Rhind Mathematical Papyrus (3 of 8)

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A short way to pi

The famous Egyptian Horus eye series 1 = '2 '4 '8 '16 '32 '64 (...) can be developed by means of a stairway:

1 = '1

1 = '2 '2

1 = '2 '4 '4

1 = '2 '4 '8 '8

1 = '2 '4 '8 '16 '16

1 = '2 '4 '8 '16 '32 '32

1 = '2 '4 '8 '16 '32 '64 '64

.................................

Stairway approximating 1:

'2 '2x2

'2 '2x2 '2x2x2

'2 '2x2 '2x2x2 '2x2x2x2

'2 '2x2 '2x2x2 '2x2x2x2 '2x2x2x2x2

'2 '2x2 '2x2x2 '2x2x2x2 '2x2x2x2x2 '2x2x2x2x2x2

..............................................................

Resulting series:

1 = '2 '2x2 '2x2x2 '2x2x2x2 '2x2x2x2x2 '2x2x2x2x2x2 ...

1 = '2 '4 '8 '16 '32 '64 ...

One eye of the Horus falcon was the moon; his other eye was the sun. If there had been a second Horus eye series it might well have been this one:

1 = '1

1 = '1x2 '2

1 = '1x2 '2x3 '3

1 = '1x2 '2x3 '3x4 '4

1 = '1x2 '2x3 '3x4 '4x5 '5

1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6

1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6x7 '7

...........................................

'1x2

'1x2 '2x3

'1x2 '2x3 '3x4

'1x2 '2x3 '3x4 '4x5

'1x2 '2x3 '3x4 '4x5 '5x6

'1x2 '2x3 '3x4 '4x5 '5x6 '6x7

'1x2 '2x3 '3x4 '4x5 '5x6 '6x7 '7x8

....................................... stairway approximating 1

1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6x7 '7x8 '8x9 '9x10 ...

1 = '2 '6 '12 '20 '30 '42 '56 '72 '90 ...

The resulting series has a fascinating sub-series

1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6x7 '7x8 '8x9 ...

'1x2 '2x3 '5x6 '6x7 ... = pi/4

which can be transformed as follows

'1x2 '2x3 = '1x3 '5x6 '6x7 = '5x7 ...

'1x3 '5x7 '7x9 '11x13 '15x17 '19x21 '23x25 ... = pi/8

'2 = '2

'2 = '1x3 '6

'2 = '1x3 '3x5 '10

'2 = '1x3 '3x5 '5x7 '14

'2 = '1x3 '3x5 '5x7 '7x9 '18

..................................

'2 = '1x3 '3x5 '5x7 '7x9 '9x11 '11x13 '13x15 '15x17 ...

'1x3 '5x7 '9x11 '13x15 ... = pi/8

or into the famous series

pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 - 1/15 ...

which was already known to the Indian mathematician Madhavan (c.1340-1425 AD), long before it was rediscovered by Gregory and Leibniz.

Duplations and right parallelepipeds

"1 = '1 '2 '2 quadruple 1-2-2-3

"2 = '2 '3 '6 quadruple 2-3-6-7

"3 = '3 '4 '12 quadruple 3-4-12-13

"4 = '4 '5 '20 quadruple 4-5-20-21

"5 = '5 '6 '30 quadruple 5-6-30-31

"6 = '6 '7 '42 quadruple 6-7-42-43

"7 = '7 '8 '56 quadruple 7-8-56-57

.......................................

As an example you may consider the quadruple 2-3-6-7: if a right parallelepiped measures 2 by 3 by 6 palms, the cubic diagonal measures exactly 7 palms or 1 royal cubit.

General form of the above quadruples: a --- a+1 --- aa+a --- aa+a+1. By choosing '2 '10 for the number a and by multiplying the resulting numbers by a factor of 25 you obtain the quadruple 15-40-24-49: if a right parallelepiped measures 15 by 40 by 24 palms, the cubic diagonal measures exactly 49 palms or 7 royal cubits.

1 = '1

1 = '1x2 '2

1 = '1x2 '2x3 '3

1 = '1x2 '2x3 '3x4 '4

1 = '1x2 '2x3 '3x4 '4x5 '5

1 = '1x2 '2x3 '3x4 '4x5 '5x6 '6x7 and so on

1 = '2 '6 '12 '20 '30 '42 '56 '72 '90 '110 '132 ...

"1 = '1x1 '1x2 '1x2 = '1 '2 '2 quadruple 1-2-2-3

"2 = '1x2 '1x3 '2x3 = '2 '3 '6 quadruple 2-3-6-7

"6 = '2x3 '2x5 '3x5 = '6 '10 '15 quadruple 6-10-15-19

"12 = '3x4 '3x7 '4x7 = '12 '21 '28 quadruple 12-21-28-37

"20 = '4x5 '4x9 '5x9 = '20 '36 '45 quadruple 20-36-45-61

"30 = '5x6 '5x11 '6x11 = '30 '55 '66 quadruple 30-55-66-91

..............................................................

