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Euclid · Elements §13.prop.13#1

Construction of a Tetrahedron in a Given Sphere

Passage 306 of 316 · Greek

Summary

As the first step in constructing a regular tetrahedron (pyramid) inscribed in a given sphere, the base equilateral triangle and the height are determined using a semicircle and a perpendicular line of specific ratios, and the equality of all edges of the pyramid is proven.

§13.prop.13#1πυραμίδα συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος δυνάμει ἡμιολία ἐστὶ τῆς πλευρᾶς τῆς πυραμίδος.
construct a pyramid and comprehend it in a given sphere, and to prove that the diameter of the sphere is one and a half times in square of the side of the pyramid.
Ἐκκείσθω ἡ τῆς δοθείσης σφαίρας διάμετρος ἡ ΑΒ, καὶ τετμήσθω κατὰ τὸ Γ σημεῖον, ὥστε διπλασίαν εἶναι τὴν ΑΓ τῆς ΓΒ· καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ ἤχθω ἀπὸ τοῦ Γ σημείου τῇ ΑΒ πρὸς ὀρθὰς ἡ ΓΔ, καὶ ἐπεζεύχθω ἡ ΔΑ· καὶ ἐκκείσθω κύκλος ὁ ΕΖΗ ἴσην ἔχων τὴν ἐκ τοῦ κέντρου τῇ ΔΓ, καὶ ἐγγεγράφθω εἰς τὸν ΕΖΗ κύκλον τρίγωνον ἰσόπλευρον τὸ ΕΖΗ· καὶ εἰλήφθω τὸ κέντρον τοῦ κύκλου τὸ Θ σημεῖον, καὶ ἐπεζεύχθωσαν αἱ ΕΘ, ΘΖ, ΘΗ· καὶ ἀνεστάτω ἀπὸ τοῦ Θ σημείου τῷ τοῦ ΕΖΗ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΘΚ, καὶ ἀφῃρήσθω ἀπὸ τῆς ΘΚ τῇ ΑΓ εὐθείᾳ ἴση ἡ ΘΚ, καὶ ἐπεζεύχθωσαν αἱ ΚΕ, ΚΖ, ΚΗ. καὶ ἐπεὶ ἡ ΚΘ ὀρθή ἐστι πρὸς τὸ τοῦ ΕΖΗ κύκλου ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ τοῦ ΕΖΗ κύκλου ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας.
Let the diameter of the given sphere be set out, namely AB, and let it be cut at the point G so that AG is double of GB; and let the semicircle ADB be described on AB, and let GD be drawn from the point G at right angles to AB, and let DA be joined; and let a circle EZH be set out having its radius equal to DG, and let there be inscribed in the circle EZH an equilateral triangle EZH; and let the center of the circle be taken, namely the point Th, and let ETh, ThZ, ThH be joined; and let ThK be set up from the point Th at right angles to the plane of the circle EZH, and let ThK, equal to the straight line AG, be cut off from ThK, and let KE, KZ, KH be joined. And since KTh is at right angles to the plane of the circle EZH, therefore it will also make right angles with all the straight lines which touch it and are in the plane of the circle EZH.
ἅπτεται δὲ αὐτῆς ἑκάστη τῶν ΘΕ, ΘΖ, ΘΗ· ἡ ΘΚ ἄρα πρὸς ἑκάστην τῶν ΘΕ, ΘΖ, ΘΗ ὀρθή ἐστιν.
And each of ThE, ThZ, ThH touches it; therefore ThK is at right angles to each of ThE, ThZ, ThH.
καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΑΓ τῇ ΘΚ, ἡ δὲ ΓΔ τῇ ΘΕ, καὶ ὀρθὰς γωνίας περιέχουσιν, βάσις ἄρα ἡ ΔΑ βάσει τῇ ΚΕ ἐστιν ἴση.
And since AG is equal to ThK, and GD to ThE, and they contain right angles, therefore the base DA is equal to the base KE.
διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ΚΖ, ΚΗ τῇ ΔΑ ἐστιν ἴση· αἱ τρεῖς ἄρα αἱ ΚΕ, ΚΖ, ΚΗ ἴσαι ἀλλήλαις εἰσίν.
For the same reasons, indeed, each of KZ, KH is also equal to DA; therefore the three KE, KZ, KH are equal to one another.
καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΑΓ τῆς ΓΒ, τριπλῆ ἄρα ἡ ΑΒ τῆς ΒΓ. ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΔ πρὸς τὸ ἀπὸ τῆς ΔΓ, ὡς ἑξῆς δειχθήσεται.
And since AG is double of GB, therefore AB is triple of BG. And as AB is to BC, so is the square on AD to the square on DG, as will be proved afterwards.
τριπλάσιον ἄρα τὸ ἀπὸ τῆς ΑΔ τοῦ ἀπὸ τῆς ΔΓ. ἔστι δὲ καὶ τὸ ἀπὸ τῆς ΖΕ τοῦ ἀπὸ τῆς ΕΘ τριπλάσιον, καί ἐστιν ἴση ἡ ΔΓ τῇ ΕΘ· ἴση ἄρα καὶ ἡ ΔΑ τῇ ΕΖ. ἀλλὰ ἡ ΔΑ ἑκάστῃ τῶν ΚΕ, ΚΖ, ΚΗ ἐδείχθη ἴση·
Therefore the square on AD is triple of the square on DG. But the square on ZE is also triple of the square on ETh, and DG is equal to ETh; therefore DA is also equal to EZ.
καὶ ἑκάστη ἄρα τῶν ΕΖ, ΖΗ, ΗΕ ἑκάστῃ τῶν ΚΕ, ΚΖ, ΚΗ ἐστιν ἴση·
But DA was proved equal to each of KE, KZ, KH; therefore each of EZ, ZH, HE is also equal to each of KE, KZ, KH; therefore the four triangles EZH, KEZ, KZH, KEH are equilateral.
ἰσόπλευρα ἄρα ἐστὶ τὰ τέσσαρα τρίγωνα τὰ ΕΖΗ, ΚΕΖ, ΚΖΗ, ΚΕΗ. πυραμὶς ἄρα συνέσταται ἐκ τεσσάρων τριγώνων ἰσοπλεύρων, ἧς βάσις μέν ἐστι τὸ ΕΖΗ τρίγωνον, κορυφὴ δὲ τὸ Κ σημεῖον.
Therefore a pyramid has been constructed out of four equilateral triangles, of which the base is the triangle EZH, and the vertex is the point K.
δεῖ δὴ αὐτὴν καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ δεῖξαι, ὅτι ἡ τῆς σφαίρας διάμετρος ἡμιολία ἐστὶ δυνάμει τῆς πλευρᾶς τῆς πυραμίδος.
It is now necessary to comprehend it in the given sphere and to prove that the diameter of the sphere is one and a half times in square of the side of the pyramid.

