§13.prop.16#2δεῖ δὴ αὐτὸ καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ δεῖξαι, ὅτι ἡ τοῦ εἰκοσαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάσσων.
It is now necessary to comprehend it in the given sphere, and to prove that the side of the icosahedron is the irrational straight line called minor.
ἐπεὶ γὰρ ἑξαγώνου ἐστὶν ἡ ΦΧ, δεκαγώνου δὲ ἡ ΧΩ, ἡ ΦΩ ἄρα ἄκρον καὶ μέσον λόγον τέμηται κατὰ τὸ Χ, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστιν ἡ ΦΧ·
For since PhiChi is the side of a hexagon, and ChiOmega of a decagon, therefore PhiOmega is cut in extreme and mean ratio at Chi, and its greater segment is PhiChi; therefore, as OmegaPhi is to PhiChi, so is PhiChi to ChiOmega.
ἔστιν ἄρα ὡς ἡ ΩΦ πρὸς τὴν ΦΧ, οὕτως ἡ ΦΧ πρὸς τὴν ΧΩ. ἴση δὲ ἡ μὲν ΦΧ τῇ ΦΕ, ἡ δὲ ΧΩ τῇ ΦΨ· ἔστιν ἄρα ὡς ἡ ΩΦ πρὸς τὴν ΦΕ, οὕτως ἡ ΕΦ πρὸς τὴν ΦΨ. καί εἰσιν ὀρθαὶ αἱ ὑπὸ ΩΦΕ, ΕΦΨ γωνίαι·
And PhiChi is equal to PhiE, and ChiOmega to PhiPsi; therefore, as OmegaPhi is to PhiE, so is EPhi to PhiPsi.
ἐὰν ἄρα ἐπιζεύξωμεν τὴν ΕΩ εὐθεῖαν, ὀρθὴ ἔσται ἡ ὑπὸ ΨΕΩ γωνία διὰ τὴν ὁμοιότητα τῶν ΨΕΩ, ΦΕΩ τριγώνων.
And the angles OmegaPhiE, EPhiPsi are right; therefore, if we join the straight line EOmega, the angle PsiEOmega will be right because of the similarity of the triangles PsiEOmega, PhiEOmega.
διὰ τὰ αὐτὰ δὴ ἐπεί ἐστιν ὡς ἡ ΩΦ πρὸς τὴν ΦΧ, οὕτως ἡ ΦΧ πρὸς τὴν ΧΩ, ἴση δὲ ἡ μὲν ΩΦ τῇ ΨΧ, ἡ δὲ ΦΧ τῇ ΧΠ, ἔστιν ἄρα ὡς ἡ ΨΧ πρὸς τὴν ΧΠ, οὕτως ἡ ΠΧ πρὸς τὴν ΧΩ. καὶ διὰ τοῦτο πάλιν ἐὰν ἐπιζεύξωμεν τὴν ΠΨ, ὀρθὴ ἔσται ἡ πρὸς τῷ Π γωνία· τὸ ἄρα ἐπὶ τῆς ΨΩ γραφόμενον ἡμικύκλιον ἥξει καὶ διὰ τοῦ Π. καὶ ἐὰν μενούσης τῆς ΨΩ περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, ἥξει καὶ διὰ τοῦ Π καὶ τῶν λοιπῶν σημείων τοῦ εἰκοσαέδρου, καὶ ἔσται σφαίρᾳ περιειλημμένον τὸ εἰκοσάεδρον.
For the same reasons, since as OmegaPhi is to PhiChi, so is PhiChi to ChiOmega, and OmegaPhi is equal to PsiChi, and PhiChi to ChiP, therefore, as PsiChi is to ChiP, so is PChi to ChiOmega. And because of this, again, if we join PPsi, the angle at P will be right; therefore the semicircle described on PsiOmega will also pass through P. And if, while PsiOmega remains fixed, the semicircle is carried round and restored again to the same position from which it began to be moved, it will also pass through P and the remaining points of the icosahedron, and the icosahedron will be comprehended in a sphere.
λέγω δή, ὅτι καὶ τῇ δοθείσῃ.
I say then, that it is also comprehended in the given sphere.
τετμήσθω γὰρ ἡ ΦΧ δίχα κατὰ τὸ Α#.
For let PhiChi be bisected at Α#.
