# File:Squaring the circle.djvu

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## Summary

 Description Research article describing an approximate squaring of the circle using ruler and compass (exact squaring is impossible with just those two). Date 1913 Source Photocopy of reprint of vol. V (1913) of the Journal of the Indian Mathematical Society, page 132, found in "Ramanujan - Collected Papers". Author Srinivasa Ramanujan Permission (Reusing this file) Public domain because author has been deceased for more than 70 years.

## Übersetzung in die deutsche Sprache

Journal der indischen Mathematical Society, v. 1913, S. 132

1. Ziehe einen Kreis mit PQR um Mittelpunkt O, wovon der Durchmesser PR sei. (Ergänzung: Q noch nicht bestimmt)
2. Halbiere PO bei H, der Punkt T ist die Dreiteilung von OR nahe R.
3. Zeichne TQ senkrecht zur PR und die Sehne RS = TQ.
4. Verbinde P mit S und zeichne die Parallelen TN und OM zu RS.
5. Lege eine Sehne PK = PM und ziehe die Tangente PL = MN.
6. Verbinde R mit L, R mit K und K mit L.
7. Schneide RC = RH.
8. Zeichne CD parallel zu KL, trifft auf RL bei D.

Dann ist das Quadrat über RD etwa gleich dem Kreis PQR

Für ${\displaystyle \textstyle RS^{2}={\frac {5}{36}}\cdot d^{2}}$,

wobei ${\displaystyle d}$ der Durchmesser des Kreises ist.

Somit ${\displaystyle \textstyle PS^{2}={\frac {31}{36}}\cdot d^{2}}$

Da PL und PK gleich sind mit MN und PM

ist somit ${\displaystyle \textstyle PK^{2}={\frac {31}{144}}\cdot d^{2}}$ und ${\displaystyle \textstyle PL^{2}={\frac {31}{324}}\cdot d^{2}}$

Damit ist ${\displaystyle \textstyle RK^{2}=PR^{2}-PK^{2}={\frac {113}{144}}\cdot d^{2}}$

und ${\displaystyle \textstyle RL^{2}=PR^{2}+PL^{2}={\frac {355}{324}}\cdot d^{2}}$

Aber ${\displaystyle \textstyle {\frac {RK}{RL}}={\frac {RC}{RD}}={\frac {3}{2}}\cdot {\sqrt {\frac {355}{113}}}}$,

und ${\displaystyle RC={\frac {3}{4}}\cdot d}$

Deshalb ${\displaystyle \textstyle RD={\frac {d}{2}}\cdot {\sqrt {\frac {355}{113}}}}$ dies ist ${\displaystyle r\cdot {\sqrt {\pi }}}$ sehr nahe. Ergänzung: Abweichung = 7,52...E-8 [LE]

Bemerkung

Wenn die Fläche des Kreises 140.000 Quadratmeilen ist, dann ist RD etwa einen Zoll größer als die wahre Länge.

Ergänzung zur Bemerkung

1 Meile [mile] = 1.609,344 m

${\displaystyle \textstyle {\sqrt {140.000}}}$ mile2 = 374,165738... mile

RD = 1,772453926158302 m

${\displaystyle \textstyle 374,165738...\cdot {\frac {1.609,344}{1,772453926158302...}}=339.733{,}1668...\;[o.D.]}$

Abweichung${\displaystyle \textstyle =7{,}52...E-8\;m\cdot 339.733{,}1668...\cdot 1000\approx 25{,}55\;mm>1\;Zoll=25{,}4\;mm}$

Ergänzung

RD2 = 3,14159292035398 (sechs Nachkommastellen sind gleich denen von ${\displaystyle \pi }$)

(-) ${\displaystyle \pi }$ = 3,14159265358979

Abweichnung zu ${\displaystyle \pi }$ = 2,66764190737234E-7 mm

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