________________
(c)
परिशिष्ट-३ A NOTE ON RAMANUJAN'S PROOF RAMANUJAN'S proof omits many obvious to a virtuoso like him, but not to many of his readers even if familiar with high school geometry. The fuller proof is as follows:
Let d and I be the diameter and radius of the circle. RS2 = TQ2 = PT. TR = 5/6.d.1/6.d = 5/36.d2 Therefore PS2 = PR2 – RS2 = d2 – 5/36.d2 = 31/36-d2 PM PQ 1 PS – PR = 2 Therefore PM2 = 31/144.d2 PM PO
3 MNOT = 2/3r = Therefore MN2 = 4/9 PM2 = 31/324 RL2 = RP2 + PL= d2 + MN2 = 355/324 d2 RK2 = RP2 - PK2 = d2 - PM2 = 113/144 d2 RC RK RD RL Therefore
RD2 = RL2
* RR2 RC2
= 355/324 x 144/113 x 9/16.d= 355/113 x d2|4 Hence RD2 = 355/115 r2
SQUARING THE CIRCLE (Journal of the Indian Mathematical Society x, 1913, 132) 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
RS2 = .d?, where d is the diameter of the circle. Therefore
PS2 = }da. But PL and PK are equal to MN and PM respectively.
