Rhind Mathematical Papyrus (5 of 8) / © 1979-2001 by Franz Gnaedinger, Zurich, fg(a)seshat.ch, fgn(a)bluemail.ch / www.seshat.ch

Rhind 1 / Rhind 2 / Rhind 3 / Rhind 4 / Rhind 5 / Rhind 6 / Rhind 7 / Rhind 8

RMP 24 to 30 - Sacred Triangle 3-4-5 (cube 1-1-1)

These problems are written in a column taking up the whole height of the sheet it is written on, and may concern the Sacred Triangle 3-4-5, the inscribed circle, the cube 1-1-1, and half a cube.

Imagine a circle inscribed in a rectangular triangle. How long is the diameter? Use this very simple formula: diameter = sum of legs minus hypotenuse. Consider the case of the Sacred Triangle measuring 3-4-5 palms. The diameter of the inscribed circle measures

3 plus 4 minus 5  =  2 palms

RMP 24     A plus '7 A equals 19

Imagine a Sacred Triangle measuring 9 units (base), 12 units (height), 15 units (slope), 3 units (radius of the inscribed circle), and 6 units (diameter of the inscribed circle):

base + height + radius = 19 palms

Base, height and slope measure 7 '8 - 9 '2 - 11 '2 '4 '8 palms. The diameter of the inscribed circle measures 19 fingers.

RMP 25     A plus '2 A equals 16

height + diameter = 16 palms

Base, height, slope and diameter measure 8 - 10 "3 - 13 '3 5 '3 palms. The periphery measures 24 palms.

RMP 26     A plus '4 A equals 15

height + radius = 15 palms

Base, height, slope and diameter measures 9 - 12 - 15 - 6 palms.

RMP 27     A plus '5 A equals 21

slope + diameter = 21 palms

Base, height, slope and diameter measure 42 - 56 - 70 - 28 fingers. The diameter of the inscribed circle measures 1 royal cubit. The height of the triangle measures 2 royal cubits, and the periphery measures 6 royal cubits.

Calculate the circumference of the inscribed circle using the formula circumference = diameter x re and compare it to the periphery of the Sacred Triangle. The ratio equals re to 6 (about 1/2 or 23/44 or 67/128 or 111/212). Calculate the area of the inscribed circle using the formula '4 x diameter x diameter x re and compare it to the area of the Sacred Triangle. You will find the same ratio: re to 6. Imagine a sphere in a cube, calculate the volume of the sphere using the formula '6 x diameter x diameter x diameter x re and compare it to the volume of the cube. The ratio equals re to 6 again. Calculate the surface of the sphere using the formula diameter x diameter x re and compare it to the surface of the cube. The ratio equals re to 6.

Compare the Sacred Triangle 3-4-5 palms to the cube 1-1-1 palms. Both the periphery of the triangle and the sum of the edges of the cube measure 12 palms. Both the area of the triangle and the surface of the cube measure 6 square palms. And both the area of the circle in the triangle and the surface of the sphere in the cube measures re square palms.

RMP 29     10 A times 1 '4 '10 equals 13 '2

The diagonal of a square measures 10 palms. The side of the square equals the circumference of a circle inscribed in a Sacred Triangle. Calculate the periphery of this triangle.

1       1       2

2       3       4

5       7      10

12      17      24

29      41      58

70      99     140

Multiply 10 palms by '99 x 140, then by '22 x 7, then by 6:

10 x '99 x 140 x '22 x 7 x 6 = 10 x 1 '4 '10 = 13 '2

The periphery of the Sacred Triangle measures 13 '2 palms. Base, height, slope and diameter measure 3 '4 '8 - 4 '2 - 5 '2 '8 - 2 '4 palms.

RMP 30     A times "3 '10 equals 10

Imagine a Sacred Triangle whose periphery measures 10 royal cubits or 70 palms Calculate the distance of the center of the inscribed circle from the corner between base and slope in palms:

The periphery of the triangle measures 10 cubits or 70 palms. Base, height and slope measure 17 '2 - 23 '3 - 29 '6 palms. Radius and diameter of the inscribed circle measure 5 '2 '3 and 11 "3 palms. The distance in question is the slope of a rectangular triangle whose rise and run measure 5 '2 '3 and 11 "3 palms, corresponding to the diagonal of a double square measuring 5 '2 '3 by 11 "3 palms.

