Het Stomachion - KNAW Onderwijsprijs

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Bijlage 3: Engelse vertaling van de Arabische tekst http://www.hs-‐ augsburg.de/~harsch/graeca/Chronologia/S_ante03/Archimedes/arc_ost2.html http://en.wikipedia.org/wiki/Ostomachion We draw a [rectangular] parallelogram ABGD, we bisect BG in E and draw EZ perpendicular to BG [to intersect AD], we draw the diagonals AG, BZ [intersecting AG at L], and ZG, we also bisect BE in H, and draw HT perpendicular to BE [to intersect BZ], then we put the ruler at point H and -‐ looking to point A -‐ we draw HK [to intersect BZ], then bisect AL in M, and draw BM. So the A-‐E rectangle is divided into seven parts. Now we bisect DG in N, ZG in C, we draw EC and attaching the ruler to the points B and C we draw CO [to intersect DG], furthermore CN. Thus the rectangle ZG is also divided in seven parts, but in another way than the first one. Therefore, the whole square has fourteen parts. We now demonstrate that each of the fourteen parts is in rational relationship to the whole square. Because ZG is the diagonal of the rectangle Z-‐G, the triangle DZG is half of this rectangle, that means 1/4 of the square. But the triangle GNC is 1/4 of triangle DZG, because, if we extend the line EC, it comes to point D, and that means triangle GDC has half area of the triangle DZG and is equal to the two triangles GNC and DNC taken together; that means triangle GNC is 1/16 of the square. If we presume that line OC is orientated to point B, as we have drawn it before, so the line NC is parallel to BG, which is the side of the square and of the triangle OBG, so we get the proportion BG : NC = GO : NO. But BG is four times NC, and in the same way GO four times NO; therefore is GN three times NO, and triangle GNC = 3 ONC. However, as we have shown, triangle GNC is 1/16 of the square, that means triangle ONC = 1/48 of the square. Furthermore, as triangle GDZ = 1/4 of the square, and therefore GNC = 1/16 of that triangle and NCO = 1/48 of that, it remains for the quadrilateral DOCZ = 1/6 of the square’s area. According to the proposition that line NC [extended] intersects [ZE at] point F, and GE is parallel to CF, [and labelling the intersection of AG and CE as Q,] we get the proportion EC : CF = EQ : CQ = GQ : FQ. Because EQ = 2 CQ and GQ = 2 FQ, triangle EQG is double to the two triangles GCQ and EFQ. It is clear, that triangle EGZ = 2 times triangle EFG, because ZE = 2 FE. As the triangle EGZ = 1/4 of the square, that means triangle EFG = 1/8 of the square. This triangle is three times as big as each of the two triangles EFQ and GCQ, so each of these 36
two triangles = 1/24 of the square A-‐G. And the triangle EGQ is double to each of the two triangles EFQ and GCQ, so it is = 1/12 of the square. Furthermore because ZF = EF, triangle ZFG = triangle EFG. If we now take away triangle GCQ (= triangle EFQ), it leaves quadrilateral FQCZ (= triangle EGQ), therefore quadrilateral FQCZ = 1/12 of the square A-‐G. If an Ostomachion were to be imposed onto a 12-‐unit square, this diagram shows the area of each piece. We have now divided the rectangle Z-‐G in 7 parts, and go on to divide the other rectangle. Because BZ and EC are two parallel diagonals, and ZF = EF, therefore triangle ZLF = EFQ, and also triangle ZLF = 1/24 of the square A-‐G. Because BH = HE, triangle BEZ is four times the triangle BHT, because each of them is rectangular. As triangle BEZ = 1/4 of the square ABGD, triangle BHT = 1/16 of that. According to our proposition the line HK [extended] intersects point A, so we get the proportion AB : HT = BK : KT. Because AB = 2 HT, and BK = 2 KT and BT = 3 KT, triangle BHT is three times the triangle KHT. However, because triangle BHT = 1/16 of the whole square, triangle KHT = 1/48 of that. Triangle BKH is double the triangle KHT, so = 1/24 of the square. Further, as BL = 2 ZL, and AL = 2 LF, triangle ABL is twice the triangle ALZ, and ALZ double the triangle ZLF. However, because triangle ZLF = 1/24 of the whole square, triangle ALZ = 1/12 of that, so triangle ABL = 1/6. But triangle ABM = triangle BML, so each of these two triangles = 1/12 of the square. It leaves the pentagon LFEHT = 7/48 of the entire square. We have now also divided the square AE into 7 sections, therefore, the whole figure ABGD in 14 parts. Each of these fourteen parts is in rational relationship to the whole, and that is what we wanted." 37
Page 1 and 2: Het Stomachion Raadsel van de puzze
Page 3 and 4: Voorwoord Als leerling in 6 VWO is
Page 5 and 6: Techniek De Schroef van Archimedes
Page 7 and 8: 1907 - 1930 - De Palimpsest raakt z
Page 9 and 10: Nu volgde het zichtbaar maken van d
Page 11 and 12: Griekse tekst en Nederlandse vertal
Page 13 and 14: 5 10 15 20 25 Griekse tekst en vert
Page 15 and 16: 50 55 60 65 70 ἡ γὰρ γὰρ
Page 17 and 18: Discussie over de tekst Archimedes
Page 19 and 20: Afbeelding 5: illustratie van Netz
Page 21 and 22: Combinatoriek Combinatoriek is een
Page 23 and 24: Puzzelstuk (afbeelding 9) Oppervlak
Page 25 and 26: E (afbeelding 16): teken je de zes
Page 27 and 28: Oppervlakten De oppervlakte van ied
Page 29 and 30: Bronvermelding 1. Tekst Heiberg Dec
Page 31: ἈΡΧ ΙΜΉΔΟ Υ Σ ΣΤΟΜΆ
Page 34: Bijlage 2: Het onleesbare manuscrip

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