PROPOSITION 18.
To set out the sides of the five figures and to compare them with one another.
Let
AB, the diameter of the given sphere, be set out, and let it be cut at
C so that
AC is equal to
CB, and at
D so that
AD is double of
DB; let the semicircle
AEB be described on
AB, from
C,
D let
CE,
DF be drawn at right angles to
AB, and let
AF,
FB,
EB be joined.
Then, since
AD is double of
DB, therefore
AB is triple of
BD.
Convertendo, therefore,
BA is one and a half times
AD.
But, as
BA is to
AD, so is the square on
BA to the square on
AF, [
V. Def. 9,
VI. 8] for the triangle
AFB is equiangular with the triangle
AFD; therefore the square on
BA is one and a half times the square on
AF.
But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. [
XIII. 13]
And
AB is the diameter of the sphere; therefore
AF is equal to the side of the pyramid.
Again, since
AD is double of
DB, therefore
AB is triple of
BD.
But, as
AB is to
BD, so is the square on
AB to the square on
BF; [
VI. 8,
V. Def. 9] therefore the square on
AB is triple of the square on
BF.
But the square on the diameter of the sphere is also triple of the square on the side of the cube. [
XIII. 15]
And
AB is the diameter of the sphere; therefore
BF is the side of the cube.
And, since
AC is equal to
CB, therefore
AB is double of
BC.
But, as
AB is to
BC, so is the square on
AB to the square on
BE; therefore the square on
AB is double of the square on
BE.
But the square on the diameter of the sphere is also double of the square on the side of the octahedron. [
XIII. 14]
And
AB is the diameter of the given sphere; therefore
BE is the side of the octahedron.
Next, let
AG be drawn from the point
A at right angles to the straight line
AB, let
AG be made equal to
AB, let
GC be joined, and from
H let
HK be drawn perpendicular to
AB.
Then, since
GA is double of
AC, for
GA is equal to
AB, and, as
GA is to
AC, so is
HK to
KC, therefore
HK is also double of
KC.
Therefore the square on
HK is quadruple of the square on
KC; therefore the squares on
HK,
KC, that is, the square on
HC, is five times the square on
KC.
But
HC is equal to
CB; therefore the square on
BC is five times the square on
CK.
And, since
AB is double of
CB, and, in them,
AD is double of
DB, therefore the remainder
BD is double of the remainder
DC.
Therefore
BC is triple of
CD; therefore the square on
BC is nine times the square on
CD.
But the square on
BC is five times the square on
CK; therefore the square on
CK is greater than the square on
CD; therefore
CK is greater than
CD.
Let
CL be made equal to
CK, from
L let
LM be drawn at right angles to
AB, and let
MB be joined.
Now, since the square on
BC is five times the square on
CK, and
AB is double of
BC, and
KL double of
CK, therefore the square on
AB is five times the square on
KL.
But the square on the diameter of the sphere is also five times the square on the radius of the circle from which the icosahedron has been described. [
XIII. 16, Por.]
And
AB is the diameter of the sphere; therefore
KL is the radius of the circle from which the icosahedron has been described; therefore
KL is a side of the hexagon in the said circle. [
IV. 15, Por.]
And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, [
XIII. 16, Por.] and
AB is the diameter of the sphere, while
KL is a side of the hexagon, and
AK is equal to
LB, therefore each of the straight lines
AK,
LB is a side of the decagon inscribed in the circle from which the icosahedron has been described.
And, since
LB belongs to a decagon, and
ML to a hexagon, for
ML is equal to
KL, since it is also equal to
HK, being the same distance from the centre, and each of the straight lines
HK,
KL is double of
KC, therefore
MB belongs to a pentagon. [
XIII. 10]
But the side of the pentagon is the side of the icosahedron; [
XIII. 16] therefore
MB belongs to the icosahedron.
