File:Squaring the circle.djvu

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

Es sei PQR ein Kreis mit dem Mittelpunkt O, von dem PR der Durchmesser ist. Halbiere PO in H, und T sei der Punkt aus der Dreiteilung von OR nahe R. Zeichne TQ senkrecht zu PR und setze die Sehne RS = TQ.

Verbinde P mit S und zeichne OM und TN parallel zu RS. Setze eine Sehne PK = PM und zeichne die Tangente PL = MN. Verbinde R mit L, R mit K und K mit L. Abschnitt RC = RH. Zeichne CD parallel zu KL, [CD] trifft auf RL in D.

Dann ist das Quadrat über RD annähernd gleich dem Kreis PQR.

Denn ${\displaystyle RS^{2}={\frac {5}{36}}d^{2},}$

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

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

Aber ${\displaystyle PL}$ und ${\displaystyle PK}$ sind gleich ${\displaystyle MN}$ bzw. ${\displaystyle PM.}$

Somit ${\displaystyle PK^{2}={\frac {31}{144}}d^{2},}$ und ${\displaystyle PL^{2}={\frac {31}{324}}d^{2}.}$

Folglich ${\displaystyle RK^{2}=PR^{2}-PK^{2}={\frac {113}{144}}d^{2},}$

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

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

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

Darum ${\displaystyle RD={\frac {d}{2}}{\sqrt {\frac {355}{113}}}=r{\sqrt {\pi }},}$ nahezu gleich.

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

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 ${\displaystyle \pi }$ ≈ 2,667E-7 mm

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