APPENDIX.¶
I. THE CONTENTS OF THE SO-CALLED BOOK XIV. BY HYPSICLES.¶
This supplement to Euclid’s Book XIII. is worth reproducing for the sake not only of the additional theorems proved in it but of the historical notices contained in the preface and in one or two later passages. Where I translate literally from the Greek text, I shall use inverted commas; except in such passages I reproduce the contents in briefer form.
I have already quoted from the Preface (Vol. I. pp. 5—6), but I will repeat it here.
“Basilides of Tyre, O Protarchus, when he came to Alexandria and met my father, spent the greater part of his sojourn with him on account of the bond between them due to their common interest in mathematics. And on one occasion, when looking into the tract written by Apollonius about the comparison of the dodecahedron and icosahedron inscribed in one and the same sphere, that is to say, on the question what ratio they bear to one another, they came to the conclusion that Apollonius’ treatment of it in this book was not correct; accordingly, as I understood from my father, they proceeded to amend and rewrite it. But I myself afterwards came across another book published by Apollonius, containing a demonstration of the matter in question, and I was greatly attracted by his investigation of the problem. Now the book published by Apollonius is accessible to all; for it has a large circulation in a form which seems to have been the result of later careful elaboration.
“For my part, I determined to dedicate to you what I deem to be necessary by way of commentary, partly because you will be able, by reason of your proficiency in all mathematics and particularly in geometry, to pass an expert judgment upon what I am about to write, and partly because, on account of your intimacy with my father and your friendly feeling towards myself, you will lend a kindly ear to my disquisition. But it is time to have done with the preamble and to begin my treatise itself.
[Prop. 1.] “The perpendicular drawn from the centre of any circle to the side of the pentagon inscribed in the same circle is half the sum of the side of the hexagon and of the side of the decagon inscribed in the same circle.”
Let ABC be a circle, and BC the side of the inscribed regular pentagon. Take D the centre of the circle, draw DE from D perpendicular to BC, and produce DE both ways to meet the circle in F, A.
I say that DE is half the sum of the side of the hexagon and of the side of the decagon inscribed in the same circle.
Let DC, CF be joined; make GE equal to EF, and join GC.
Since the circumference of the circle is five times the arc BFC, and half the circumference of the circle is the arc ACF, while the arc FC is half the arc BFC, therefore $(\text{arc }$ ACF $) = 5 (\text{arc }$ FC $)$ or $(\text{arc }$ AC $) = 4 (\text{arc }$ CF $)$.
Hence $\angle$ ADC $= 4 \angle$ CDF, and therefore $\angle$ AFC $= 2 \angle$ CDF.
Thus $\angle$ CGF $= \angle$ AFC $= 2 \angle$ CDF; therefore [I. 32] $\angle$ CDG $= \angle$ DCG, so that DG $= $ GC $= $ CF.
And GE $= $ EF; therefore DE $= $ EF $+ $ FC.
Add DE to each; therefore $2$ DE $= $ DF $+ $ FC.
And DF is the side of the regular hexagon, and FC the side of the regular decagon, inscribed in the same circle.
Therefore etc.
“Next it is manifest from the theorem [12] in Book XIII. that the perpendicular drawn from the centre of the circle to the side of the equilateral triangle [inscribed in it] is half of the radius of the circle.
[Prop. 2.] “The same circle circumscribes both the pentagon of the dodecahedron and the triangle of the icosahedron inscribed in the same sphere.
“This is proved by Aristaeus in his work entitled Comparison of the five figures. But Apollonius proves in the second edition of his comparison of the dodecahedron with the icosahedron that, as the surface of the dodecahedron is to the surface of the icosahedron, so also is the dodecahedron itself to the icosahedron, because the perpendicular from the centre of the sphere to the pentagon of the dodecahedron and to the triangle of the icosahedron is the same.
“But it is right that I too should prove that
[Prop. 2] The same circle circumscribes both the pentagon of the dodecahedron and the triangle of the icosahedron inscribed in the same sphere.
“For this I need the following
Lemma.
“If an equilateral and equiangular pentagon be inscribed in a circle, the sum of the squares on the straight line subtending two sides and on the side of the pentagon is five times the square on the radius.”
Let ABC be a circle, AC the side of the pentagon, D the centre; draw DF perpendicular to AC and produce it to B, E; join AB, AE.
