Chapter 25

Volumes in Euclid’s Element

Definitions (Euclid XI.12). A pyramid is a solid figure, contained byplanes, which is constructed from one plane to one point.

Notation:(V, ABC . . .K).(XI.13). A prism is a solid figure contained by planes two of which,namely those which are opposite, are equal, similar, and parallel, whilethe rest are parallelograms. A right triangular prism is a prism whose twobases are congruent right trianglesABC andA′B′C ′ in which the linesAA′, BB′, CC ′ are perpendicular to the planes ofABC andA′B′C ′.

Notation:(ABC, A′B′C ′).

Proposition 25.1(Euclid XII.3). Any pyramid which has a triangularbase is divided into two pyramids equal and similar to one another, sim-ilar to the whole and having triangular bases, and into two equal prisms;and the two prisms are greater than the half of the whole pyramid.

Heath’s outline.We will denote a pyramid with vertexD and baseABCby D(ABC) or D−ABC and the triangular prism with trianglesGCF ,HLK for bases by(GCF, HLK). The following are the steps of theproof.

I. To prove pyramidH(AEG) equal and similar to pyramidD(HKL).Since sides of△DAB are bisected atH, E, K,

HE//DB, and HK//AB.

HenceHK = EB = EA,

andHE = KB = DK.

302 Volumes in Euclid’s Element

A B

D

H K

C

G

F

E

L

A B

D

H K

C

G

F

E

L

A B

D

H K

C

G

F

E

L

Therefore (1)△HAE and△DHK are equal and similar.Similarly (2)△HAG and△DHL are equal and similar.Again, LH, HK are respectively parallel toGA, AE in a different

plane; therefore∠GAE = ∠LHK.And LH, HK are respectively equal toGA, AE.Therefore (3)△GAE and△LHK are equal and similar.Similarly (4)△HGE and△DLK are equal and similar.Therefore1 the pyramidsH(AEG) andD(HKL) are equal and sim-

ilar.II. To prove the pyramidD(HKL) similar to the pyramidD(ABC). 2

(1)△DHK and△DAB are equiangular and therefore similar.Similarly (2)△DLH and△DCA are similar, as also(3)△DLK and△DCB.Again, BA, AC are respectively parallel toKH, HL in a different

plane; therefore∠BAC = ∠KHL. And BA : AC = KH : HL.Therefore (4)△BAC and△KHL are similar.Consequently the pyramidD(ABC) is similar to the pyramidD(HKL),

and therefore also to the pyramidH(AEG).III. To prove prism(GCF, HLK) equal to prism(HGE, KFB).The prisms may be regarded as having the same height (the distance

between the planesHKL, ABC) and having for bases(1)△CGF and(2) the parallelogramEBFG, which is the double of△CGF .Therefore, by XI.39,3 the prisms are equal.IV. To prove the prisms greater than the small pyramids. Prism (HGE, KFB)

1Eucl. XI, Definition 10: Equal and similar solid figures are those contained by similar planes equal inmultitude and in magnitude.

2Eucl. XI, Definition 9: Similar solid figures are those contained bysimilar planes equal in multitude.3If there be two prisms of equal height, and one have a parallelogram as base and the other a triangle, and

if the parallelogram be double of the triangle, the prisms will be equal.

303

is clearly greater than pyramidK(EFB) and therefore greater than pyra-mid H(AEG).

Therefore each of the prisms is greater than each of the smallpyra-mids; and the sum of the two prisms is greater than the sum of the twosmall pyramids, which, with the two prisms, make up the wholepyra-mid.

Proposition 25.2(Euclid XII.4). If there be two pyramids of the sameheight which have triangular bases, and each of them be divided intotwo pyramids equal to one another and similar to the whole, and intotwo equal prisms, then, as the base of the one pyramid is to thebaseof the other pyramid, so will all the prisms in one pyramid be to all theprisms, being equal in multitude, in the other pyramid.

A

B

C

G

P

M

N

KO

LD

E

F

H

S

T

U

QV

R

Proposition 25.3 (Euclid XII.5(6)). Pyramids which are of the sameheight and have triangular (polygonal) bases are to anotheras the bases.

Proposition 25.4(Euclid XII.7). Any prism which has a triangular baseis divided into three pyramids equal to one another which have triangu-lar bases.

Porism

From this it is manifest that any pyramid is a third part of theprismwhich has the same base with it and equal height.

