Commentary on Proposition XIII.16¶
Back to Proposition XIII.16.
Euclid’s method is
to find the pentagons in the two parallel circular sections of the sphere, the sides of which form ten (five in each circle) of the edges of the icosahedron,
to find the two points which are the poles of the two circular sections,
to prove that the triangles formed by joining the angular points of the pentagons which are nearest to one another two and two are equilateral,
to prove that the triangles of which the poles are the vertices and the sides of the pentagons the bases are also equilateral,
that all the angular points other than the poles lie on a sphere the diameter of which is the straight line joining the poles,
that this sphere is of the same size as the given sphere,
that, if the diameter of the sphere is rational, the edge of the icosahedron is the minor irrational straight line.
I have drawn another figure which will perhaps show the pentagons, and the position of the poles with regard to them, more clearly than does Euclid’s figure.
If AB is the diameter of the given sphere, divide AB at C so that AC $= 4$ CB; draw CD at right angles to AB meeting the semicircle on AB in D. Join BD.
BD is the radius of the circular sections containing the pentagons.
[If $r$ is the radius of the sphere, since AB $:$ BC $= $ AB $^2 : $ AB $\cdot$ BC $= $ AB $^2 : $ BD $^2$, while AB $= 5$ BC, it follows that AB $^2 = 5$ BD $^2$, or
$$\text{(radius of section)}^2 = \frac{4}{5} r^2.$$
Thus [XIII. 10, note] $\text{(side of pentagon)}^2 = \frac{r^2}{5} (10 - 2\sqrt{5})$.]
Inscribe the regular pentagon EFGHK in the circle EFGHK of radius equal to BD. Bisect the arcs EF, FG, …, so forming a decagon in the circle. Joining successive points of bisection, we obtain another regular pentagon LMNOP.
LMNOP is one of the pentagons containing five edges of the icosahedron.
The other circle and inscribed pentagon are obtained by drawing perpendiculars from E, F, G, H, K to the plane of the circle, as EQ, FR, GS, HT, KU, and making each of these perpendiculars equal to the radius of the circle, or, as Euclid says, the side of the regular hexagon in it.
QRSTU is the second pentagon (of course equal to the first) containing five edges of the icosahedron.
Joining each angular point of one of the two pentagons to the two nearest angular points in the other pentagon, we complete ten triangles each of which has for one side a side of one or other of the two pentagons.
V, W are the centres of the two circles, and VW is of course perpendicular to the planes of both.
Produce VW in both directions, making VX and WZ both equal to a side of the regular decagon in the circle (as EL ). Joining X, Z to the angular points of the corresponding pentagons, we
have five more triangles formed with the sides of each pentagon as bases, ten more triangles in all.
Now we come to the proof.
Taking two adjacent perpendiculars, EQ, KU, to the plane of the circle EFGHK, we see that they are parallel as well as equal; therefore QU, EK are equal and parallel.
Similarly for QR, EF etc.
Thus the pentagons have their sides equal.
To prove that the triangles QPL etc., are equilateral, we have, e.g. QL $^2 = $ LE $^2 + $ EQ $^2$
$$= (\text{side of decagon})^2 + (\text{side of hexagon})^2$$
$$= (\text{side of pentagon})^2, \qquad \text{[XIII. 10]}$$
i.e. QL $= (\text{side of pentagon in circle})$ $= $ LP. Similarly QP $= $ LP, and $\triangle$ QPL is equilateral.
So for the other triangles between the two pentagons.
Since VW, EQ are equal and parallel, VE, WQ are equal and parallel. Thus WQ is equal to the side of a regular hexagon in the circles.
Now the angle ZWQ is right; therefore ZQ $^2 = $ ZW $^2 + $ WQ $^2$
$$= (\text{side of decagon})^2 + (\text{side of hexagon})^2$$
$$= (\text{side of pentagon})^2. \qquad \text{[XIII. 10]}$$
Thus ZQ, ZR, ZS, ZT, ZU are all equal to QR, RS etc. ; and the triangles with Z as vertex and bases QR, RS etc. are equilateral.
Similarly for the triangles with X as vertex and LM, MN etc. as bases.
Hence the figure is an icosahedron, being contained by twenty equal equilateral triangles.
To prove that all the vertices of the icosahedron lie on the sphere which has XZ for diameter.