καὶ ἐπεὶ εὐθεῖα γραμμὴ ἡ ΦΩ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Χ, καὶ τὸ ἔλασσον αὐτῆς τμῆμά ἐστιν ἡ ΩΧ, ἡ ἄρα ΩΧ προσλαβοῦσα τὴν ἡμίσειαν τοῦ μείζονος τμήματος τὴν ΧΑ# πενταπλάσιον δύναται τοῦ ἀπὸ τῆς ἡμισείας τοῦ μείζονος τμήματος· πενταπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΩΑ# τοῦ ἀπὸ τῆς Α#Χ. καί ἐστι τῆς μὲν ΩΑ# διπλῆ ἡ ΩΨ, τῆς δὲ Α#Χ διπλῆ ἡ ΦΧ· πενταπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΩΨ τοῦ ἀπὸ τῆς ΧΦ. καὶ ἐπεὶ τετραπλῆ ἐστιν ἡ ΑΓ τῆς ΓΒ, πενταπλῆ ἄρα ἐστὶν ἡ ΑΒ τῆς ΒΓ. ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΔ· πενταπλάσιον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΒΔ. ἐδείχθη δὲ καὶ τὸ ἀπὸ τῆς ΩΨ πενταπλάσιον τοῦ ἀπὸ τῆς ΦΧ. καί ἐστιν ἴση ἡ ΔΒ τῇ ΦΧ·
And since the straight line PhiOmega is cut in extreme and mean ratio at Chi, and its lesser segment is OmegaChi, therefore OmegaChi with the half of the greater segment ChiΑ# is five times in square the half of the greater segment; therefore the square on OmegaΑ# is five times the square on Α#Chi. And OmegaPsi is double of OmegaΑ#, and PhiChi is double of Α#Chi; therefore the square on OmegaPsi is five times the square on ChiPhi. And since AC is quadruple of GB, therefore AB is quintuple of BC. But as AB is to BC, so is the square on AB to the square on BD; therefore the square on AB is five times the square on BD. And the square on OmegaPsi was also proved to be five times the square on PhiChi.
ἑκατέρα γὰρ αὐτῶν ἴση ἐστὶ τῇ ἐκ τοῦ κέντρου τοῦ ΕΖΗΘΚ κύκλου· ἴση ἄρα καὶ ἡ ΑΒ τῇ ΨΩ. καί ἐστιν ἡ ΑΒ ἡ τῆς δοθείσης σφαίρας διάμετρος·
And DB is equal to PhiChi; for each of them is equal to the radius of the circle EZHThK; therefore AB is also equal to PsiOmega.
καὶ ἡ ΨΩ ἄρα ἴση ἐστὶ τῇ τῆς δοθείσης σφαίρας διαμέτρῳ.
And AB is the diameter of the given sphere; therefore PsiOmega is also equal to the diameter of the given sphere.
τῇ ἄρα δοθείσῃ σφαίρᾳ περιείληπται τὸ εἰκοσάεδρον.
Therefore the icosahedron is comprehended in the given sphere.
λέγω δή, ὅτι ἡ τοῦ εἰκοσαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάττων.
I say then, that the side of the icosahedron is the irrational straight line called minor.
ἐπεὶ γὰρ ῥητή ἐστιν ἡ τῆς σφαίρας διάμετρος, καί ἐστι δυνάμει πενταπλασίων τῆς ἐκ τοῦ κέντρου τοῦ ΕΖΗΘΚ κύκλου, ῥητὴ ἄρα ἐστὶ καὶ ἡ ἐκ τοῦ κέντρου τοῦ ΕΖΗΘΚ κύκλου· ὥστε καὶ ἡ διάμετρος αὐτοῦ ῥητή ἐστιν.
For since the diameter of the sphere is rational, and is in square five times the radius of the circle EZHThK, therefore the radius of the circle EZHThK is also rational; so that its diameter is also rational.
ἐὰν δὲ εἰς κύκλον ῥητὴν ἔχοντα τὴν διάμετρον πεντάγωνον ἰσόπλευρον ἐγγραφῇ, ἡ τοῦ πενταγώνου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάττων.
And if an equilateral pentagon is inscribed in a circle which has its diameter rational, the side of the pentagon is the irrational straight line called minor.
ἡ δὲ τοῦ ΕΖΗΘΚ πενταγώνου πλευρὰ ἡ τοῦ εἰκοσαέδρου ἐστίν. ἡ ἄρα τοῦ εἰκοσαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάττων.
And the side of the pentagon EZHThK is the side of the icosahedron; therefore the side of the icosahedron is the irrational straight line called minor.
Πόρισμα
ἐκ δὴ τούτου φανερόν, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει πενταπλασίων ἐστὶ τῆς ἐκ τοῦ κέντρου τοῦ κύκλου, ἀφʼ οὗ τὸ εἰκοσάεδρον ἀναγέγραπται, καὶ ὅτι ἡ τῆς σφαίρας διάμετρος σύγκειται ἔκ τε τῆς τοῦ ἑξαγώνου καὶ δύο τῶν τοῦ δεκαγώνου τῶν εἰς τὸν αὐτὸν κύκλον ἐγγραφομένων.
Porism From this it is manifest that the diameter of the sphere is in square five times the radius of the circle from which the icosahedron has been described, and that the diameter of the sphere is composed of the side of the hexagon and two of the sides of the decagon inscribed in the same circle.
ὅπερ ἔδει δεῖξαι.
Which it was required to prove.