# Squaring the circle

Squaring the circle  (1913)
by Srinivasa Ramanujan

Squaring the circle

(Journal of the Indian Mathematical Society, v, 1913, 138)

Let $PQR$ be a circle with centre $O$ , of which a diameter is $PR$ . Bisect $PO$ at $H$ and let $T$ be the point of trisection of $OR$ nearer $R$ . Draw $TQ$ perpendicular to $PR$ and place the chord $RS=TQ$ .

Join $PS$ , and draw $OM$ and $TN$ parallel to $RS$ . Place a chord $PK=PM$ , and draw the tangent $PL=MN$ . Join $RL$ , $RK$ and $KL$ . Cut off $RC=RH$ . Draw $CD$ parallel to $KL$ , meeting $RL$ at $D$ .

Then the square on $RD$ will be equal to the circle $PQR$ approximately.

 For $RS^{2}={\frac {5}{36}}d^{2}$ ,
where $d$ is the diameter of the circle.
 Therefore $PS^{2}={\frac {31}{36}}d^{2}$ .

But $PL$ and $PK$ are equal to $MN$ and $PM$ respectively.

 Therefore $PK^{2}={\frac {31}{144}}d^{2}$ , and $PL^{2}={\frac {31}{324}}d^{2}$ .
 Hence $RK^{2}=PR^{2}-PK^{2}={\frac {113}{144}}d^{2}$ , and $RL^{2}=PR^{2}+PL^{2}={\frac {355}{324}}d^{2}$ .
 But ${\frac {RK}{RL}}={\frac {RC}{RD}}={\frac {3}{2}}{\sqrt {\frac {113}{355}}}$ , and $RC={\frac {3}{4}}d$ .
 Therefore $RD={\frac {d}{2}}{\sqrt {\frac {355}{113}}}=r{\sqrt {\pi }}$ , very nearly.

Note.—If the area of the circle be $140,000$ square miles, then $RD$ is greater than the true length by about an inch. This work was published before January 1, 1926, and is in the public domain worldwide because the author died at least 100 years ago.