I say that BA $^2 + $ AC $^2 = 5 $ DE $^2$.
For, since BE $= 2 $ ED, BE $^2 = 4 $ ED $^2$.
And BE $^2 = $ BA $^2 + $ AE $^2$; therefore BA $^2 + $ AE $^2 + $ ED $^2 = 5 $ ED $^2$.
But AC $^2 = $ DE $^2 + $ EA $^2$; [Eucl. XIII. 10] therefore BA $^2 + $ AC $^2 = 5 $ DE $^2$.
“This being proved, it is required to prove that the same circle circumscribes both the pentagon of the dodecahedron and the triangle of the icosahedron inscribed in the same sphere.”
Let AB be the diameter of the sphere, and let a dodecahedron and an icosahedron be inscribed.
Let CDEFG be one pentagon of the dodecahedron, and KLH one triangle of the icosahedron.
I say that the radii of the circles circumscribing them are equal.
Join DG; then DG is the side of a cube inscribed in the sphere. [Eucl. XIII. 17]
Take a straight line MN such that AB $^2 = 5 $ MN $^2$.
Now the square on the diameter of the sphere is five times the square on the radius of the circle from which the icosahedron is described. [XIII. 16, Por.]
Therefore MN is equal to the radius of the circle passing through the five vertices of the icosahedron which form a pentagon.
Cut MN in extreme and mean ratio at O, MO being the greater segment.
Therefore MO is the side of the decagon in the circle with radius MN. [XIII. 9 and 5, converse]
Now $5 $ MN $^2 = $ AB $^2 = 3 $ DG $^2$. [XIII. 15]
But $3 $ DG $^2 : 3 $ CG $^2 = 5 $ MN $^2 : 5 $ MO $^2$ (since, if DG is cut in extreme and mean ratio, the greater segment is equal to CG, and, if two straight lines are cut in extreme and mean ratio, their segments are in the same ratio: see lemma later, pp. 518–9).
And $5 $ MO $^2 + 5 $ MN $^2 = 5 $ KL $^2$. [This follows from XIII. 10, since KL is, by the construction of XIII. 16, the side of the regular pentagon in the circle with radius equal to MN, that is, the circle in which MN is the side of the inscribed hexagon and MO the side of the inscribed decagon.]
Therefore $5 $ KL $^2 = 3 $ CG $^2 + 3 $ DG $^2$.
But $5 $ KL $^2 = 15 (\text{radius of circle about }$ KLH $)^2$, [XIII. 12] and $3 $ DG $^2 + 3 $ CG $^2 = 15 (\text{radius of circle about }$ CDEFG $)^2$. [Lemma above]
Therefore the radii of the two circles are equal.
[Prop. 3.] “If there be an equilateral and equiangular pentagon and a circle circumscribed about it, and if a perpendicular be drawn from the centre to one side, then 30 times the rectangle contained by the side and the perpendicular is equal to the surface of the dodecahedron.”
Let ABCDE be the pentagon, F the centre of the circle, FG the perpendicular on a side CD.
I say that $30 $ CD $cdot $ FG $= 12 (\text{area of pentagon})$.
Let CF, FD be joined.
Then, since CD $cdot $ FG $= 2 (triangle $ CDF $)$, $5 $ CD $cdot $ FG $= 10 (triangle $ CDF $)$, whence $30 $ CD $cdot $ FG $= 12 (\text{area of pentagon})$.
Similarly we can prove that,
[Prop. 4] If ABC be an equilateral triangle in a circle, D the centre, and DE perpendicular to BC, $30 $ BC $cdot $ DE $= (\text{surface of icosahedron})$.
For DE $cdot $ BC $= 2 (triangle $ DBC $)$; therefore $3 $ DE $cdot $ BC $= 6 (triangle $ DBC $) = 2 (triangle $ ABC $)$, whence $30 $ DE $cdot $ BC $= 20 (triangle $ ABC $)$.
It follows that [Prop. 5]
$$\text{(surface of dodecahedron)} : \text{(surface of icosahedron)} = \text{(side of pentagon)} \cdot \text{(its perpendicular)} : \text{(side of triangle)} \cdot \text{(its perp.)}.$$
“This being clear, we have next to prove that,
[Prop. 6] As the surface of the dodecahedron is to the surface of the icosahedron, so is the side of the cube to the side of the icosahedron.”