"3 = '1x3 '1x4 '3x4 = '3 '4 '12 quadruple 3-4-12-13

"15 = '3x5 '3x8 '5x8 = '15 '24 '40 quadruple 15-24-40-49

"35 = '5x7 '5x12 '7x12 = '35 '60 '84 quadruple 35-60-84-109

...............................................................

"5 = '1x5 '1x6 '5x6 = '5 '6 '30 quadruple 5-6-30-31

"21 = '3x7 '3x10 '7x10 = '21 '30 '70 quadruple 21-30-70-79

..............................................................

General form: "ab = 'ab 'aa+ab 'ab+bb, quadruple ab -- aa+ab -- ab+bb -- aa+ab+bb

Every duplation "a = 'b 'c 'd ... constitutes a division of 2, and every division of 2 generates a right parallelepiped with an easily calculable cubic diagonal:

2 : a = "a quadruple 2 - a - "a - a "a

2 : 1 = "1 quadruple 2 - 1 - "1 - 1 "1 or 2-1-2-3

2 : 3 = "3 quadruple 2 - 3 - "3 - 3 "3 or 6-9-2-11

2 : 5 = "5 quadruple 2 - 5 - "5 - 5 "5 or 10-25-2-27

2 : 7 = "7 quadruple 2 - 7 - "7 - 7 "7 or 14-49-2-51

............................................................

2 : 1'1 = 1 2 - 1'1 - 1 - 2'1 or 2-2-1-3

2 : 1'3 = 1'2 2 - 1'3 - 1'2 - 2'2'3 or 12-8-9-17

2 : 1'5 = 1"3 2 - 1'5 - 1"3 - 2"3'5 or 30-18-25-43

2 : 1'7 = 1'2'4 2 - 1'7 - 1'2'4 - 2'2'4'7 or 56-32-49-81

If the floor of a chamber measures 49 by 56 palms or 7 by 8 royal cubits, and if the height of the chamber measures 32 palms, the cubic diagonal will measure exactly 81 palms. The volume of the chamber measures exactly 2x2x2x2x2x2x2 = 256 cube cubits. If a right parallelepiped measures 4 by 8 by 8 royal cubits, the volume will again measure 256 cube cubits, and the cubic diagonal will measure exactly 12 royal cubits, according to the basic quadruple 1-2-2-3.

RHIND MATHEMATICAL PAPYRUS, problems no. 7-38

In my opinion, many problems of the Rhind Mathematical Papyrus may be read on several levels of learning and understanding:

level A) how to handle unit fractions series

level B) geometrical applications

level C) theorems

RMP 32 - a magic parallelepiped

Ahmes divides 2 by 1 1/3 1/4 and obtains 1 1/6 1/12 1/114 1/228:

2 divided by 1 '3 '4 equals 1 '6 '12 '114 '228

By following Ahmes, the young pupils learn how to handle unit fraction series, while advanced learners may imagine a right parallelepiped of the following measurements:

height 2 units

length 1 '3 '4 units

width 1 '6 '12 '114 '228 units

How long is the cubic diagonal? Simply

1 '3 '4 units plus 1 '6 '12 '114 '228 units

1 1 plus '3 '6 plus '4 '12 plus '114 '228 units

2 '2 '3 '76 units

Divide 2 by any number a and you obtain b (2:a = b). Use these numbers as measurements for a right parallelepiped. It will be a 'magic parallelepiped' with the following properties:

height 2 units

width or length a units

length or width b units

area base / top ab = 2 square units

volume 2ab = 4 cubic units

cubic diagonal a+b units

RMP 34 - an easy way to measure a granary

Let us imagine a granary in the shape of a magic parallelepiped:

volume 500 cubic cubits / capacity 750 khars or 15,000 hekat

inner height 2 units or 10 cubits or 70 palms or 280 fingers

length x width 2 square units or 50 square cubits or 2,450

square palms or 39,200 square fingers

Here are four exact solutions, one of them provided by Ahmes' equation 1 '2 '4 times 5 '2 '7 '14 equals 10 in RMP 34:

inner height 10 royal cubits

inner length 10 royal cubits

inner width 5 royal cubits

cubic diagonal 15 royal cubits

inner height 280 fingers (10 royal cubits or 2 units)

inner length 245 fingers (1 '2 '4 units)

inner width 160 fingers (5 '2 '7 '14 royal cubits)

cubic diagonal 405 fingers (RMP 34)

inner height 280 fingers

inner length 224 fingers

inner width 175 fingers

cubic diagonal 399 fingers

inner height 280 fingers or 70 palms

inner length 200 fingers or 50 palms

inner width 196 fingers or 49 palms

cubic diagonal 396 fingers or 99 palms

If height, length and width measure 70, 50 and 49 palms, the cubic diagonal measures exactly 99 palms. Now consider a very good approximation:

inner height 280 fingers

inner length 198 fingers

inner width 198 fingers

diagonal practically 396 fingers

Such a granary can be measured simply using a rope:

height length width

o----------------------o---------------o---------------o

10 royal cubits 198 fingers 198 fingers

o----------------------o-------------------------------o

diagonal base cubic diagonal

RMP 33 - a wooden container in the shape of a cube

37 divided by 1 "3 '2 '7 equals 16 '56 '679 '776

Let me invent a task related to these numbers. A wooden container in the shape of a cube has an inner edge measuring 37 fingers, while the outer edge measures 41 fingers. How long are its diagonals?