Notes

  1. §13.prop.13#1πυραμίδα συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν — The infinitives συστήσασθαι, περιλαβεῖν, and δεῖξαι are used to state the objective of the proposition (problem), functioning as infinitives of purpose or command ("to construct...", "to comprehend...").
  2. §13.prop.13#1ἀφῃρήσθω ἀπὸ τῆς ΘΚ τῇ ΑΓ εὐθείᾳ ἴση ἡ ΘΚ — The first ΘΚ refers to the perpendicular ray set up from the point Θ, while the subject ΘΚ refers to the specific segment ΘΚ cut off from that ray (equal in length to ΑΓ). In Greek geometry, the same letter combination often designates both an indefinite ray and a segment defined on it.
  3. §13.prop.13#1ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΔ πρὸς τὸ ἀπὸ τῆς ΔΓ — This proportion is based on the properties of the right-angled triangle ΑΔΒ inscribed in the semicircle ΑΔΒ. When the perpendicular ΔΓ is drawn to the hypotenuse ΑΒ, the relation ΑΒ : ΒΓ = ΑΔ^2 : ΔΓ^2 holds by similarity (cf. Elements VI.8). As the text states, this will be proved in detail later in this proposition.

Cite this passage

Euclid, Elements §13.prop.13#1. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:13.prop.13%231

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