|
|
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
A-B = B-C = C-D = D-A = 9 royal cubits
A-a = a-b = b-B = B-c = ... = g-h = h-A = 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 *
|
|
