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
   

Edition

convex
1 . .
2ἀποδείκνυται ἡ πρότασις, ὅτι ἡ τοῦ
3πενταγώνου πλευρδύναται
4τήν τε τοῦ ἑξα-
5γώνου το γ κύκλωι τῶι αβ
6καὶ τὴν τοῦ δεκαγώνου. ἔστωσαν αἱ αγ βγ 1marg.:ἰσοπλεύρου
7πλευραὶ δεκαγνου. ἀλλ'δεκα
8γώνου ἡ γβ περιφέρεια δ1ίχα τετμήσθω.
9δῆλον, ὅτι ἡ ὑπὸ τῶν λγβ γωνία τῆι
10ὑπὸ τῶν γβλ, κοινῆι τοῦ τε αβγ καὶ τοῦ βκλ
11τριγώνου ἐστὶν ἴση. ἔστιν ρ'ς ἡ αβ
12πρὸς βγ, οὕτως βγ πρὸς βλ. τὸ ἄρ'ὑπὸ τἠς ῶν αβ βλ
13ἴσον τῶι ἀπὸ τῆς βγ. πάλιν ἡ α . .
14 λωι. αιδὲ γη βη εἰκοσαγώνοθ

1 marg.:

Translation

convex
1...
2The theorem is proved, that the square over the side of the
3pentagon is equal to the sum of the squares
4over the sides of the hexa-
5gon in the circle AB
6and of the decagon. Let ΑΓ and ΒΓ be sides of the 1marg.:equilateral
7decagon. Let the arc over the decagon side
8ΓB be bisected.
9It is clear, since the angle ΛΓΒ is equal to
10the angle ΓΒΛ common to the triangles ΑΒΓ and ΒΚ (read Γ)Λ.
11Thus the ratio ΑΒ
12 to ΒΓ is equal to the ratio ΒΓ to ΒΛ. Thus the rectangle formed by the sides ΑΒ ΒΛ
13is equal in area to the square over the side ΒΓ ...
14but the stretches ΓΗ and ΒΗ are sides of an icosagon ... (after Müller/Mau 1960)

1 marg.:
 
 
TEI-XML-File https://p612399.webspaceconfig.de/xml/elephantine_erc_db_016498.tei.xml