§13.prop.16#1εἰκοσάεδρον συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν, ᾗ καὶ τὰ προειρημένα σχήματα, καὶ δεῖξαι, ὅτι ἡ τοῦ εἰκοσαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἐλάττων.
To construct an icosahedron and to comprehend it in a sphere, as in the case of the aforementioned figures, and to prove that the side of the icosahedron is the irrational straight line called minor.
Ἐκκείσθω ἡ τῆς δοθείσης σφαίρας διάμετρος ἡ ΑΒ καὶ τετμήσθω κατὰ τὸ Γ ὥστε τετραπλῆν εἶναι τὴν ΑΓ τῆς ΓΒ, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ ἤχθω ἀπὸ τοῦ Γ τῇ ΑΒ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἡ ΓΔ, καὶ ἐπεζεύχθω ἡ ΔΒ, καὶ ἐκκείσθω κύκλος ὁ ΕΖΗΘΚ, οὗ ἡ ἐκ τοῦ κέντρου ἴση ἔστω τῇ ΔΒ, καὶ ἐγγεγράφθω εἰς τὸν ΕΖΗΘΚ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον τὸ ΕΖΗΘΚ, καὶ τετμήσθωσαν αἱ ΕΖ, ΖΗ, ΗΘ, ΘΚ, ΚΕ περιφέρειαι δίχα κατὰ τὰ Λ, Μ, Ν, Ξ, Ο σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΛΜ, ΜΝ, ΝΞ, ΞΟ, ΟΛ, ΕΟ. ἰσόπλευρον ἄρα ἐστὶ καὶ τὸ ΛΜΝΞΟ πεντάγωνον, καὶ δεκαγώνου ἡ ΕΟ εὐθεῖα.
Let the diameter AB of the given sphere be laid down, and let it be cut at C so that AC is quadruple of GB, and let the semicircle ADB be described on AB, and let a straight line GD be drawn from C at right angles to AB, and let DB be joined. And let a circle EZHThK be laid down whose radius is equal to DB, and let an equilateral and equiangular pentagon EZHThK be inscribed in the circle EZHThK, and let the circumferences EZ, ZH, HTh, ThK, KE be bisected at the points L, M, N, Xi, O, and let LM, MN, NXi, XiO, OL, EO be joined. Therefore the pentagon LMNOXi is also equilateral, and the straight line EO is the side of a decagon.
καὶ ἀνεστάτωσαν ἀπὸ τῶν Ε, Ζ, Η, Θ, Κ σημείων τῷ τοῦ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς γωνίας εὐθεῖαι αἱ ΕΠ, ΖΡ, ΗΣ, ΘΤ, ΚΥ ἴσαι οὖσαι τῇ ἐκ τοῦ κέντρου τοῦ ΕΖΗΘΚ κύκλου, καὶ ἐπεζεύχθωσαν αἱ ΠΡ, ΡΣ, ΣΤ, ΤΥ, ΥΠ, ΠΛ, ΛΡ, ΡΜ, ΜΣ, ΣΝ, ΝΤ, ΤΞ, ΞΥ, ΥΟ, ΟΠ. καὶ ἐπεὶ ἑκατέρα τῶν ΕΠ, ΚΥ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν, παράλληλος ἄρα ἐστὶν ἡ ΕΠ τῇ ΚΥ. ἔστι δὲ αὐτῇ καὶ ἴση· αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπιζευγνύουσαι ἐπὶ τὰ αὐτὰ μέρη εὐθεῖαι ἴσαι τε καὶ παράλληλοί εἰσιν.
And let straight lines EP, ZR, HS, TT, KY, equal to the radius of the circle EZHThK, be set up from the points E, Z, H, Th, K at right angles to the plane of the circle, and let PR, RS, ST, TY, YP, PL, LR, RM, MS, SN, NT, TXi, XiY, YO, OP be joined. And since each of EP, KY is at right angles to the same plane, therefore EP is parallel to KY. And it is also equal to it; and the straight lines joining equal and parallel straight lines on the same sides are equal and parallel.
ἡ ΠΥ ἄρα τῇ ΕΚ ἴση τε καὶ παράλληλός ἐστιν.
Therefore PY is equal and parallel to EK.
πενταγώνου δὲ ἰσοπλεύρου ἡ ΕΚ· πενταγώνου ἄρα ἰσοπλεύρου καὶ ἡ ΠΥ τοῦ εἰς τὸν ΕΖΗΘΚ κύκλον ἐγγραφομένου.