(i) C(ABD) = (C, DEB) = (D, EBC) by XII.5.(ii) (D, EBC) = (D, ECF ).Thus, (C, ABD) = (D, EBC) = (D, ECF ). Since these three

pyramids together make up the prism, each of them is one thirdof theprism.

304 Volumes in Euclid’s Element

B A

C

E D

F

B A

C

E D

F

B A

C

E D

F

Figure 25.1:

Proposition 25.5(Euclid XII.8). Similar pyramids which have triangu-lar bases are in the triplicate ratio of their correspondingsides.

Proposition 25.6(Euclid XII.9). In equal pyramids which have trian-gular bases the bases are reciprocally proportional to the heights; andthose pyramids in which the bases are reciprocally proportional to theheights are equal.

Proposition 25.7(Euclid XII.10). Any cone is a third part of the cylinderwhich has the same base with it and equal height.

Proposition 25.8(Euclid XII.11). Cones and cylinders which are of thesame height are to one another as their bases.

Proposition 25.9 (Euclid XII.12). Similar cones and cylinders are toone another in the triplicate ratio of the diameters in theirbases.

Proposition 25.10(Euclid XII.13). If a cylinder be cut by a plane whichis parallel to its opposite planes, then as the cylinder is tothe cylinder,so will the axis be to the axis.

Proposition 25.11(Euclid XII.14). Cones and cylinders which are onequal bases are to one another as their heights.

Proposition 25.12(Euclid XII.15). In equal cones and cylinders thebases are reciprocally proportional to the heights; and those cones andcylinders in which the bases are reciprocally proportionalto the heightsare equal.

Proposition 25.13(Euclid XII.18). Spheres are to one another in thetriplicate ratio of their respective diameters.

Chapter 26

Archimedes’ calculation of thevolume of a sphere

Proposition 26.1(Method, Proposition 2). (1) Any sphere is (in respectof solid content) four times the cone with base equal to a great circle ofthe sphere and height equal to its radius.

(2) The cylinder with base equal to a great circle of the sphere andheight equal to the diameter is11

2times the sphere.

G

N

F

L

M

E

B DK

W C Y

V A X

O P

S

R

Q

H

(1) LetABCD be a great circle of a sphere, andAC, BD diametersat right angles to one another.

306 Archimedes’ calculation of the volume of a sphere

Let a circle be drawn aboutBD as diameter and in a plane perpen-dicular toAC, and on this circle as base let a cone be described withA as vertex. Let the surface of this cone be produced and then cut by aplane throughC parallel to its base; the section will be a circle onEFas diameter. On this circle as base let a cylinder be erected with heightand axisAC, and produceCA to H, makingAH equal toCA.

Let CH be regarded as the bar of a balance,A being its middle point.Draw any straight lineMN in the plane of the circleABCD and

parallel toBD. Let MN meet the circle inO, P , the diameterAC in S,and the straight linesAE, AF in Q, R respectively. JoinAO.

ThroughMN draw a plane at right angles toAC; this plane will cutthe cylinder in a circle with diameterMN , the sphere in a circle withdiameterOP , and the cone in a circle with diameterQR.

Now, sinceMS = AC, andQS = AS,

MS · SQ = CA · AS = AO2 = OS2 + SQ2.

And, sinceHA = AC,

HA : AS =CA : AS

=MS : SQ

=MS2 : MS · SQ

=MS2 : (OS2 + SQ2)

=MN2 : (OP 2 + QR2)

=(circle, diameter MN) : (circle, diameter OP + circle, diameter QR).

That is,HA : AS = (circle in cylinder) : (circle in sphere + circle incone).

Therefore the circle in the cylinder, placed where it is, is in equilib-rium, aboutA, with the circle in the sphere together with the circle inthe cone, if both the latter circles are placed with their centers of gravityatH.

Similarly for the three corresponding sections made by a plane per-pendicular toAC and passing through any other straight line in the par-allelogramLF parallel toEF .

If we deal in the same way with all the sets of three circles in whichplanes perpendicular toAC cut the cylinder, the sphere and the cone,and which make up those solids respectively, it follows thatthe cylinder,

307

in the place where it is, will be in equilibrium about A with the sphereand the cone together, when both are placed with their centers of gravityatH.

Therefore, sinceK is the centre of gravity of the cylinder,HA : AK = (cylinder): (sphere + coneAEF ).

But HA = 2AK; therefore cylinder = 2 (sphere + coneAEF ). Nowcylinder = 3 (coneAEF ); 1 therefore coneAEF = 2 (sphere). But, sinceEF = 2BD, coneAEF = 8 (coneABD); therefore sphere = 4 (coneABD).

