PROPOSITION 17.
To construct a dodecahedron and comprehend it in a sphere,
like the aforesaid figures, and to prove that the side of the dodecahedron is the irrational straight line called apotome.
Let
ABCD,
CBEF, two planes of the aforesaid cube at right angles to one another, be set out, let the sides
AB,
BC,
CD,
DA,
EF,
EB,
FC be bisected at
G,
H,
K,
L,
M,
N,
O respectively, let
GK,
HL,
MH,
NO be joined, let the straight lines
NP,
PO,
HQ be cut in extreme and mean ratio at the points
R,
S,
T respectively, and let
RP,
PS,
TQ be their greater segments; from the points
R,
S,
T let
RU,
SV,
TW be set up at right angles to the planes of the cube towards the outside of the cube, let them be made equal to
RP,
PS,
TQ, and let
UB,
BW,
WC,
CV,
VU be joined.
I say that the pentagon
UBWCV is equilateral, and in one plane, and is further equiangular.
For let
RB,
SB,
VB be joined.
Then, since the straight line
NP has been cut in extreme and mean ratio at
R, and
RP is the greater segment, therefore the squares on
PN,
NR are triple of the square on
RP. [
XIII. 4]
But
PN is equal to
NB, and
PR to
RU; therefore the squares on
BN,
NR are triple of the square on
RU.
But the square on
BR is equal to the squares on
BN,
NR; [
I. 47] therefore the square on
BR is triple of the square on
RU; hence the squares on
BR,
RU are quadruple of the square on
RU.
But the square on
BU is equal to the squares on
BR,
RU; therefore the square on
BU is quadruple of the square on
RU; therefore
BU is double of
RU.
But
VU is also double of
UR, inasmuch as
SR is also double of
PR, that is, of
RU; therefore
BU is equal to
UV.
Similarly it can be proved that each of the straight lines
BW,
WC,
CV is also equal to each of the straight lines
BU,
UV.
Therefore the pentagon
BUVCW is equilateral.
I say next that it is also in one plane.
For let
PX be drawn from
P parallel to each of the straight lines
RU,
SV and towards the outside of the cube, and let
XH,
HW be joined; I say that
XHW is a straight line.
For, since
HQ has been cut in extreme and mean ratio at
T, and
QT is its greater segment, therefore, as
HQ is to
QT, so is
QT to
TH.
But
HQ is equal to
HP, and
QT to each of the straight lines
TW,
PX; therefore, as
HP is to
PX, so is
WT to
TH.
And
HP is parallel to
TW, for each of them is at right angles to the plane
BD; [
XI. 6] and
TH is parallel to
PX, for each of them is at right angles to the plane
BF. [
id.]
But if two triangles, as
XPH,
HTW, which have two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining straight lines will be in a straight line; [
VI. 32] therefore
XH is in a straight line with
HW.
But every straight line is in one plane; [
XI. 1] therefore the pentagon
UBWCV is in one plane.
I say next that it is also equiangular.
For, since the straight line
NP has been cut in extreme and mean ratio at
R, and
PR is the greater segment, while
PR is equal to
PS, therefore
NS has also been cut in extreme and mean ratio at
P, and
NP is the greater segment; [
XIII. 5] therefore the squares on
NS,
SP are triple of the square on
NP. [
XIII. 4]
But
NP is equal to
NB, and
PS to
SV; therefore the squares on
NS,
SV are triple of the square on
NB; hence the squares on
VS,
SN,
NB are quadruple of the square on
NB.
But the square on
SB is equal to the squares on
SN,
NB; therefore the squares on
BS,
SV, that is, the square on
BV —for the angle
VSB is right—is quadruple of the square on
NB; therefore
VB is double of
BN.
But
BC is also double of
BN; therefore
BV is equal to
BC.
And, since the two sides
BU,
UV are equal to the two sides
BW,
WC, and the base
BV is equal to the base
BC, therefore the angle
BUV is equal to the angle
BWC. [
I. 8]
Similarly we can prove that the angle
UVC is also equal to the angle
BWC; therefore the three angles
BWC,
BUV,
UVC are equal to one another.
But if in an equilateral pentagon three angles are equal to one another, the pentagon will be equiangular, [
XIII. 7] therefore the pentagon
BUVCW is equiangular.
And it was also proved equilateral; therefore the pentagon
BUVCW is equilateral and equiangular, and it is on one side
BC of the cube.
Therefore, if we make the same construction in the case of each of the twelve sides of the cube, a solid figure will have been constructed which is contained by twelve equilateral and equiangular pentagons, and which is called a dodecahedron.
It is then required to comprehend it in the given sphere, and to prove that the side of the dodecahedron is the irrational straight line called apotome.
For let
XP be produced, and let the produced straight line be
XZ; therefore
PZ meets the diameter of the cube, and they bisect one another, for this has been proved in the last theorem but one of the eleventh book. [
XI. 38]
Let them cut at
Z; therefore
Z is the centre of the sphere which comprehends the cube, and
ZP is half of the side of the cube.
Let
UZ be joined.
Now, since the straight line
NS has been cut in extreme and mean ratio at
P, and
NP is its greater segment, therefore the squares on
NS,
SP are triple of the square on
NP. [
XIII. 4]
But
NS is equal to
XZ, inasmuch as
NP is also equal to
PZ, and
XP to
PS.
But further
PS is also equal to
XU, since it is also equal to
RP; therefore the squares on
ZX,
XU are triple of the square on
NP.
But the square on
UZ is equal to the squares on
ZX,
XU; therefore the square on
UZ is triple of the square on
NP.
But the square on the radius of the sphere which comprehends the cube is also triple of the square on the half of the side of the cube, for it has previously been shown how to construct a cube and comprehend it in a sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube. [
XIII. 15]
But, if whole is so related to whole, so is half to half also; and
NP is half of the side of the cube; therefore
UZ is equal to the radius of the sphere which comprehends the cube.
And
Z is the centre of the sphere which comprehends the cube; therefore the point
U is on the surface of the sphere.
Similarly we can prove that each of the remaining angles of the dodecahedron is also on the surface of the sphere; therefore the dodecahedron has been comprehended in the given sphere.
I say next that the side of the dodecahedron is the irrational straight line called apotome.
For since, when
NP has been cut in extreme and mean ratio,
RP is the greater segment, and, when
PO has been cut in extreme and mean ratio,
PS is the greater segment, therefore, when the whole
NO is cut in extreme and mean ratio,
RS is the greater segment.
[Thus, since, as
NP is to
PR, so is
PR to
RN, the same is true of the doubles also, for parts have the same ratio as their equimultiples; [
V. 15] therefore as
NO is to
RS, so is
RS to the sum of
NR,
SO.
But
NO is greater than
RS; therefore
RS is also greater than the sum of
NR,
SO; therefore
NO has been cut in extreme and mean ratio, and
RS is its greater segment.]
But
RS is equal to
UV; therefore, when
NO is cut in extreme and mean ratio,
UV is the greater segment.
And, since the diameter of the sphere is rational, and the square on it is triple of the square on the side of the cube, therefore
NO, being a side of the cube, is rational.
[But if a rational line be cut in extreme and mean ratio, each of the segments is an irrational apotome.]
Therefore
UV, being a side of the dodecahedron, is an irrational apotome. [
XIII. 6]
PORISM.
From this it is manifest that, when the side of the cube is cut in extreme and mean ratio, the greater segment is the side of the dodecahedron. Q. E. D.