Squaring the circle - constructing, with only compass and unmarked straightedge, a square whose area equals that of a given circle - is one of the 3 classical problems of Greek antiquity. It was finally proven impossible in 1882, when Lindemann showed that π is transcendental: since every straightedge-and-compass construction produces only algebraic lengths, no construction can ever produce $r\sqrt{\pi}$​ exactly. Ramanujan knew this - note the word approximately in the text. What he does in the paper instead is a construction so accurate that the impossibility becomes "almost invisible". This is one of Ramanujan's earliest publications, appearing the same year he sent his famous first letter to G. H. Hardy (January 16, 1913). At the time Ramanujan was a clerk at the Madras Port Trust, and the JIMS was essentially his only outlet. He published his first paper there in 1911 and famously posed problems in its question section. $PR$ is a diameter, so by Thales' theorem the inscribed angle $\angle PSR$ is a right angle. Pythagoras then gives $PS^2 = PR^2 - RS^2 = d^2 - \frac{5}{36}d^2$. Both come from similar triangles inside $\triangle PSR$. Since $OM \parallel RS$ and $O$ is the midpoint of $PR$, $M$ is the midpoint of $PS$, so $PM = \frac{1}{2}PS$ and $PK^2 = \frac{1}{4}PS^2$. Since $TN \parallel RS$ with $PT = \frac{5}{6}PR$, we get $PN = \frac{5}{6}PS$, hence $MN = PN - PM = \frac{1}{3}PS$ and $PL^2 = \frac{1}{9}PS^2$. The fractions $\frac{1}{4}$ and $\frac{1}{9}$ are exactly what's needed to make $144 = 4\times 36$ and $324 = 9\times 36$ appear as denominators. This follows from the intersecting-chords relation (the "geometric mean" theorem): since $TQ \perp PR$ with $Q$ on the circle, $TQ^2 = PT \cdot TR = \frac{5d}{6}\cdot\frac{d}{6} = \frac{5d^2}{36}$, and $RS$ was placed equal to $TQ$. A circle of area $140{,}000$ sq. miles (roughly the size of Japan) has $r\sqrt{\pi} = \sqrt{A} \approx 374.17$ miles. The relative error of $RD$ is $\frac{1}{2}\cdot\frac{355/113 - \pi}{\pi} \approx 4.2\times 10^{-8}$, so the absolute error is about $1.6\times 10^{-5}$ miles, almost exactly one inch. A continent-sized circle, squared to within the width of a thumb. Since $CD \parallel KL$, triangles $RCD$ and $RKL$ are similar, so $\frac{RC}{RD} = \frac{RK}{RL} = \sqrt{\frac{113/144}{355/324}} = \frac{18}{12}\sqrt{\frac{113}{355}}$. The final two steps of the construction ($RC = RH$ and the parallel $CD$) are just a similarity trick to divide out the unwanted factor of $\frac{3}{2}$. This approximation is far older than the paper: it was discovered by the Chinese mathematician and astronomer Zu Chongzhi in the 5th century AD, where it is known as *Milü* ("close ratio"). It was independently rediscovered in 16th-century Europe (Valentin Otho, 1573, and Adriaan Anthonisz ). Ramanujan's contribution is not the fraction itself but the "economical" compass-and-straightedge construction of its square root. Putting it together: $RD = RC \cdot \frac{RL}{RK} = \frac{3d}{4}\cdot\frac{2}{3}\sqrt{\frac{355}{113}} = r\sqrt{\frac{355}{113}}$. The fraction $\frac{355}{113} = 3.14159292\ldots$ agrees with $\pi = 3.14159265\ldots$ to 6 decimal places (relative error $\approx 8.5\times 10^{-8}$). Its accuracy is not an accident: the continued fraction of $\pi$ is $[3; 7, 15, 1, 292, \ldots]$, and $\frac{355}{113}$ is the convergent obtained by truncating just before the enormous term $292$. It is the best rational approximation to $\pi$ with denominator below $16{,}604$.