Let ABC be the circle circumscribing the pentagon of the dodecahedron and the triangle of the icosahedron, and let CD be the side of the triangle, AC that of the pentagon.
Let E be the centre, and EF, EG perpendiculars to CD, AC.
Produce EG to meet the circle in B and join BC.
Set out H equal to the side of the cube inscribed in the same sphere.
I say that $(text{surface of dodecahedron}) : (text{surface of icosahedron}) = $ H $: $ CD.
For, since the sum of EB, BC is divided at B in extreme and mean ratio, and BE is the greater segment, [XIII. 9] and EG $= \frac{1}{2}($ EB $+ $ BC $)$, [Prop. 1] while EF $= \frac{1}{2}$ BE, [see p. 513 above] therefore, if EG is divided in extreme and mean ratio, the greater segment is equal to EF [that is to say, since EB is the greater segment of EB $+ $ BC divided in extreme and mean ratio, $\frac{1}{2}$ EB is the greater segment of $\frac{1}{2}($ EB $+ $ BC $)$ similarly divided].
But, if H is also divided in extreme and mean ratio, the greater segment is equal to CA. [XIII. 17, Por.]
Therefore H $: $ CA $= $ EG $: $ EF, or FE $cdot $ H $= $ CA $cdot $ EG.
And, since H $: $ CD $= $ FE $cdot $ H $: $ FE $cdot $ CD, and FE $cdot $ H $= $ CA $cdot $ EG, therefore H $: $ CD $= $ CA $cdot $ EG $: $ FE $cdot $ CD $= (\text{surface of dodecahedron}) : (\text{surf. of icos.})$. [Prop. 5]
Another proof of the same theorem.
Preliminary.
Let ABC be a circle and AB, AC sides of an inscribed regular pentagon. Join BC; take D the centre of the circle, join AD and produce it to meet the circle at E. Join BD.
Let DF be made equal to $\frac{1}{2}$ AD, and CH equal to $\frac{1}{3}$ CG.
I say that $\text{rect. }$ AF, BH $= (\text{area of pentagon})$.
For, since AD $= 2 $ DF, AF $= frac{3}{2} $ AD.
And, since GC $= 3 $ HC, GC $= frac{3}{2} $ GH.
Therefore FA $: $ AD $= $ CG $: $ GH, so that AF $cdot $ GH $= $ AD $cdot $ CG $= $ AD $cdot $ BG $= 2 (triangle $ ABD $)$.
Therefore $5 $ AF $cdot $ GH $= 10 (triangle $ ABD $) = 2 (\text{area of pentagon})$.
And GH $= 2 $ HC; therefore $5 $ AF $cdot $ HC $= (\text{area of pentagon})$, or AF $cdot $ BH $= (\text{area of pentagon})$.
Proof of theorem.
This being clear, let the circle be set out which circumscribes the pentagon of the dodecahedron and the triangle of the icosahedron inscribed in the same sphere.
Let ABC be the circle, and AB, AC two sides of the pentagon; join BC.
Take E the centre of the circle, join AE and produce it to F.
Let AE $= 2 $ EG, KC $= 3 $ CH.
Through G draw DM at right angles to AF meeting the circle at D, M; DM is then the side of the inscribed equilateral triangle.
Join AD, AM, which are equal to DM.
Now, since AG $cdot $ BH $= (\text{area of pentagon})$, and AG $cdot $ GD $= (\text{area of triangle})$, therefore BH $: $ GD $= (\text{area of pentagon}) : (\text{area of triangle})$, and $12 $ BH $: 20 $ GD $= (\text{surface of dod.}) : (\text{surface of icos.})$.
But $12 $ BH $= 10 $ BC, since BH $= 5 $ HC, and BC $= 6 $ HC; and $20 $ GD $= 10 $ DM; therefore
$$\text{(surface of dodecahedron)} : \text{(surface of icosahedron)} = \text{(side of cube)} : \text{(side of icosahedron)}.$$
“Next we have to prove that,
[Prop. 7] If any straight line whatever be cut in extreme and mean ratio, then, as is (1) the straight line the square on which is equal to the sum of the squares on the whole line and on the greater segment to (2) the straight line the square on which is equal to the sum of the squares on the whole line and on the lesser segment, so is (3) the side of the cube to (4) the side of the icosahedron.”