(2) ThroughB, D drawV BW , XDY parallel toAC; and imagine acylinder which hasAC for axis and the circles onV X, WY as diametersfor bases.

Then cylinderV Y = 2 (cylinderV D) = 6 (coneABD) = 32

(sphere),from above.

From this theorem, to the effect that a sphere is four times asgreatas the cone with a great circle of the sphere as base and with heightequal to the radius of the sphere, I conceived the notion thatthe surfaceof any sphere is four times as great as a great circle in it; for, judgingfrom the fact that any circle is equal to a triangle with base equal to thecircumference and height equal to the radius of the circle, Iapprehendedthat in like manner, any sphere is equal to a cone with base equal to thesurface of the sphere and height equal to the radius.

1Eucl. XII. 10.

308 Archimedes’ calculation of the volume of a sphere

Proposition 26.2(Method, Proposition 7). Any segment of a sphere hasto the cone with the same base and height the ratio which the sum of theradius of the sphere and the height of the complementary segment has tothe height of the complementary segment.

G

N

F

L

M

E

B DK

W C Y

V A X

O P

S

R

Q

H

D BFE G

segmentBAD : coneABD =1

2AC + GC : GC.

Chapter 27

Cavalieri’s principle

If two solids between two horizontal planes have equal sectional areas atevery level, then they have equal volumes.

Consider two solids of the same height:(i) a hemisphere of radiusa (with a horizontal base),(ii) a cylinder of base radiusa and heighta, with the right circular coneon the top face and apex the center of the base removed.

x xa

x

a a

The bases of the two solids are on the same horizontal plane. At eachlevelx above the base,(i) the section of the hemisphere is a circle of radius

√a2 − x2,

(ii) the section of the other solid is an annulus of outer radiusa and innerradiusx.

These two sections clearly have equal areas. Therefore, by Cavalieri’sprinciple, the two solids have equal volumes. The volume of the cylinderwith cone removed is

πa2 · a − 1

3πa2 · a =

2

3πa3.

Therefore, the volume of a sphere of radiusa is 43πa3.

Chapter 28

Fermat: Area under the graphof y = xn

To find the area under the graph ofy = xn, above thex-axis, and be-tween the two vertical linesx = a andx = b, Fermat introducedN − 1mean proportions betweena andb, namely,

a < ar < ar2 < · · ·arN−1 < arN = b,

for r = N

√ba. The region is divided intoN small regions. The widths of

the intervals form a geometric progression of common ratior. The val-ues of at the points of division form a geometric progressionof commonratiorn.

Therefore if we consider the rectangles with the left boundaries assides, these form a geometric progression ofN terms with common ratiorn+1 and first terman+1(r − 1). This has sum

an+1(r − 1) · (rn+1)N − 1

rn+1 − 1

= an+1(r − 1) · (rN)n+1 − 1

rn+1 − 1

= an+1((rN)n+1 − 1) · r − 1

rn+1 − 1

= ((arN)n+1 − an+1) · 1

1 + r + r2 + · · ·+ rn

= (bn+1 − an+1) · 1

1 + r + r2 + · · ·+ rn.

312 Fermat: Area under the graph ofy = xn

On the other hand, if we consider the rectangles with the right bound-aries as sides, these form a geometric progression ofN terms with thesame common ratiorn+1 but first terman+1(r − 1)rn. This has sum

(bn+1 − an+1)rn

1 + r + r2 + · · · + rn.

The true area is between these two:

bn+1 − an+1

1 + r + r2 + · · ·+ rn<

∫ b

a

xndx <(bn+1 − an+1)rn

1 + r + r2 + · · · + rn.

SincelimN→∞ r = 1, we have

limN→∞

bn+1 − an+1

1 + r + r2 + · · ·+ rn= lim

N→∞

(bn+1 − an+1)rn

1 + r + r2 + · · ·+ rn=

bn+1 − an+1

n + 1.

Therefore, we conclude that∫ b

a

xndx =bn+1 − an+1

n + 1.

Exercise

Apply the same method to find the area under the graph ofy = 1x, above

thex-axis, and between the vertical linesx = a andx = b.