Now, since
FB is a side of the cube, let it be cut in extreme and mean ratio at
N, and let
NB be the greater segment; therefore
NB is a side of the dodecahedron. [
XIII. 17, Por.]
And, since the square on the diameter of the sphere was proved to be one and a half times the square on the side
AF of the pyramid, double of the square on the side
BE of the octahedron and triple of the side
FB of the cube, therefore, of parts of which the square on the diameter of the sphere contains six, the square on the side of the pyramid contains four, the square on the side of the octahedron three, and the square on the side of the cube two.
Therefore the square on the side of the pyramid is fourthirds of the square on the side of the octahedron, and double of the square on the side of the cube; and the square on the side of the octahedron is one and a half times the square on the side of the cube.
The said sides, therefore, of the three figures, I mean the pyramid, the octahedron and the cube, are to one another in rational ratios.
But the remaining two, I mean the side of the icosahedron and the side of the dodecahedron, are not in rational ratios either to one another or to the aforesaid sides; for they are irrational, the one being minor [
XIII. 16] and the other an apotome [
XIII. 17].
That the side
MB of the icosahedron is greater than the side
NB of the dodecahedron we can prove thus.
For, since the triangle
FDB is equiangular with the triangle
FAB, [
VI. 8] proportionally, as
DB is to
BF, so is
BF to
BA. [
VI. 4]
And, since the three straight lines are proportional, as the first is to the third, so is the square on the first to the square on the second; [
V. Def. 9,
VI. 20, Por.] therefore, as
DB is to
BA, so is the square on
DB to the square on
BF; therefore, inversely, as
AB is to
BD, so is the square on
FB to the square on
BD.
But
AB is triple of
BD; therefore the square on
FB is triple of the square on
BD.
But the square on
AD is also quadruple of the square on
DB, for
AD is double of
DB; therefore the square on
AD is greater than the square on
FB; therefore
AD is greater than
FB; therefore
AL is by far greater than
FB.
And, when
AL is cut in extreme and mean ratio,
KL is the greater segment, inasmuch as
LK belongs to a hexagon, and
KA to a decagon; [
XIII. 9] and, when
FB is cut in extreme and mean ratio,
NB is the greater segment; therefore
KL is greater than
NB.
But
KL is equal to
LM; therefore
LM is greater than
NB.
Therefore
MB, which is a side of the icosahedron, is by far greater than
NB which is a side of the dodecahedron. Q. E. D.
I say next that
no other figure,
besides the said five figures,
can be constructed which is contained by equilateral and equiangular figures equal to one another.
For a solid angle cannot be constructed with two triangles, or indeed planes.
With three triangles the angle of the pyramid is constructed, with four the angle of the octahedron, and with five the angle of the icosahedron; but a solid angle cannot be formed by six equilateral and equiangular triangles placed together at one point, for, the angle of the equilateral triangle being two-thirds of a right angle, the six will be equal to four right angles: which is impossible, for any solid angle is contained by angles less than four right angles. [
XI. 21]
For the same reason, neither can a solid angle be constructed by more than six plane angles.
By three squares the angle of the cube is contained, but by four it is impossible for a solid angle to be contained, for they will again be four right angles.
By three equilateral and equiangular pentagons the angle of the dodecahedron is contained; but by four such it is impossible for any solid angle to be contained, for, the angle of the equilateral pentagon being a right angle and a fifth, the four angles will be greater than four right angles: which is impossible.
Neither again will a solid angle be contained by other polygonal figures by reason of the same absurdity.
Therefore etc. Q. E. D.
LEMMA.
But that
the angle of the equilateral and equiangular pentagon is a right angle and a fifth we must prove thus.
Let
ABCDE be an equilateral and equiangular pentagon, let the circle
ABCDE be circumscribed about it, let its centre
F be taken, and let
FA,
FB,
FC,
FD,
FE be joined.
Therefore they bisect the angles of the pentagon at
A,
B,
C,
D,
E.
And, since the angles at
F are equal to four right angles and are equal, therefore one of them, as the angle
AFB, is one right angle less a fifth; therefore the remaining angles
FAB,
ABF consist of one right angle and a fifth.
But the angle
FAB is equal to the angle
FBC; therefore the whole angle
ABC of the pentagon consists of one right angle and a fifth. Q. E. D.