Let AHB be the circle circumscribing both the pentagon of the dodecahedron and the triangle of the icosahedron inscribed in the same sphere, C the centre of the circle, and CB any radius divided at D in extreme and mean ratio, CD being the greater segment.
CD is then the side of the decagon inscribed in the circle. [XIII. 9 and 5, converse]
Let E be the side of the icosahedron, F that of the dodecahedron, and G that of the cube, inscribed in the sphere.
Then E, F are the sides of the equilateral triangle and pentagon inscribed in the circle, and, if G is divided in extreme and mean ratio, the greater segment is equal to F. [XIII. 17, Por.]
Thus E $^2 = 3 $ BC $^2$, [XIII. 12] and CB $^2 + $ BD $^2 = 3 $ CD $^2$. [XIII. 4]
Therefore E $^2 : $ CB $^2 = ($ CB $^2 + $ BD $^2) : $ CD $^2$, or E $^2 : ($ CB $^2 + $ BD $^2) = $ CB $^2 : $ CD $^2 = $ G $^2 : $ F $^2$.
Therefore, alternately and inversely, G $^2 : $ E $^2 = $ F $^2 : ($ CB $^2 + $ BD $^2)$.
But F $^2 = $ BC $^2 + $ CD $^2$; for the square on the side of the pentagon is equal to the sum of the squares on the sides of the hexagon and decagon inscribed in the same circle. [XIII. 10]
Therefore G $^2 : $ E $^2 = ($ BC $^2 + $ CD $^2) : ($ CB $^2 + $ BD $^2)$, which is the result required.
It has now to be proved that
[Prop. 8] (Side of cube) : (side of icosahedron) = (content of dodecahedron) : (content of icosahedron).
Since equal circles circumscribe the pentagon of the dodecahedron and the triangle of the icosahedron inscribed in the same sphere, and in a sphere equal circular sections are equally distant from the centre, the perpendiculars from the centre of the sphere to the faces of the two solids are equal; in other words, the pyramids with the centre as vertex and the pentagons of the dodecahedron and the triangles of the icosahedron respectively as bases are of equal height.
Therefore the pyramids are to one another as their bases.
Thus
$$\text{(12 pentagons)} : \text{(20 triangles)} = \text{(12 pyramids on pentagons)} : \text{(20 pyramids on triangles)},$$
or
$$\text{(surface of dodecahedron)} : \text{(surface of icosahedron)} = \text{(content of dod.)} : \text{(content of icos.)}.$$
Therefore
$$\text{(content of dodecahedron)} : \text{(content of icosahedron)} = \text{(side of cube)} : \text{(side of icosahedron)}. \quad \text{[Prop. 6]}$$
Lemma.
If two straight lines be cut in extreme and mean ratio, the segments of both are in one and the same ratio.
Let AB be cut in extreme and mean ratio at C, AC being the greater segment; and let DE be cut in extreme and mean ratio at F, DF being the greater segment.
I say that AB $: $ AC $= $ DE $: $ DF.
Since AB $cdot $ BC $= $ AC $^2$, and DE $cdot $ EF $= $ DF $^2$, AB $cdot $ BC $: $ AC $^2 = $ DE $cdot $ EF $: $ DF $^2$, and $4 $ AB $cdot $ BC $: $ AC $^2 = 4 $ DE $cdot $ EF $: $ DF $^2$.
Componendo, $(4 $ AB $cdot $ BC $+ $ AC $^2) : $ AC $^2 = (4 $ DE $cdot $ EF $+ $ DF $^2) : $ DF $^2$, or $( $ AB $+ $ BC $)^2 : $ AC $^2 = ($ DE $+ $ EF $)^2 : $ DF $^2$; [II. 8] therefore $( $ AB $+ $ BC $) : $ AC $= ($ DE $+ $ EF $) : $ DF.
Componendo, $( $ AB $+ $ BC $+ $ AC $) : $ AC $= ($ DE $+ $ EF $+ $ DF $) : $ DF, or $2 $ AB $: $ AC $= 2 $ DE $: $ DF; that is, AB $: $ AC $= $ DE $: $ DF.
