
Re: Ariadne's thread
Posted:
Jan 27, 2012 3:22 AM


RMP 48
Problem no. 48 of the Rhind Mathematical Papyrus contains a famous drawing of a square with an inscribed octagon (ASCII drawing adopted to the proportional font):
A a b B h  c g  d D f e C
Square ABCD  irregular octagon abcdefgh
AB = BC = CD = DA = 9 royal cubits
Aa = ab = bB = Bc = ... = gh = hA = 3 royal cubits
grid 3+3+3 by 3+3+3 royal cubits
area square 9x9 = 81 square cubits
area octagon 9x9  2x3x3 = 63 square cubits
A circle inscribed in the square would have about the same area as the octagon. This yields the value 3 '9 for pi:
'4 x 9 rc x 9 rc x 3 '9 = 63 square cubits
63 square cubits are about 8 by 8 royal cubits. From this we derive a well known formula: if the diameter of a circle measures 9 units and if the side of a square measures 8 units the circle and the square have roughly the same area. This formula yields '81 of 256 for pi, or nearly '6 of 19 or 3 '6, according to a crosswise multiplication:
256 x 6 = 1536  81 x 19 = 1539
So we found the values 3 '9 and 3 '6. The equally simple values in between are 3 '8 and 3 '7. These values are part of the first pi sequence
4/1 (plus 3/1) 7/2 10/3 13/4 16/5 * 19/6 22/7 25/8 28/9 *