Chapter 29

Pascal: Summation of powers ofnumbers in arithmeticprogression

Pascal outlined a method of finding the sums of powers of integers inarithmetic progression, namely,

Σk := ak + (a + d)k + · · ·+ (a + (n − 1)d)k

for givena, d, andn. Clearly,Σ0 = n andΣ1 = n(2a+(n−1)d)2

. In general,

(a + jd + d)k+1 = (a + jd)k+1 +

(

k + 1

1

)

(a + jd)kd + · · ·+

(

k + 1

k

)

(a + jd)dk + dk+1

for j = 0, 1, . . . , n − 1. Combining the equations

(a + d)k+1 = ak+1 +(k + 1

1

)

akd + · · · +(k + 1

k

)

adk + dk+1,

(a + 2d)k+1 = (a + d)k+1 +(k + 1

1

)

(a + d)kd + · · · +(k + 1

k

)

(a + d)dk + dk+1,

...

(a + (n − 1)d)k+1 = (a + (n − 2)d)k+1 +(k + 1

1

)

(a + (n − 2)d)kd + · · · +(k + 1

k

)

(a + (n − 2)d)dk + dk+1,

(a + nd)k+1 = (a + (n − 1)d)k+1 +(k + 1

1

)

(a + (n − 1)d)kd + · · · +(k + 1

k

)

(a + (n − 1)d)dk + dk+1,

we obtain, after cancelling common terms,

(a + nd)k+1 = ak+1 +

(k + 1

1

)

Σkd + · · ·+(

k + 1

k

)

Σ1dk + n · dk+1.

314 Pascal: Summation of powers of numbers in arithmetic progression

KnowingΣ1, Σ2, . . . ,Σk−1, one can use this to determineΣk.Fora = d = 1, denoteΣk by Sk. Thus,

Sk := 1k + 2k + · · ·nk.

We have

S1 =1

2n(n + 1) =

1

2n2 +

1

2n,

S2 =1

6n(n + 1)(2n + 1) =

1

3n3 +

1

2n2 +

1

6n,

S3 =1

4n2(n + 1)2 =

1

4n4 +

1

2n3 +

1

4n2

...

Sk is a polynomial of degreek + 1 in n:

Sk =1

k + 1nk+1 +

1

2nk + lower order terms.

This is enough to give

limn→∞

1k + 2k + · · ·+ nk

nk+1=

1

k + 1,

solving the quadrature problem ofxn (for a positive integral powern):∫ a

0

xkdx =ak+1

k + 1.

Chapter 30

Pascal: On the sines of aquadrant of a circle

CA

B

D

RR

E

E

K

I

Proposition I

The sum of the sines of any arc of a quadrant is equal to the portion of the base betweenthe extreme sines, multiplied by the radius.

Preparation for the proof. Let any arc BP be divided into an infinite number of partsby the points D from which we draw the sines PO, DI etc. Let us take in the otherquadrant of the circle the segment AQ, equal to AO (which measures the distancebetween the extreme sines of the arc, BA, PO). Let AQ be divided into an infinitenumber of equal parts by the points H , at which the ordinates HL will be drawn.

Proof of Proposition I. I say that the sum of the sines DI (each of them multipliedof course by one of the equal small arcs DD) is equal to the segment AO multiplied bythe radius AB. Indeed, let us draw at all the point D the tangent DE, each of whichintersects its neighbor at the points E; if we drop the perpendiculars ER it is clear thateach sine DI multiplied by the tangent EE is equal to each distance RR multiplied

i

AI H

D L

G N

O

P

by the radius AB. Therefore, all the quadrilaterals formed by the sines DI and theirtangents EE (which are all equal to each other) are equal to tall the quadrilaterals,formed by all the portions RR with the radius AB; that is, (since one of the tangents EE

multiplies each of the sines, and since the radius AB multiplies each of the distances),the sum of the sines DI , each of them multiplied by one of the tangents EE, is equal tothe sum of the distances RR, each multiplied by AB. But each tangent EE is equal toeach one of the equal arcs DD. Therefore the sum of the sines multiplied by one of theequal small arcs is equal to the distances AO multiplied by the radius.

Note: It should not cause surprise when I say that all the distances RR are equal toAO and likewise that each tangent EE is equal to each of the small arcs DD, since itis well known that, even though this equality is not true when the number of the sinesis finite, nevertheless the equality is true when the number is infinite; because then thesum of all the equal tangents EE differs from the entire arc BD, or from the sum of allthe equal arcs DD, by less than any given quantity; similarly the sum of the RR formthe entire AC.