1 1 2

2 3 4

5 7 10

12 17 24

29 41 58

A face of the cube measures 41 by 41 fingers, and the diagonal measures practically 58 fingers.

1 1 3

2 4 6

1 2 3

3 5 9

8 14 24

4 7 12

11 19 33

30 52 90

15 26 45

41 71 123

112 194 336

56 97 168

The cube measures 41 by 41 by 41 fingers, the cubic diagonal measures practically 71 fingers. The cavity measures 37 by 37 by 37 fingers. How long is the cubic diagonal? Use the numbers 97 and 168. Multiply 37 fingers by 168/97 and you obtain the diagonal in fingers. Divide 37 fingers by 97/42 = 1 "3 '2 '7 and you obtain the diagonal in palms. Ahmes found the number 16 '56 '679 '776, hence the diagonal measures practically 16 '56 '679 '776 palms or about 16 palms. And the volume? A surprise: it measures practically 1 "3 '2 '7 cubic cubits:

RMP 38 - transforming a square hekat into a cylinder

Let me begin with the subdivision of the royal cubit rod of Amenhotep I:

1 royal cubit (52.5 cm) = 7 palms (7.5 cm) = 28 fingers (1.875 cm) = 56 Re marks = 84 Shu marks = 112 Tefnut marks = 140 Geb marks = 168 Nut marks = 196 Osiris marks= 224 Isis marks = 252 Seth marks = 280 Nephtys marks = 308 Horus marks = 336 Imsety marks = 364 Hapy marks = 392 Duamutf marks = 420 Qhebsenuf marks = 468 Thoth marks

Now for RMP 38. Ahmes comes up with a funny equation: 1 hekat x 3 '7 x '22 x 7 = 1 hekat

The hekat is a measure of capacity. 30 hekat equal 1 cubic cubit. 1 hekat may be defined as a right parallelepiped with the following measurements:

'2 royal cubit x '3 royal cubit x '5 royal cubit

28 Re marks x 28 Shu marks x 28 Geb marks

In Qhebsenuf marks:

210 Qm x 140 Qm x 84 Qm cubic diagonal exactly 266 Qm

according to the quadruple 6-10-15-19

A hekat in the shape of a right parallelepiped is well-defined. How about other shapes? Let us look again at Ahmes' equation:

1 hekat x 3 '7 x '7 x 22 = 1 hekat

The numbers 3 '7 and '7 of 22 remind us of pi and 1/pi. How about a hekat in the shape of a cylinder? I replace the left hekat with the above definition:

210 Qm x 140 Qm x 84 Qm x 3 '7 x '22 x 7 = 1 hekat

Now I transform my equation:

'4 x 105 Qm x 105 Qm x 3 '7 x '11 x 3136 Qm = 1 hekat

The first term

'4 x 105 Qm x 105 Qm x 3 '7 or '4 x 7 f x 7 f x 3 '7

can be regarded as the area of a circle whose diameter measures 105 Qhebsenuf marks or 7 fingers, while the second term

'11 x 3136 Qm = 19 '165 fingers

can be regarded as the height of a cylinder. I leave out the small fraction '165 and keep the height 19 fingers. Now I can define my cylindrical hekat and quadruple-hekat simply:

diameter 7 fingers 14 fingers

circumference 22 fingers 44 fingers

height 19 fingers 19 fingers

volume 1 hekat 4 hekat

The mistakes are tiny.

RMP 37 - a cone

Ahmes finds a volume that measures '4 '32 hekat or 90 ro. 1 cubic cubit equals 30 hekat. 1 hekat equals 320 ro. 90 ro can be given as a cone with these measurements (mistake '4 ro):

diameter base 7 fingers circumference 22 fingers height 4 palms

RMP 35 - a triangular pyramid

Another volume or capacity measures 96 ro = '100 of a cubic cubit. 96 ro can be given as a triangular pyramid whose base and height measure 13 and 9 fingers (mistake '131 ro).

1 1 3

2 4 6

1 2 3

3 5 9

8 14 24

4 7 12

11 19 33

30 52 90

15 26 45

41 71 123

112 194 336

56 97 168

If the side of an equilateral triangle measures 194 parts, its height measures practically 168 parts. Check my above solution using these numbers. You will find the equation

195 x 195 equals 194 x 196 plus 1

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