Rhind Mathematical Papyrus (6 of 8) / © 1979-2001 by Franz Gnaedinger, Zurich, fg@seshat.ch / www.seshat.ch

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RMP 46 and 47

Two square containers measure

10 by 10  by  3 '3  royal cubits

10 by 10  by  5     royal cubits

I am assuming that the floor of each container measures 10 by 10 royal cubits. Now let me inscribe an octagon in that square:

A    a      b   B

h               c

g               d

D    f      e   C

Square ABCD   AB = BC = CD = DA = 10 royal cubits

inscribed octagon abcdefgh   ab=bc=cd=de=ef=fg=gh=ha

A side of the square measures 10 royal cubits or 70 palms or 280 fingers. How long is the side of the regular octagon within the frame of the square?

1       1       2

2       3       4

5       7      10

12      17      24

29      41      58

70      99     140

169      ...     ...

side of octagon    side of circumscribed square

7             5  +   7  +   5   =   12

10             7  +  10  +   7   =   24

17            12  +  17  +  12   =   41

24            17  +  12  +  17   =   58

41            29  +  41  +  29   =   99

58            41  +  58  +  41   =  140

By doubling the numbers of the last line we find:

side of octagon  116    side of square  82+116+82 = 280

The side of the square measures 10 royal cubits or 70 palms or 280 fingers, hence the side of the inscribed octagon   measures practically 116 fingers or 4 cubits 1 palm.

Now let me go a step further and transform the square containers into granaries of the same volume but in the shape of prisms. The floor of these containers is given by the above octagon. The area of an octagon is smaller than that of the square. Accordingly, the granaries will be taller. How do we calculate their heights? Again by means of the above number pattern:

height of granary based on square    10  24  58  140  ...

height of granary based on octagon   12  29  70  169  ...

The former square containers were 3 '3 and 5 royal cubits tall. The new granaries on the base of the octagon are taller. By using the first numbers 10 and 12 we find the heights:

3 '3 royal cubits x '10 x 12  =  4 royal cubits

5 royal cubits    x '10 x 12  =  6 royal cubits

Very simple numbers. - Now let us consider only the second granary. Wishing to get a more accurate volume I use the numbers 140 and 169:

former height  5 royal cubits or 140 fingers

new height  140f x '140 x 169 = 169f = 6 cubits 1 finger

Calculating the area of the inner wall (right parallelepiped and prism):

base  10 c x 10 c     height  5 royal cubits

periphery  4 x 10 c  =  40 royal cubits

area of wall  40 c x 5 c  =  200 square cubits

side of octagon  116 fingers or 4 cubits 1 palm

height  169 fingers or 6 cubits 1 finger

periphery  8 x 116 fingers = 928 fingers or 33 cubits 1 palm

area of wall  =  33 '7 c x 6 '28 c  =  200 '28 '196 cc

The walls of the two granaries have practically the same area. Even better: they have exactly the same area (the minor error in the above result is due to the margins of error in the values '41 of 58 and '140 of 169 used for the calculation of the octagon).

Imagine a pair of ideal granaries with vertical walls. One granary is based on a square. The other granary is based on the regular octagon inscribed in the square. If the granaries have the same capacity, their walls have the same area. Or if the inner walls have the same area, the two granaries have the same volume.

RMP 48

Problem no. 48 of the Rhind Mathematical Papyrus contains a famous drawing of a square with an inscribed octagon:

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 generates the value 3 '9 for re:

'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 generates the value '81 of 256 for re, 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

RMP 49

A rectangle measures 2 by 10 khet or 200 by 1,000 royal cubits while its area measures 200,000 square cubits. Can you transform the rectangle into a regular octagon of about the same area?

circumscribed square  492 by 492 royal cubits

partition  12 times 12+17+12 royal cubits

grid  144+204+144 by 144+204+144 royal cubits

side of octagon  204 royal cubits

area  200,592 square cubits

RMP 50

Ahmes calculates the area of a circle whose diameter measures 9 khet = 900 royal cubits. By using his well known formula he obtains 8 by 8 khet = 64 square khet = 64 aroures or setat = 640,000 square cubits.

Advanced learners may try to solve a more demanding task: transforming the circle into a regular octagon of the same area using the following extended number pattern:

1       1       2       2

2       3       4       6

5       7      10      14

12      17      24      34

29      41      58      82

70      99     140     198

169     239     338     478

The squared side of a regular octagon and the area of the same octagon maintain a relation that can be approximated by means of the above numbers:

side x side   12  17   29   41   70   99  ...  square cubits

area octagon  58  82  140  198  338  478  ...  square cubits

The number re may be chosen from the following sequence:

3  (plus 22)  25  47  69  91  113  135  157  179  201  223

1  (plus  7)   8  15  22  29   36   43   50   57   64   71

245  267  289  311  333  355  377  399

78   85   92   99  106  113  120  127

Two values contain the number 99: '99 of 478 and '99 of 311. Now the area of a regular octagon and the one of a circle may be defined like this:

side x side x '99 x 478    radius x radius x '99 x 311

The octagon and the circle have the same area, therefore:

side x side x '99 x 478  =  radius x radius x '99 x 311

The diameter of the circle measures 9 khet or 900 royal cubits while the radius measures 450 royal cubits. Now we obtain:

side x side  =  450 cubits x 450 cubits x 311 x '478

side x side  =  practically 131,752 square cubits

By consulting a table of square numbers you will find

362 x 362  =  131,044  ---       708 less than 131,752

363 x 363  =  131,769  ---  only  17 more than 131,752

364 x 364  =  132,496  ---       744 more than 131,752

The number 363 is a good solution to our problem. Hence a circle of the diameter 9 khet and a regular octagon of the side length 363 royal cubits have practically the same area.

grid  770+1089+770 by 770+1089+770  '3 royal cubits

RMP 51

Ahmes calculates the area of a triangle whose base measures 4 khet and its height 10 khet, generating an area of 20 square khet = 20 aroures = 200,000 square cubits. This area can be transformed into a regular octagon (see RMP 49):

grid  144+204+144 by 144+204+144 royal cubits

side  204 royal cubits   area  200,592 square cubits

The drawing of RMP 51 shows a triangle a base measuring 4 khet, while its height measurement is given as 13 khet. In this case the area measures 260,000 square cubits, while the regular octagon of roughly the same area has the following measurements:

grid  164+232+164 by 164+232+164 royal cubits

side  232 royal cubits   area  259,808 square cubits

RMP 52

This problem concerns a trapezoid whose base measures 6 khet, its upper side 4 khet and its height 20 khet. Its area is 100 aroures or 1,000,000 square cubits. Transform that area into a regular octagon by using an alternative number pattern:

2       1       4

3       5       6

8      11      16

19      27      38

46      65      92

322     455     644

grid  322+455+322 by 322+455+322 royal cubits

side octagon  455 royal cubits   area  1,000,433 sc

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