And EK is the side of an equilateral pentagon; therefore PY is also the side of an equilateral pentagon inscribed in the circle EZHThK.
διὰ τὰ αὐτὰ δὴ καὶ ἑκάστη τῶν ΠΡ, ΡΣ, ΣΤ, ΤΥ πενταγώνου ἐστὶν ἰσοπλεύρου τοῦ εἰς τὸν ΕΖΗΘΚ κύκλον ἐγγραφομένου· ἰσόπλευρον ἄρα τὸ ΠΡΣΤΥ πεντάγωνον.
For the same reasons, each of PR, RS, ST, TY is also the side of an equilateral pentagon inscribed in the circle EZHThK; therefore the pentagon PRSTY is equilateral.
καὶ ἐπεὶ ἑξαγώνου μέν ἐστιν ἡ ΠΕ, δεκαγώνου δὲ ἡ ΕΟ, καί ἐστιν ὀρθὴ ἡ ὑπὸ ΠΕΟ, πενταγώνου ἄρα ἐστὶν ἡ ΠΟ· ἡ γὰρ τοῦ πενταγώνου πλευρὰ δύναται τήν τε τοῦ ἑξαγώνου καὶ τὴν τοῦ δεκαγώνου τῶν εἰς τὸν αὐτὸν κύκλον ἐγγραφομένων.
And since PE is the side of a hexagon, and EO of a decagon, and the angle PEO is right, therefore PO is the side of a pentagon; for the square on the side of the pentagon is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the same circle.
διὰ τὰ αὐτὰ δὴ καὶ ἡ ΟΥ πενταγώνου ἐστὶ πλευρά.
For the same reasons, OY is also the side of a pentagon.
ἔστι δὲ καὶ ἡ ΠΥ πενταγώνου· ἰσόπλευρον ἄρα ἐστὶ τὸ ΠΟΥ τρίγωνον.
And PY is also the side of a pentagon; therefore the triangle POY is equilateral.
διὰ τὰ αὐτὰ δὴ καὶ ἕκαστον τῶν ΠΛΡ, ΡΜΣ, ΣΝΤ, ΤΞΥ ἰσόπλευρόν ἐστιν.
For the same reasons, each of PLR, RMS, SNT, TXiY is also equilateral.
καὶ ἐπεὶ πενταγώνου ἐδείχθη ἑκατέρα τῶν ΠΛ, ΠΟ, ἔστι δὲ καὶ ἡ ΛΟ πενταγώνου, ἰσόπλευρον ἄρα ἐστὶ τὸ ΠΛΟ τρίγωνον.
And since each of PL, PO was proved to be the side of a pentagon, and LO is also the side of a pentagon, therefore the triangle PLO is equilateral.
διὰ τὰ αὐτὰ δὴ καὶ ἕκαστον τῶν ΛΡΜ, ΜΣΝ, ΝΤΞ, ΞΥΟ τριγώνων ἰσόπλευρόν ἐστιν.
For the same reasons, each of the triangles LRM, MSN, NTXi, XiYO is also equilateral.
εἰλήφθω τὸ κέντρον τοῦ ΕΖΗ ΘΚ κύκλου τὸ Φ σημεῖον· καὶ ἀπὸ τοῦ Φ τῷ τοῦ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς ἀνεστάτω ἡ ΦΩ, καὶ ἐκβεβλήσθω ἐπὶ τὰ ἕτερα μέρη ὡς ἡ ΦΨ, καὶ ἀφῃρήσθω ἑξαγώνου μὲν ἡ ΦΧ, δεκαγώνου δὲ ἑκατέρα τῶν ΦΨ, ΧΩ, καὶ ἐπεζεύχθωσαν αἱ ΠΩ, ΠΧ, ΥΩ, ΕΦ, ΛΦ, ΛΨ, ΨΜ. καὶ ἐπεὶ ἑκατέρα τῶν ΦΧ, ΠΕ τῷ τοῦ κύκλου ἐπιπέδῳ πρὸς ὀρθάς ἐστιν, παράλληλος ἄρα ἐστὶν ἡ ΦΧ τῇ ΠΕ. εἰσὶ δὲ καὶ ἴσαι· καὶ αἱ ΕΦ, ΠΧ ἄρα ἴσαι τε καὶ παράλληλοί εἰσιν.
