Edition
τῆι ηζἰσόπλευρον ἄρα τὸ ζρη. διὰ δὲ ταῦτα καὶ τὰ ησθ|
2 | | θτι ιυκ κφζ. ἴση δὲ καὶ ἡ ρσ τῆ ι ηθ καὶ ἴσας ἑδείξαμεν|
3 | | τὰς ρη σθ. ἰσόπλευρον ἄρ' ἑστὶν ἕκαστον τῶν μεταξὺ τριγώνων.
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4 | | ἴσας γὰρ καὶ παραλλήλους τῆς χλ τὰς ζμ ην θξ ιο κπ ἐγράψαμεν. ἑξα-|
5 | | γώνου ἄρα καὶ ἡ χρ, ἀλλὰ καὶ ἡ ψχδεκαγώνου, ὥστ' εἶναι τὰς ψρ ψσ|
6 | | πενταγώνου. ἴση ἄρ' ἡ φρ τῆι ρσ, διὸ
ἰσόπλευρά ἐστι τὰ ψφρ ψρσ|
7 | | τρίγωνα. διὰ ταῦτα δὲ καὶ ἔκαστον τῶν τὸ ψ ὡς κορυφὴν|
8 | | ἐχόντων ἰσόπλευρά ἐστιν. πάλιν
ἐκβεβλήσθω
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9 | | ἐπ' εὐθείας τῆι λψ ἡ λω, ἴση τῆι τοῦ δεκαγώνου πλευρᾶι.
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10 | | ἤχθω δὴ ἀπὸ τοῦ λ κέντρου ἡ λζ ἑξαγώνου, δεκαγώνου δ' ἦν ἡ|
11 | | λω, πενταγώνου ἄρ' ἡ ζω, διὰ ταῦτα δ' ἰσόπλευρόν ἐστι|
12 | | τὰ τρίγωνον τὸ ωζη, διὰ ταὐτὰ
τὸ ζηθικω στερεὸν σχἠμα ἐκ|
13 | | πρὸς τῶι ω τὰς κορυφὰς ἐχόντων πέντε|
14 | | τριγώνων ἰσοπλεύρων πεποίηται.|
15 | | δύο δεκαγώνου πλευραὶ καὶ μία ἑξαγώνου τῆς|
16 | | δοθείσης σφαίρας
τὴν διάμετρον συμπληροῦσιν. καὶ ἡ τῆς|
17 | | δοθείσης σφαίρας διάμετρος δυνάμει πενταπλασίων τῆς ἡμι-|
18 | | διαμέτρου
τοῦ κύκλου ἐστιν τοῦ τὸ πεντάγωνον|
19 | | περιλαμβανομένου, ἀφ' οὗ τὸ εἰκοσάεδρον ἀναγέγραπται.
Translation
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1 | | ΗΖ ... So the triangle ZΡH is equilateral. Thus also the triangles ΗΣΘ,|
2 | | ΘΤΙ, ΙΥΚ, ΚΦΖ. But equal is also ΡΣ to the line ΗΘ, and as equal we have shown|
3 | | the lines ΡΗ and ΣΘ. So each of the triangles lying between (scil. the circles) is equilateral, |
4 | | because we had drawn the perpendiculars ΖΜ, ΗΝ, ΘΞ, ΙΟ, ΚΠ equal and parallel to the line XΛ. |
5 | | That is why XP is also (scil. equal to one) hexagonal side. But also ΨΧ was (scil. equal to a) decagonal side, so that ΨΡ and ΨΣ |
6 | | are pentagonal sides. Equal was <also> ΦΡ to side ΡΣ, therefore ΨΦΡ and ΨΡΣ are equilateral|
7 | | triangles. Therefore each of the triangles whose common apex is Ψ is also|
8 | | equilateral. Furthermore, let the distance|
9 | | ΛΩ be drawn as an extension of the distance ΛΨ, equal to the side of the decagon. |
11 | | Let the radius ΛΖ equal to the hexagon side be drawn from the centre of the circle Λ, but (scil. equal) to the decagon side was|
12 | | ΛΩ , so ΖΩ is a pentagon side. Therefore equilateral is|
13 | | the triangle ΩΖΗ. For the same reason, the spatial figure ΖΗΘΙΚΩ is formed by|
14 | | five equilateral triangles having their apex in Ω. |
15 | | Two decagonal sides and one hexagonal side together|
16 | | give the diameter of the given sphere cf. Euclid El. XIII 16 corollary, |
17 | | and the diameter of the given sphere squared is five times |
18 | | the radius of the given circle squared,|
19 | | drawn around the pentagon on which the icosahedron was built. (after Müller/Mau 1960)
1 cf. Euclid El. XIII 16 corollary
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