Pascal’s formula Translation∑

DI · EE = AB∑

RR∫ β

αr sin θd(rθ) = r2(cos α − cos β)

∑DI2

· EE = AB∑

DI · RR∫ β

αr2 sin2 θd(rθ) = r

∫ β

αr sin θd(r cos θ)

∑DI3

· EE = AB∑

DI2· RR

∫ β

αr3 sin3 θd(rθ) = r

∫ β

αr2 sin2 θd(r cos θ)

ccci

Supplement 25: Hilbert’s third problem: The equality of the vol-umes of two tetrahedra of equal bases and equal altitudes

In two letters to Gerling, Gauss1 expresses his regret that certain theo-rems of solid geometry depend upon the method of exhaustion,i.e., inmodem phraseology, upon the axiom of continuity (or upon theaxiom ofArchimedes). Gauss mentions in particular the theorem of Euclid, thattriangular pyramids of equal altitudes are to each other as their bases.Now the analogous problem in the plane has been solved: Gerling alsosucceeded in proving the equality of volume of symmetrical polyhedraby by dividing them into congruent parts. Nevertheless, it seems to meprobable that a general proof of this kind for the theorem of Euclid justmentioned is impossible, and it should be our task to give a rigorousproof of its impossibility. This would be obtained, as soon as we suc-ceeded inspecifying two tetrahedra of equal bases and equal altitudeswhich can in no way he split up into congruent tetrahedra, andwhichcannot be combined with congruent tetrahedra to form two polyhedrawhich themselves could be split up into congruent tetrahedra. 2

Combinatorial equivalence of polyhedra

We say that two polyhedra are combinatorially equivalent ifthey can bedivided into congruent parts. A polyhedron has a dihedral angle for eachtwo faces intersecting in an edge. We can therefore speak of its sum ofdihedral angles. If two polyhedraP with dihedral anglesθ1, . . . , θp andQ with dihedral anglesϕ1, . . . , ϕq are combinatorially equivalent, thenthere are positive integersk1, . . . , kp, kp+1 andm1, . . . , mq, mq+1 suchthat

k1θ1 + · · · kpθp + kp+1π = m1ϕ1 + · · ·+ m1ϕq + mq+1π.

Max Dehn’s proof of non-equivalence of the cube and regular tetra-hedron

Each dihedral angle of a cube is clearlyπ2. On the other hand, each

dihedral angle of a regular tetrahedron isθ = arccos 13. If the two are

1Werke, vol. 8, pp.241 and 244.2Since this was written Herr Dehn has succeeded in proving this impossibility. See his note[hUber

raumgleiche Polyheder in Nach. d. K. Geselsch. d Wiss. zu Gottingen, 1900, and a paper soon to appear inMath. Annalen [vol. 55, pp. 465–178].

cccii Pascal: On the sines of a quadrant of a circle

θ

combinatorially equivalent, then there are positive integersk andm suchthatkθ = m · π

2. This means thatcos kθ = 0 or±1. This is a contradic-

tion sincecos kθ cannot be an integer ifcos θ = 13.

Two tetrahedra which are not combinatorially equivalent 3

A

BC

D

A′

B′

C′

D′

P Q

The tetrahedronP hasAC = BC = DC and∠BCD = π2, and

AC perpendicular to the plane ofBCD. It has three dihedral anglesequal toπ

2and three equal toarccos 1√

3. This solid is not combinatorially

equivalent to a cube.The tetrahedronQ has dihedral angles equal toπ

2, π

2, π

6, π

4, arccos 1√

6

andarccos√

23. This is combinatorially equivalent to a trianglular prism

with baseB′C ′D′, which in turn is combinatorially equivalent to a cube.This shows that the solidsP andQ are not combinatorially equivalent.

3I learned this from my friend Professor M. K. Siu of the University of Hong Kong.

ccciii

Supplement 26: Archimedes’ calculation of the surface areaof asphere

Proposition 0.1 (SCI 33). The surface of any sphere is equal to fourtimes the greatest circle in it.

Lower bounds of surface of sphere

Proposition 0.2(SCI 21). A regular polygon of an even number of sidesbeing inscribed in a circle, asABC . . . A′ . . . C ′B′A, so thatAA′ is adiameter, if two angular points next but one to each other, asBB′, bejoined, and the other lines parallel toBB′ and joining pairs of angularpoints be drawn, asCC ′, DD′ . . . , then

(BB′ + CC ′ + · · · ) : AA′ = A′B : BA.