VW being equal to the side of a regular hexagon, and WZ to the side of a regular decagon inscribed in the same circle, VZ is divided at W in extreme and mean ratio. [XIII. 9]
Therefore ZV $:$ VW $= $ VW $:$ WZ, or, since VW $= $ VE, WZ $= $ VX, ZV $:$ VE $= $ VE $:$ VX.
Thus E lies on the semicircle on ZX as diameter. [VI. 8]
Similarly for all the other vertices of the icosahedron.
Hence the sphere with diameter XZ circumscribes it.
To prove XZ $= $ AB.
Since VZ is divided in extreme and mean ratio at W, and VW is bisected at A', A'Z $^2 = 5$ A'W $^2$. [XIII. 3]
Taking the doubles of A'Z, A'W, we have XZ $^2 = 5$ VW $^2$ $= 5$ BD $^2$ $= $ AB $^2$. [see under (1) above] That is, XZ $= $ AB.
[If $r$ is the radius of the sphere, VW $= $ BD $= \frac{2}{\sqrt{5}} r$, VX $= (\text{side of decagon in circle of radius }$ BD $)$ $= \frac{1}{2}$ BD $(\sqrt{5} - 1)$ [XIII. 9, note]
$$= \frac{r}{\sqrt{5}}(\sqrt{5} - 1).$$
Consequently XZ $= $ VW $+ 2$ VX
$$= \frac{2}{\sqrt{5}} r + \frac{2}{\sqrt{5}} r (\sqrt{5} - 1)$$
$$= 2r.]$$
The radius of the circle EFGHK is equal to $\frac{2}{\sqrt{5}} r$, and is therefore “rational” in Euclid’s sense. Hence the side of the inscribed pentagon is the irrational straight line called minor. [XIII. 11] [The side of this pentagon is the edge of the icosahedron, and its value is (note on XIII. 10) $\frac{1}{2}$ BD $\sqrt{10 - 2\sqrt{5}}$
$$= \frac{r}{\sqrt{5}} \sqrt{10 - 2\sqrt{5}}$$
$$= \frac{r}{5} \sqrt{10(5 - \sqrt{5})}.]$$
Pappus’ solution. This solution (Pappus, III. pp. 150—6) differs considerably from that of Euclid. Whereas Euclid uses two circular sections of the sphere (those circumscribing the pentagons of his construction), Pappus finds four parallel circular sections each passing through three of the vertices of the icosahedron; two of the circles are small circles circumscribing two opposite triangular faces respectively, and the other two circles are between these two circles, parallel to them and equal to one another.
Analysis. Suppose the problem solved, the vertices of the icosahedron being A, B, C; D, E, F; G, H, K; L, M, N. Since the straight lines BA, BC, BF, BG, BE drawn from B to the surface of the sphere are equal, A, C, F, G, E are in one plane. And AC, CF, FG, GE, EA are equal; therefore ACFGE is an equilateral and equiangular pentagon. So are the figures KEBCD, DHFBA, AKLGB, AKNHC, and CHMGB. Join EF, KH.
Now AC will be parallel to EF (in the pentagon ACFGE) and to KH (in the pentagon AKNHC), so that EF, KH are also parallel; and further KH is parallel to LM (in the pentagon LKDHM).
Similarly BC, ED, GH, LN are all parallel ; and likewise BA, FD, GK, MN are all parallel.
Since BC is equal and parallel to LN, and BA to MN, the circles ABC, LMN are equal and parallel.
Similarly the circles DEF, KGH are equal and parallel; for the triangles inscribed in them are equal (since each of the sides in both is the chord subtending an angle of equal pentagons), and their sides are parallel respectively.
Now in the equal and parallel circles DEF, KGH the chords EF, KH are equal and parallel, and on opposite sides of the centres; therefore FK is a diameter of the sphere [Pappus' lemma, pp. 136—8], and the angle FEK is right [Pappus' lemma, p. 138, 20—26].
[The diameter FK is not actually drawn in the figure.]
In the pentagon GEACF, if EF be divided in extreme and mean ratio, the greater segment is equal to AC. [Eucl. XIII. 8]
Therefore EF $:$ AC = (side of hexagon) : (side of decagon in same circle). [XIII. 9]
And EF $^2 + $ AC $^2 = $ EF $^2 + $ EK $^2 = d^2$, where $d$ is the diameter of the sphere.
