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.

Line 2

Let the diameter AB of the given sphere be set out, and

Line 3

let it be cut at C so that AC is quadruple of CB,

Line 4

let the semicircle ADB be described on AB,

Line 5

let the straight line CD be drawn from C at right angles to AB,

Line 6

and let DB be joined;

Line 7

let the circle EFGHK be set out and let its radius be equal to DB,

Line 8

let the equilateral and equiangular pentagon EFGHK be inscribed in the circle EFGHK,

Line 9

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.

Line 10

Therefore the pentagon LMNOP is also equilateral,
and the straight line EP belongs to a decagon.

Line 11

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,

Line 12

let QR, RS, ST, TU, UQ, QL, LR, RM, MS, SN, NT, TO, OU, UP, PQ be joined.

Line 13

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.

Line 14

But EK belongs to an equilateral pentagon;
therefore QU also belongs to the equilateral pentagon inscribed in the circle EFGHK.

Line 15

For the same reason
each of the straight lines QR, RS, ST, TU also belongs to the equilateral pentagon
inscribed in the circle EFGHK;

Line 16

therefore the pentagon QRSTU is equilateral.

Line 17

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]

Line 18

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.

Line 19

For the same reason
each of the triangles QLR, RMS, SNT, TOU is also equilateral.

Line 20

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.

Line 21

For the same reason each of the triangles LRM, MSN, NTO, OUP is also equilateral.

Line 22

Let the center 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,

Line 23

and let QZ, QW, UZ, EV, LV, LX, XM be joined.

Line 24

Now, since each of the straight lines VW, QE is at right angles to the plane 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.

Line 25

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]

Line 26

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.3]
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]

Line 27

But QU also belongs to a pentagon;
therefore the triangle QUZ is equilateral.

Line 28

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.

Line 29

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]

Line 30

For the same reason,
if we join MV, which belongs to a hexagon,
MX is also inferred to belong to a pentagon.

Line 31

But LM also belongs to a pentagon;
therefore the triangle LMX is equilateral.

Line 32

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.

Line 33

Therefore an icosahedron has been constructed which is contained by twenty
equilateral triangles.

Line 34

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.

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But VW is equal to VE, and WZ to VX;
therefore, as ZV is to VE, so is EV to VX.

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And the angles ZVE, EVX are right;

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therefore, if we join the straight line EZ, the angle XEZ will be right because
of the similarity of the triangles XEZ, VEZ.

Line 38

For the same reason,
since, as ZV is to VW, so is VW to WZ,

Line 39

and ZV is equal to XW, and VW to WQ,

Line 40

therefore, as XW is to WQ, so is QW to WZ.

Line 41

And for this reason again,
if we join QX, the angle at Q will be right;
[VI.8]

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therefore the semicircle described on XZ will also pass through Q.
[III.31]

Line 43

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.

Line 44

I say next that it is also comprehended in the given sphere.
For let VW be bisected at A'.

Line 45

Then, since the straight line VZ has been cut in extreme and mean ratio at W,
and ZW is its lesser segment,

Line 46

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 A'W.

Line 47

And ZX is double of ZA', and VW double of A'W;

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therefore the square on ZX is five times the square on WV.

Line 49

And, since AC is quadruple of CB,
therefore AB is five times BC.

Line 50

But, as AB is to BC, so is the square on AB to the square on BD;
therefore the square on AB is five times the square on BD.
[VI.8]
[V. def.9]

Line 51

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.

Line 52

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 is also rational;
hence its diameter is also rational.

Line 53

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.

Line 54

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 hexagon and two of the sides of
the decagon inscribed in the same circle.

`
Q.E.D.`