| Ostr. Berlin P. 12002 (More information about the Object) |
| Publication No. |
lit. math |
| Modern Title |
Geometrische Aufgaben |
| Summary |
geometrical exercise: mathematical proof of the equality and ratio of the squares over the sides of various polygons (pentagons, hexagons, decagons) in a circle (cf. Euclid Stoicheia XIII 10); corrections in lines 12 and 14 |
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Edition
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1 | | . .|
2 | | ἀποδείκνυται ἡ πρότασις, ὅτι ἡ τοῦ|
3 | | πενταγώνου πλευρὰ δύναται|
4 | | τήν τε τοῦ ἑξα-|
5 | | γώνου το
ἐγ κύκλωι τῶι αβ
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6 | | καὶ τὴν τοῦ δεκαγώνου. ἔστωσαν
αἱ αγ βγ marg.:ἰσοπλεύρου|
7 | | πλευραὶ δεκαγώνου. ἀλλ'ἡ δεκα|
8 | | γώνου ἡ γβ περιφέρεια δ1ίχα τετμήσθω.
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9 | | δῆλον, ὅτι ἡ ὑπὸ
τῶν λγβ γωνία τῆι |
10 | | ὑπὸ τῶν γβλ, κοινῆι
τοῦ τε αβγ καὶ τοῦ βκλ|
11 | | τριγώνου ἐστὶν ἴση. ἔστιν ἄρ'ὡς ἡ αβ|
12 | | πρὸς βγ, οὕτως βγ πρὸς βλ. τὸ ἄρ'ὑπὸ τἠς ῶν αβ βλ|
13 | | ἴσον τῶι ἀπὸ τῆς βγ. πάλιν ἡ α . .|
14 | |
λωι. ἡαιδὲ γη
βη εἰκοσαγώνοθ
1 marg.:
Translation
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1 | | ...|
2 | | The theorem is proved, that the square over the side of the |
3 | | pentagon is equal to the sum of the squares|
4 | | over the sides of the hexa-|
5 | | gon in the circle AB|
6 | | and of the decagon. Let ΑΓ and ΒΓ be sides of the marg.:equilateral |
7 | | decagon. Let the arc over the decagon side |
8 | | ΓB be bisected.|
9 | | It is clear, since the angle ΛΓΒ is equal to |
10 | | the angle ΓΒΛ common to the triangles ΑΒΓ and ΒΚ (read Γ)Λ.|
11 | | Thus the ratio ΑΒ|
12 | | to ΒΓ is equal to the ratio ΒΓ to ΒΛ. Thus the rectangle formed by the sides ΑΒ ΒΛ|
13 | | is equal in area to the square over the side ΒΓ ...|
14 | | but the stretches ΓΗ and ΒΗ are sides of an icosagon ... (after Müller/Mau 1960)
1 marg.:
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| TEI-XML-File |
https://p612399.webspaceconfig.de/xml/elephantine_erc_db_016498.tei.xml |