PROPOSITION 16.
To construct an icosahedron and comprehend it in a sphere,
like the aforesaid figures; and to prove that the side of the icosahedron is the irrational straight line called minor.
Let the diameter
AB of the given sphere be set out, and let it be cut at
C so that
AC is quadruple of
CB, let the semicircle
ADB be described on
AB, let the straight line
CD be drawn from
C at right angles to
AB, and let
DB be joined; let the circle
EFGHK be set out and let its radius be equal to
DB, let the equilateral and equiangular pentagon
EFGHK be inscribed in the circle
EFGHK, let the circumferences
EF,
FG,
GH,
HK,
KE be bisected at the points
L,
M,
N,
O,
P, and let
LM,
MN,
NO,
OP,
PL,
EP be joined.
Therefore the pentagon
LMNOP is also equilateral, and the straight line
EP belongs to a decagon.
Now from the points
E,
F,
G,
H,
K let the straight lines
EQ,
FR,
GS,
HT,
KU be set up at right angles to the plane of the circle, and let them be equal to the radius of the circle
EFGHK, let
QR,
RS,
ST,
TU,
UQ,
QL,
LR,
RM,
MS,
SN,
NT,
TO,
OU,
UP,
PQ be joined.
Now, since each of the straight lines
EQ,
KU is at right angles to the same plane, therefore
EQ is parallel to
KU. [
XI. 6]
But it is also equal to it; and the straight lines joining those extremities of equal and parallel straight lines which are in the same direction are equal and parallel. [
I. 33]
Therefore
QU is equal and parallel to
EK.
But
EK belongs to an equilateral pentagon; therefore
QU also belongs to the equilateral pentagon inscribed in the circle
EFGHK.
For the same reason each of the straight lines
QR,
RS,
ST,
TU also belongs to the equilateral pentagon inscribed in the circle
EFGHK; therefore the pentagon
QRSTU is equilateral.
And, since
QE belongs to a hexagon, and
EP to a decagon, and the angle
QEP is right, therefore
QP belongs to a pentagon; for the square on the side of the pentagon is equal to the square on the side of the hexagon and the square on the side of the decagon inscribed in the same circle. [
XIII. 10]
For the same reason
PU is also a side of a pentagon.
But
QU also belongs to a pentagon; therefore the triangle
QPU is equilateral.
For the same reason each of the triangles
QLR,
RMS,
SNT,
TOU is also equilateral.
And, since each of the straight lines
QL,
QP was proved to belong to a pentagon, and
LP also belongs to a pentagon, therefore the triangle
QLP is equilateral.
For the same reason each of the triangles
LRM,
MSN,
NTO,
OUP is also equilateral.
Let the centre of the circle
EFGHK the point
V, be taken; from
V let
VZ be set up at right angles to the plane of the circle, let it be produced in the other direction, as
VX, let there be cut off
VW, the side of a hexagon, and each of the straight lines
VX,
WZ, being sides of a decagon, and let
QZ,
QW,
UZ,
EV,
LV,
LX,
XM be joined.
Now, since each of the straight lines
VW,
QE is at right angles to the plane of the circle, therefore
VW is parallel to
QE. [
XI. 6]
But they are also equal; therefore
EV,
QW are also equal and parallel. [
I. 33]
But
EV belongs to a hexagon; therefore
QW also belongs to a hexagon.
And, since
QW belongs to a hexagon, and
WZ to a decagon, and the angle
QWZ is right, therefore
QZ belongs to a pentagon. [
XIII. 10]
For the same reason
UZ also belongs to a pentagon, inasmuch as, if we join
VK,
WU, they will be equal and opposite, and
VK, being a radius, belongs to a hexagon; [
IV. 15, Por.] therefore
WU also belongs to a hexagon.
But
WZ belongs to a decagon, and the angle
UWZ is right; therefore
UZ belongs to a pentagon. [
XIII. 10]
But
QU also belongs to a pentagon; therefore the triangle
QUZ is equilateral.
For the same reason each of the remaining triangles of which the straight lines
QR,
RS,
ST,
TU are the bases, and the point
Z the vertex, is also equilateral.
Again, since
VL belongs to a hexagon, and
VX to a decagon, and the angle
LVX is right, therefore
LX belongs to a pentagon. [
XIII. 10]
For the same reason, if we join
MV, which belongs to a hexagon,
MX is also inferred to belong to a pentagon.
But
LM also belongs to a pentagon; therefore the triangle
LMX is equilateral.
Similarly it can be proved that each of the remaining triangles of which
MN,
NO,
OP,
PL are the bases, and the point
X the vertex, is also equilateral.
Therefore an icosahedron has been constructed which is contained by twenty equilateral triangles.
It is next required to comprehend it in the given sphere, and to prove that the side of the icosahedron is the irrational straight line called minor.
For, since
VW belongs to a hexagon, and
WZ to a decagon, therefore
VZ has been cut in extreme and mean ratio at
W, and
VW is its greater segment; [
XIII. 9] therefore, as
ZV is to
VW, so is
VW to
WZ.
But
VW is equal to
VE, and
WZ to
VX; therefore, as
ZV is to
VE, so is
EV to
VX.
And the angles
ZVE,
EVX are right; therefore, if we join the straight line
EZ, the angle
XEZ will be right because of the similarity of the triangles
XEZ,
VEZ.
For the same reason, since, as
ZV is to
VW, so is
VW to
WZ, and
ZV is equal to
XW, and
VW to
WQ, therefore, as
XW is to
WQ, so is
QW to
WZ.
And for this reason again, if we join
QX, the angle at
Q will be right; [
VI. 8] therefore the semicircle described on
XZ will also pass through
Q. [
III. 31]
And if,
XZ remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through
Q and the remaining angular points of the icosahedron, and the icosahedron will have been comprehended in a sphere.
I say next that it is also comprehended in the given sphere.
For let
VW be bisected at
A'.
Then, since the straight line
VZ has been cut in extreme and mean ratio at
W, and
ZW is its lesser segment, therefore the square on
ZW added to the half of the greater segment, that is
WA', is five times the square on the half of the greater segment; [
XIII. 3] therefore the square on
ZA' is five times the square on .
And
ZX is double of
ZA', and
VW double of ; therefore the square on
ZX is five times the square on
WV.
And, since
AC is quadruple of
CB, therefore
AB is five times
BC.
But, as
AB is to
BC, so is the square on
AB to the square on
BD; [
VI. 8,
V. Def. 9] therefore the square on
AB is five times the square on
BD.
But the square on
ZX was also proved to be five times the square on
VW.
And
DB is equal to
VW, for each of them is equal to the radius of the circle
EFGHK; therefore
AB is also equal to
XZ.
And
AB is the diameter of the given sphere; therefore
XZ is also equal to the diameter of the given sphere.
Therefore the icosahedron has been comprehended in the given sphere
I say next that the side of the icosahedron is the irrational straight line called minor.
For, since the diameter of the sphere is rational, and the square on it is five times the square on the radius of the circle
EFGHK, therefore the radius of the circle
EFGHK is also rational; hence its diameter is also rational.
But, if an equilateral pentagon be inscribed in a circle which has its diameter rational, the side of the pentagon is the irrational straight line called minor. [
XIII. 11]
And the side of the pentagon
EFGHK is the side of the icosahedron.
Therefore the side of the icosahedron is the irrational straight line called minor.
PORISM.
From this it is manifest that the square on the diameter of the sphere is five times the square on the radius of the circle from which the icosahedron has been described, and that the diameter of the sphere is composed of the side of the hexagon and two of the sides of the decagon inscribed in the same circle. Q. E. D.