Let the point Phi be taken as the center of the circle EZHThK; and let PhiOmega be set up from Phi at right angles to the plane of the circle, and let it be produced to the other side as PhiPsi, and let PhiChi be cut off as the side of a hexagon, and each of PhiPsi, ChiOmega as the side of a decagon, and let POmega, PChi, YOmega, EPhi, LPhi, LPsi, PsiM be joined. And since each of PhiChi, PE is at right angles to the plane of the circle, therefore PhiChi is parallel to PE. And they are also equal; therefore EPhi, PChi are also equal and parallel.
ἑξαγώνου δὲ ἡ ΕΦ· ἑξαγώνου ἄρα καὶ ἡ ΠΧ. καὶ ἐπεὶ ἑξαγώνου μέν ἐστιν ἡ ΠΧ, δεκαγώνου δὲ ἡ ΧΩ, καὶ ὀρθή ἐστιν ἡ ὑπὸ ΠΧΩ γωνία, πενταγώνου ἄρα ἐστὶν ἡ ΠΩ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΥΩ πενταγώνου ἐστίν, ἐπειδήπερ, ἐὰν ἐπιζεύξωμεν τὰς ΦΚ, ΧΥ, ἴσαι καὶ ἀπεναντίον ἔσονται, καί ἐστιν ἡ ΦΚ ἐκ τοῦ κέντρου οὖσα ἑξαγώνου·
And EPhi is the side of a hexagon; therefore PChi is also the side of a hexagon. And since PChi is the side of a hexagon, and ChiOmega of a decagon, and the angle PChiOmega is right, therefore POmega is the side of a pentagon. For the same reasons, YOmega is also the side of a pentagon, since indeed, if we join PhiK, XY, they will be equal and opposite, and PhiK, being the radius, is the side of a hexagon; therefore XY is also the side of a hexagon.
ἑξαγώνου ἄρα καὶ ἡ ΧΥ. δεκαγώνου δὲ ἡ ΧΩ, καὶ ὀρθὴ ἡ ὑπὸ ΥΧΩ·
And ChiOmega is the side of a decagon, and the angle YChiOmega is right; therefore YOmega is the side of a pentagon.
πενταγώνου ἄρα ἡ ΥΩ. ἔστι δὲ καὶ ἡ ΠΥ πενταγώνου· ἰσόπλευρον ἄρα ἐστὶ τὸ ΠΥΩ τρίγωνον.
And PY is also the side of a pentagon; therefore the triangle PYOmega is equilateral.
διὰ τὰ αὐτὰ δὴ καὶ ἕκαστον τῶν λοιπῶν τριγώνων, ὧν βάσεις μέν εἰσιν αἱ ΠΡ, ΡΣ, ΣΤ, ΤΥ εὐθεῖαι, κορυφὴ δὲ τὸ Ω σημεῖον, ἰσόπλευρόν ἐστιν.
For the same reasons, each of the remaining triangles also, of which the straight lines PR, RS, ST, TY are bases, and the point Omega is the apex, is equilateral.
πάλιν, ἐπεὶ ἑξαγώνου μὲν ἡ ΦΛ, δεκαγώνου δὲ ἡ ΦΨ, καὶ ὀρθή ἐστιν ἡ ὑπὸ ΛΦΨ γωνία, πενταγώνου ἄρα ἐστὶν ἡ ΛΨ. διὰ τὰ αὐτὰ δὴ ἐὰν ἐπιζεύξωμεν τὴν ΜΦ οὖσαν ἑξαγώνου, συνάγεται καὶ ἡ ΜΨ πενταγώνου.
Again, since PhiL is the side of a hexagon, and PhiPsi of a decagon, and the angle LPhiPsi is right, therefore LPsi is the side of a pentagon. For the same reasons, if we join MPhi, which is the side of a hexagon, it is also inferred that MPsi is the side of a pentagon.
ἔστι δὲ καὶ ἡ ΛΜ πενταγώνου· ἰσόπλευρον ἄρα ἐστὶ τὸ ΛΜΨ τρίγωνον.
And LM is also the side of a pentagon; therefore the triangle LMPsi is equilateral.
ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἕκαστον τῶν λοιπῶν τριγώνων, ὧν βάσεις μέν εἰσιν αἱ ΜΝ, ΝΞ, ΞΟ, ΟΛ, κορυφὴ δὲ τὸ Ψ σημεῖον, ἰσόπλευρόν ἐστιν.
Similarly it will also be proved that each of the remaining triangles, of which MN, NXi, XiO, OL are bases, and the point Psi is the apex, is equilateral.
συνέσταται ἄρα εἰκοσάεδρον ὑπὸ εἴκοσι τριγώνων ἰσοπλεύρων περιεχόμενον.
Therefore an icosahedron has been constructed, comprehended by twenty equilateral triangles.