INLGKF

C'

C

A

E'B'

B E

D'

D

HA'

Proposition 0.3 (SCI 22). If a polygon be inscribed in a segment of acircle LAL′, so that all its sides excluding the base are equal and theirnumber is even, asLK . . .A . . .K ′L′, A being the middle point of thesegment, and if the linesBB′, CC ′ . . . , parallel to the baseLL′ andjoining pairs of angular points be drawn, then

(BB′ + CC ′ + · · ·+ LM) : AM = A′B : BA.

ccciv Pascal: On the sines of a quadrant of a circle

RHQGPF A'

KC

B

K'

B'

C'

A

M

L'

L

Proposition 0.4(SCI 23). The surface of the sphere is greater than thesurface described by the revolution of the polygon inscribed in the greatcircle about the diameter of the great circle.

Proposition 0.5 (SCI 24). If a regular polygonAB . . . A′ . . . B′A, thenumber of whose sides is a multiple of 4, be inscribed in a great circleof a sphere, and ifBB′ subtending two sides be joined, and all the otherlines parallel toBB′ ad joining pairs of angular points be drawn, thenthe surface of the figure inscribed in the sphere by the revolution of thepolygon about the diameterAA′ is equal to a circle the square of whoseradius is equal to the rectangle

BA(BB′ + CC ′ + · · · ).Proposition 0.6(SCI 25). The surface of the figure inscribed in a sphereas in the last propositions, consisting of portions of conical surfaces, isless than 4 times the greatest circle in the sphere.

Upper bounds of surface of sphere

Let a regular polygon, whose sides are a multiple of 4 in number, becircumscribed about a great circle of given sphere, asAB . . . A′ . . . B′A,and about the polygon, describe another circle, which will therefore havehe same centre as the great circle of the sphere.

Proposition 0.7(SCI 28). The surface of the figure circumscribed to thegiven sphere is greater than that of the sphere itself.

cccv

m'

m

a'a

B'

B

M'

M

A

OA'

Proposition 0.8(SCI 29). The surface of the figure circumscribed to thegiven sphere is equal to a circle the square on whose radius isequal to

AB(BB′ + CC ′ + · · · ).

Proposition 0.9(SCI 30). The surface of the figure circumscribed to thegiven sphere is greater than 4 times the great circle of the sphere.

Proposition 0.10(SCI 32). If a regular polygon with4n sides be in-scribed in a great circle of a sphere, asab . . . a′ · · · b′a, and a similarpolygonAB . . .A′ . . . B′A be described about the great circle, and ifthe polygons revolve with the great circle about the diameters aa′, AA′

respectively, so that they describe the surfaces of solid figures inscribedin and circumscribed to the sphere respectively, then the surfaces of thecircumscribed and the inscribed figures are to one another inthe dupli-cate ratio of their sides.

Proof of SCI 33

Let C be a circle equal to four times the great circle.Then if C is not equal to the surface of the sphere, it must either be

less or greater.I. SupposeC less than the surface of the sphere.

It is then possible to find two linesβ > γ such that

β : γ < surface of sphere: C.

cccvi Pascal: On the sines of a quadrant of a circle

Let δ be a mean proportional betweenβ andγ.Suppose similar regular polygons with4n sides circumscribed about

and inscribed in a great circle such that the ratio of their sides is less thanthe ratioβ : δ. 4 Let the polygons with the circle revolve together about adiameter common to all, describing solids of revolution as before. Then,

surface of outer solid: surface of inner solid

=side of outer polygon2 : side of inner polygon2

<β2 : δ2

=β : γ

<surface of sphere: C.

But this is impossible, since the surface of the circumscribed solid isgreater than that of the sphere, while the surface of the inscribed solid isless thanC.

ThereforeC is not less than the surface of the sphere.II. SupposeC greater than the surface of the sphere. Take linesβ and

γ such thatβ : γ < C : surface of sphere.

Circumscribe and inscribe to the great circle similar regular polygons,as before, such that the ratio of their sides is less than the ratio β : δ (δbeing a mean proportional ofβ andγ), and suppose solids of revolutiongenerated in the usual manner. Then in this case,

surface of outer solid: surface of inner solid

=side of outer polygon2 : side of inner polygon2

<β2 : δ2

=β : γ

<C : surface of sphere.

But this is impossible, since the surface of the circumscribed solid isgreater thanC, while the surface of the inscribed solid is less than thatof the sphere.

4Proposition 3: Given two unequal magnitudes and a circle, itis possible to inscribe a polygon in thecircle and to describe another about it so that the side of thecircumscribed polygon may have to the side ofthe inscribed polygon a ratio less than that of the greater magnitude to the less.

cccvii

ThusC is not greater than the surface of the sphere.Therefore, since it is neither greater nor less,C is equal to the surface

of the sphere.