Trije klasični problemi grške geometrije
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1 Trije klasični problemi grške geometrije Milan Hladnik Predavanja iz zgodovine matematike FMF, Univerza v Ljubljani 17. oktober 2012
2 Grčija v 5. stoletju pnš. Perzijci sredi 6. stoletja zasedli Malo Azijo, vstaja jonskih mest 499 pnš. zatrta, neuspešna pomoč Aten Perzijske vojne: poraz Perzijcev na Maratonskem polju 490 pnš., zmaga Kserksa pri Termopilah, poraz v pomorski bitki pri Salamini 480, poraz pri Plataji 179 pnš. Zlata doba Aten (Periklej, Sokrat, drugi filozofi, razvoj matematike v Atenah) Peloponeške vojne 431 pnš., prevlada Šparte od 404 pnš. do 371 pnš. Matematika se razvija tudi in predvsem v kolonijah.
3 Grčija
4 Trije klasični problemi Duplikacija kocke: konstruirati rob kocke z dvakrat večjo prostornino kot dana kocka, Trisekcija kota: razdeliti poljuben kot na tri enake dele, Kvadratura kroga: konstruirati kvadrat, ki ima enako ploščino kot dani krog. Samo z evklidskim orodjem
5 Podvojtev kocke Legenda: Dve varianti: - Kralj Minos in velikost grobnice sina Glavka - Deloški problem in povečanje oltarja boga Apolona Hipokratova redukcija: dvojno geometrijsko razmerje a:x = x : y = y : b formula x 3 = 2s 3
6 Hipokrat s Kiosa ( pnš.) Slika: Hipokrat s Kiosa
7 Metode reševanja problema duplikacije Menajhem 350 pnš.: stožnice Platon: čevljarski kotnik Apolonij: prilagoditev polmera kroga Eratosten: trije enaki pravokotniki z diagonalo Diokles: cisoida
8 Menajhemova podvojitev kocke s stožnicami 2 x = 2sy xy = 2s 2 2 y = sx 2 y = sx 2s y y 0 s x 0 s x ( a) ( b) Slika: Podvojitev kocke s stožnicami
9 Podvojitev kocke po Platonu R D D b P a C S P C U y x A B V W T ( a) ( b) Slika: Podvojitev kocke po Platonu PC/PB = PB/PA=PA/PD Točka V mora ležati na premici skozi D in P.
10 Platon ( pnš.) Slika: Filozof in matematik Platon
11 Dvojno razmerje po Apoloniju y ( a,b) b 0 a x ( B) Slika: Konstrukcija dvojnega razmerja po Apoloniju Tetiva skozi točko (a, b) mora sekati krožnico v presečiščih le-te z osema. Potem je a/y = y/x = x/b =(a+x)/(b+ y).
12 Eratostenova metoda E A B C D ( a) E F A B C D ( b) G H Slika: Konstrukcija dvojnega razmerja po Eratostenu AE : BF = BF : CG=CG: DH
13 OP = RQ Dioklova cisoida B Q Q R P 2d P 3 d 2 R O d ( d,0) O d A ( a) ( b) Slika: Dioklova cisoida in podvojitev kocke
14 Tretjinjenje kota z vstavljanjem D A F a a a G E a B C Slika: Tretjinjenje kota z vstavljanjem EF = 2AB
15 Arhimedova metoda B a C a a D O A Slika: Tretjinjenje kota po Arhimedu
16 Vietova in Newtonova metoda A x a M a C b 60 B D N Slika: Tretjinjenje kota po Vietu in Newtonu
17 Nikomedova konhoida A B C b b Konhoida b D L a O Slika: Tretjinjenje kota z Nikomedovo konhoido
18 Hipijeva trisektrisa oziroma Dinostratova kvadratrisa 1 B b b 3 0 a 1 A Slika: Hipijeva trisektrisa oziroma Dinostratova kvadratrisa
19 Trisekcija z Arhimedov spiralo Q B a P a O a A Slika: Tretjinjenje kota z Arhimedovo spiralo
20 Trisekcija kota s tomahawkom A R S T U C D B Slika: Tretjinjenje kota s tomahawkom
21 Papusova trisekcija s hiperbolo B Q a C 0 2 P 2a A Slika: Tretjinjenje kota s hiperbolo
22 Hipokratova kvadratura lune C D C A B A B ( a) ( b) Slika: Hipokratovi luni Hipokrat s Kiosa je odkril še dva primera lun, ki se dajo kvadrirati. M. Hladnik, Hipokratove lune, Presek 28 (2001/02),št. 2,
23 Rektifikacija krožnice Približne metode: (a) Pogosto πd 3d+ d 2/10 oziroma π 3+ 2/ (b) Poljski jezuit Adam Kochanski ( ): V krajišču B premera AB danega kroga načrtamo tangento, na eno stran odmerimo centralni kot 30 stopinj in od presečišča C kraka s tangento na drugo stran odmerimo tri polmere do točke D. Potem je obseg kroga približno enak 2AD.
24 Metoda Kochanskega D A 30 0 B C Slika: Približna rektifikacija
25 Nezmožnost rešitve z evklidskim orodjem (1) Duplikacija kocke (s stranico 1). Konstruirati število x = 3 2, ki reši enačbo x 3 2=0. Koren algebraične enačbe lahko konstruiramo z ravnilom in šestilom samo, če se izraža s samimi kvadratnimi koreni. Da pa se pokazati, da mora enačba tretje stopnje z racionalnimi koeficienti v tem primeru imeti vsaj en racionalen koren. (2) Trisekcija kota. Konstruirati x = cos(θ/3), če poznamo cosθ. Zaradi cosθ = 4cos 3 (θ/3) 3cos(θ/3) bi pri θ = 60 0 tak x zadoščal kubični enačbi 8x 3 6x 1=0, ki nima racionalnih korenov. (3) Kvadratura kroga. Stranica kvadrata z isto ploščino kot krog s polmerom 1 meri π. Ker je to število transcendentno, se ga ne da konstruirati samo z ravnilom in šestilom.
26 Kvadratura kroga Kako konstruirati (z evklidskim orodjem) kvadrat, ki je ploščinsko enak krogu? Klasično: Konstrukcija z evklidskim orodjem Ferdinand Lindemann 1882: Število π je transcendentno. Posledica: Klasična kvadratura kroga ni možna Moderno: V smislu teorije množic Miklos Laczkovich 1990: Poljubna dva topološka diska z merljivim robom in enako ploščino v ravnini sta enaka po razkosanju. Posledica: Moderna kvadratura kroga je možna.
27 Konstrukcije samo s šestilom ali samo z ravnilom Samo s šestilom: Lorenzo Mascheroni ( ) in (že prej 1672) Georg Mohr Samo z ravnilom: Jean Victor Poncelet ( ) Z ravnilom in fiksnim krogom: Abul Wefa ( ), Jakob Steiner ( ), Francesc Severi ( )
28 Posebna literatura H. Eves, An Introduction t the History of Mathematics, Holt, Rinehart and Winston, G.E. Martin, Geometric Constructions, UTM, Springer Wikipedia, spletni portal.
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