Thus FK, EF, AC are as the sides of the pentagon, hexagon and decagon respectively inscribed in the same circle. [XIII. 10]
But FK, the diameter of the sphere, is given; therefore EF, AC are given respectively; thus the radii of the circles EFD, ACB are given (if $r, r'$ are their radii, $r^2 = \frac{1}{3}$ EF $^2$, $r'^2 = \frac{1}{3}$ AC $^2$).
Hence the circles are given. and so are the circles KHG, LMN which are equal and parallel to them respectively.
Synthesis. If $d$ be the diameter of the sphere, set out two straight lines x, y, such that $d$, x, y are in the ratio of the sides of the pentagon, hexagon and decagon respectively inscribed in one and the same circle.
Draw (1) two equal and parallel circular sections in the sphere, with radii equal to $r$, where $r^2 = \frac{1}{3}$ x $^2$, as DEF, KGH, and (2) two equal and parallel circular sections as ABC, LMN, with radius $r'$ such that $r'^2 = \frac{1}{3}$ y $^2$.
In the circles (1) draw EF, KH as sides of inscribed equilateral triangles, parallel to one another, and on opposite sides of the centres; and in the circles (2) draw AC, LM as sides of inscribed equilateral triangles parallel to one another and to EF, KH, and so that AC, EF are on opposite sides of the centres, and likewise KH, LM.
Complete the figure. The correctness of the construction is proved as in the analysis. It follows also (says Pappus) that (diam. of sphere)$^2 = 3$ (side of pentagon in DEF)$^2$.
For, by construction, KF : FE = $p : h$, where $p, h$ are the sides of the pentagon and hexagon inscribed in the same circle DEF.
And FE : $h$ = the ratio of the side of an equilateral triangle to that of a hexagon inscribed in the same circle; that is, FE : $h = \sqrt{3} : 1$, whence KF : $p = \sqrt{3} : 1$, or KF $^2 = 3p^2$.
Another construction.
Mr H. M. Taylor has a neat construction for an icosahedron of edge $a$. Let $l$ be the length of the diagonal of a regular pentagon with side equal to $a$.
Then (figure of XIII. 8), by Ptolemy’s theorem, $l^2 = la + a^2$.
Construct a cube with edge equal to $l$. Let O be the centre of the cube.
From O draw OL, OM, ON perpendicular to three adjacent faces, and in these draw PP', QQ', RR' parallel to AB, AD, AE respectively.
Make LP, LP', MQ, MQ', NR, NR' all equal to $\frac{1}{2}a$.
Let p̄, p̄', q̄, q̄', r̄, r̄' be the reflexes of P, P', Q, Q', R, R' respectively.
Then will P, P', Q, Q', R, R', p̄, p̄', q̄, q̄', r̄, r̄' be the vertices of a regular icosahedron.
The projections of PQ on AB, AD, AE are equal to $\frac{1}{2}(l - a), \frac{1}{2}a, \frac{1}{2}l$ respectively.
Therefore PQ $^2 = \frac{1}{4}(l - a)^2 + \frac{1}{4}a^2 + \frac{1}{4}l^2$ = $\frac{1}{2}(l^2 - al + a^2)$ = $a^2$.
Therefore
PQ $= a$.
Similarly it may be proved that every other edge is equal to $a$. All the angular points lie on a sphere with radius OP, and
OP $^2 = \frac{1}{4}(a^2 + l^2)$.
Each solid pentahedral angle is composed of five equal plane angles, each of which is the angle of an equilateral triangle. Therefore the icosahedron is regular.
[$a^2 = 4$ OP $^2 - l^2$. And, from the equation $l^2 = la + a^2$, we derive
$$l = \frac{a}{2}(\sqrt{5} + 1).$$
Therefore, if $r$ be the radius of the sphere,
$$a^2 \left\{ 1 + \frac{(\sqrt{5} + 1)^2}{4} \right\} = 4r^2,$$
whence
$$\begin{aligned} a &= 4r / \sqrt{10 + 2\sqrt{5}} \\ &= 4r \sqrt{10 - 2\sqrt{5}} / \sqrt{80} \\ &= \frac{r}{\sqrt{5}} \sqrt{10 - 2\sqrt{5}} \\ &= \frac{r}{5} \sqrt{10(5 - \sqrt{5})}, \end{aligned}$$
as above.]