Summary of results.
If AB be any straight line divided at C in extreme and mean ratio, AC being the greater segment, and if we have a cube, a dodecahedron and an icosahedron inscribed in one and the same sphere, then:
(side of cube) : (side of icosahedron) = $\sqrt{(}$ AB $^2 + $ AC $^2) : \sqrt{(}$ AB $^2 + $ BC $^2)$;
(surface of dod.) : (surface of icos.) = (side of cube) : (side of icosahedron);
(content of dod.) : (content of icos.) = (surface of dod.) : (surface of icos.);
and (4) (content of dodecahedron) : (content of icos.) = $\sqrt{(}$ AB $^2 + $ AC $^2) : \sqrt{(}$ AB $^2 + $ BC $^2)$.
II. NOTE ON THE SO-CALLED “BOOK XV.”¶
The second of the two Books added to the genuine thirteen is also supplementary to the discussion of the regular solids, but is much inferior to the first, “Book XIV.” Its contents are of less interest and the exposition leaves much to be desired, being in some places obscure and in others actually inaccurate. It consists of three portions unequal in length. The first (Heiberg, Vol. v. pp. 40—48) shows how to inscribe certain of the regular solids in certain others, (a) a tetrahedron (“pyramid”) in a cube, (b) an octahedron in a tetrahedron (“pyramid”), (c) an octahedron in a cube, (d) a cube in an octahedron and (e) a dodecahedron in an icosahedron. The second portion (pp. 48—50) explains how to calculate the number of edges and the number of solid angles in the five solids respectively. The third (pp. 50—66) shows how to determine the angle of inclination between faces meeting in an edge of any one of the solids. The method is to construct an isosceles triangle with vertical angle equal to the said angle of inclination; from the middle point of any edge two perpendiculars are drawn to it, one in each of the two faces intersecting in that edge; these perpendiculars (forming an angle which is the inclination of the two faces to one another) are used to determine the two equal sides of an isosceles triangle, and the base of the triangle is easily found from the known properties of the particular solid. The rules for drawing the respective isosceles triangles are first given all together in general terms (pp. 50—52); and the special interest of the passage consists in the fact that the rules are attributed to “Isidorus our great teacher.” This Isidorus is no doubt Isidorus of Miletus, the architect of the Church of St Sophia at Constantinople (about 532 A.D.), whose pupil Eutocius also was; he is often referred to by Eutocius (Comm. on Archimedes) as ὁ Μιλήσιος μηχανικὸς Ἰσίδωρος ἡμέτερος διδάσκαλος. Thus the third portion of the Book at all events was written by a pupil of Isidorus in the sixth century. Kluge (De Euclidis elementorum libris qui feruntur XIV et XV, Leipzig, 1891) has closely examined the language and style of the three portions and conjectures that they may be the work of different authors; the first portion may, he thinks, date from the end of the third century (the time of Pappus), and the second portion too may be older than the third. Hultsch however (art. “Eukleides” in Pauly-Wissowa’s Real-Encyclopädie der classischen Altertumswissenschaft, 1907) does not think his arguments convincing.
It may be worth while to set out the particulars of Isidorus’ rules for constructing isosceles triangles with vertical angles equal respectively to the angles of inclination between faces meeting in an edge of the several regular solids. A certain base is taken, and then with its extremities as centres and a certain other straight line as radius two circles are drawn; their point of intersection determines the vertex of the particular isosceles triangle. In the case of the cube the triangle is of course right-angled; in the other cases the bases and the equal sides are as shown below.
Base of isosceles triangle |
Equal sides of isosceles triangle |
|
|---|---|---|
For the tetrahedron |
the side of a triangular face |
the perpendicular from the vertex of a triangular face to its base |
For the octahedron |
the diagonal of the square on one side of a triangular face |
ditto |
For the icosahedron |
the chord joining two non-consecutive angular points of the regular pentagon on an edge (the “pentagon of the icosahedron”) |
ditto |
For the dodecahedron |
the chord joining two non-consecutive angular points of a pentagonal face [$BC$ in the figure of Eucl. XIII. 17] |
the perpendicular from the middle point of the chord joining two non-consecutive angular points of a face to the parallel side of that face [$HX$ in the figure of Eucl. XIII. 17] |