Publication No. lit math
Modern Title Geometrische Aufgabe
Summary geometrical exercise: drawing of a regular icosahedron into a sphere of given diameter (cf. Euclid Stoicheia XIII 16)

### Edition

convex

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. 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 9 ἐπ' εὐθείας τῆι λψ ἡ λω, ἴση τῆι τοῦ δεκαγώνου πλευρᾶι. 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

convex
 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 1cf. 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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