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XIII. 18]

PROPOSITION 18

509

therefore one of them, as the angle AFB, is one right angle less a fifth;

therefore the remaining angles FAB, ABF consist of one right angle and a fifth.

But the angle FAB is equal to the angle FBC;

therefore the whole angle ABC of the pentagon consists of one right angle and a fifth.

Q. E. D.

We have seen in the preceding notes that, if be the radius of the sphere circumscribing the five solid figures,

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Euclid here exhibits the edges of all the five regular solids in one figure.

(1) Make AD equal to 2DB.

Thus

and

therefore

Thus

(2)

Therefore

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AF= √.2r=√16. r = (edge of tetrahedron).

3

AB2: BF2 = AB : BD

=3:1.

BF2 = AB2,

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(4) Draw AG perpendicular and equal to AB. Join GC, meeting the semicircle in H, and draw HK perpendicular to AB.

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It follows that KL (= √√.r) is the radius of, or the side of t hexagon in, the circle containing the pentagonal sections of the icos

And, since

But

2r= (side of hexagon) + 2 (side of decagon in same circle) [XIII.

=

AK= LB (side of decagon in the said circle).
LM = HK = KL = (side of hexagon in circle).

Therefore LM2 + LB2 (= BM2) = (side of pentagon in circle)

and

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whence

and

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BM: == √10 (5 — √5) = (edge of icosahedron).]

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(5) Cut BF (the edge of the cube) in extreme and mean ratio Then, if BN be the greater segment,

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whence

σ =

2 = 3. X,

4r2 : t2 : 02 : c2 = 6 : 4 : 3 : 2,

and the ratios between 2r, t, o, c are all rational (in Euclid's sense).

The ratios between these and the edges of the icosahedron and the dodecahedron are irrational.

(7) To prove that

i.e. that

or

(edge of icosahedron) > (edge of dodecahedron),

By similar As FDB, AFB,

But

therefore

By hypothesis,

therefore

and, a fortiori,

MB > NB.

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Now LK is the side of a hexagon, and AK the side of a decagon in the same circle;

therefore, when AL is divided in extreme and mean ratio, KL is the greater segment.

And, when BF is divided in extreme and mean ratio, BN is the greater segment.

or

Therefore, since

And therefore, a fortiori,

AL> BF,

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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. 1. 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. I.] "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."

and half the circumference of the circle is the arc

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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.

66

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."

H. E. III.

33

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