, which are therefore equiangular. Also, by similar triangles,

VA $:$ Va $=$ AB $:$ ab,

and so on. Therefore, by the ratios above,

AB $:$ ab $=$ BC $:$ bc $= \dots.$
Therefore the polygons are similar.

  1. If two pyramids of the same height be cut by planes which are at the same perpendicular distance from the vertices, the sections are as the respective bases.

For, if we place the pyramids so that the vertices coincide and the bases are in one plane, the planes of the sections will coincide.

If, e.g., the base of the second pyramid be XYZ and the section xyz, we shall have, by the argument of the last proposition,

VX $:$ Vx $=$ VY $:$ Vy $=$ VZ $:$ Vz $=$ VO $:$ Vo $=$ VA $:$ Va $= \dots,$

and XYZ, xyz will be similar.

Now

(polygon ABCDE) $:$ (polygon abcde) $=$ AB2 $:$ ab2 $=$ VA2 $:$ Va2,

and

$\triangle$ XYZ $: \triangle$ xyz $=$ XY2 $:$ xy2 $=$ VX2 $:$ Vx2 $=$ VA2 $:$ Va2.

Therefore

(polygon ABCDE) $:$ (polygon abcde) $= \triangle$ XYZ $: \triangle$ xyz.
As a particular case, if the bases of the two pyramids are equivalent, the sections are also equivalent.

  1. Two triangular pyramids which have equivalent bases and equal heights are equivalent.

Let VABC, vabc be pyramids with equivalent bases ABC, abc, which for convenience we will suppose placed in one plane, and let TA be the common height.

T X A V B C D E F G H I K L M N O P v a b c d e f g h i k l m n o p

Then, if the pyramids are not equivalent, one must be greater than the other. Let VABC be the greater; and let AX be the height of a prism on ABC as base which is equal in volume to the difference of the pyramids. Divide the height AT into equal parts such that each is less than AX, and let each part be equal to $z$. Through the points of division draw planes parallel to the bases cutting both pyramids in the sections DEF, GHI, dots and def, ghi, dots.

The sections DEF, def will then be equivalent; so will the sections GHI, ghi, and so on. [(2) above]

On the triangles ABC, DEF, GHI, dots as bases draw exterior prisms having for edges the parts AD, DG, GK, dots of the edge AV;

and on the triangles def, ghi, dots as bases draw interior prisms having for edges the parts ad, dg, dots of av. All the partial prisms will then have the same height $z$.

Now the sum of the exterior prisms of the pyramid VABC is greater than that pyramid; and the sum of the interior prisms in the pyramid vabc is less than that pyramid.

Consequently the difference between the sum of the first set of prisms and the sum of the second set of prisms is greater than the difference between the two pyramids.

Again, if we start from the bases ABC, abc, the second exterior prism DEFG is equivalent to the first interior prism defa, since their bases are equivalent and they have the same height $z$. [XI. 28 and note; XI. 32] Similarly the third exterior prism is equivalent to the second interior prism, and so on, until we arrive at the last of each.

Therefore the prism ABCD, the first exterior prism, is the difference between the sums of the exterior and interior prisms respectively.
Therefore the difference between the two pyramids is less than the prism ABCD, which should therefore be greater than the prism with base ABC and height AX.
But the prism ABCD is, by hypothesis, less than the latter prism:
which is impossible.
Consequently the pyramid VABC cannot be greater than the pyramid vabc.
Similarly it may be proved that vabc cannot be greater than VABC.
Therefore the pyramids are equivalent.

Legendre next establishes a proposition corresponding to Eucl. XII. 7, viz.

  1. Any triangular pyramid is one third of the triangular prism on the same base and of the same height,

and from this he deduces that

COR. The volume of a triangular pyramid is equal to a third of the product of its base by its height.

He has previously proved that the volume of a triangular prism is equal to the product of its base and height, since (1) the prism is half of a parallelepiped of the same height and with a parallelogram for base which is double of the base of the prism, and (2) this parallelepiped can be transformed into an equivalent rectangular parallelepiped with the same height and an equivalent base.

The theorem (4) is then extended to any pyramid in the proposition

  1. Any pyramid has for its measure the third part of the product of its base and its height, from which follow

COR. I. Any pyramid is the third part of the prism on the same base and of the same height.
COR. II. Two pyramids of the same height are to one another as their bases, and two pyramids on the same base are to one another as their heights.

The first part of the second corollary corresponds to the present proposition as extended by the next, XII. 6.