1       1       5

2       6      10

1       3       5

4       8      20

2       4      10

1       2       5

3       7      15

10      22      50

5      11      25

16      36      80

8      18      40

4       9      20

13      29      65

42      94     210

21      47     105

68     152     340

34      76     170

17      38      85

55     123     275

178     398     890

89     199     445

288     644    1440

144     322     720

72     161     360

Multiply 5 '2 '3 palms by 360 and you obtain 2100 palms. Divide 2100 palms by 161 and you obtain 13 '23 palms. Solution of RMP 30: 13 '23 (palms). Check the result. Multiply 13 '23 palms by '360 x 161, thus you obtain the radius of the circle in palms. Multiply it by 12 and you obtain the periphery of the triangle in palms. Multiply it by '7 and you obtain the periphery in royal cubits:

13 '23 x '360 x 161 x 12 x '7 = 13 '23 x "3 '10 = 10

Half a cube measures 5 '2 '3 by 11 "3 by 11 "3 palms. The diagonals of the faces measure practically 13 '23 palms and 66 fingers respectively. The cubic diagonal measures exactly 70 fingers, according to the quadruple 1-2-2-3. The four cubic diagonals together measures 10 royal cubits.

If you were told to imagine half a cube, you were informed  that the sum of the cubic diagonals measures 10 royal cubits, and you were asked to calculate the diagonal of a smaller face in palms, you would have to carry out the following multiplication

10 times '4 x 7 x '3 x 360 x '161 equals A

or to solve this equation

A times 4 x '7 x 3 x '360 x 161 equals 10

corresponding to Ahmes' equation A times "3 '10 equals 10 (RMP 30).

RHIND MATHEMATICAL PAPYRUS, problems no. 39, 40, 64 (intermezzo)

RMP 39

100 loaves of bread are distributed among ten men. 50 loaves are given to 5 men. Each man obtains 12 '2 loaves. The remaining 50 loaves are given to 6 men. Each man receives 8 '3 loaves. Now Ahmes calculates the difference

12 '2  minus  8 '3  equals 4 '6

Let me continue as follows:

'2 x 25  minus  '3 x 25  equals  '6 x 25

'2 minus '3 equals '6

'2 equals '3 plus '6

This equation belongs to a stairway:

'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 '22

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

'2 = '1x3 '3x5 '5x7 '7x9 '9x11 '11x13 '13x15 '15x17 '17x19 ...

RMP 40

100 loaves of bread are distributed as follows: 3 men obtain 23 "3 '7 loaves each, and 2 men receive 14 '4 '28 loaves each. Let me proceed in the same way as before:

23 "3 '7  minus  14 '4 '28  equals  9 '2 '42

'42 x 1000  minus  '70 x 1000  equals  '105 x 1000

'42 minus '70 equals '105

'6 minus '10 equals '15

'6 equals '10 plus '15

This equation belongs to another number pattern:

'2  =  '6   '1x3

'6  =  '10  '3x5

'10  =  '14  '5x7

'14  =  '18  '7x9

'18  =  '22  '9x11   (and so on)

RMP 64

Ten men obtain 7, 9, 11, 13, 15, 17, 19, 21, 23, 25 parts of barley, all in all ten hekat. The volume of one part can be defined as follows:

7 fingers  x  7 fingers  x  14 Qhebsenuf marks

(15 Qhebsenuf marks equal 1 finger)

15 parts can be defined as follows:  7 by 7 by 14 fingers

RHIND MATHEMATICAL PAPYRUS, problems no. 41- 60 (demanding)

RMP 41

A granary in the form of a cylinder has an inner diameter of 9 royal cubits and an inner height of 10 royal cubits.

A square of the side 8 rc and a circle of the diameter 9 rc have about the same area. So the floor of the granary measures about 64 square cubits. Multiply this area by the height 10 royal cubits and you will obtain a volume of 640 cubic cubits = 960 khar = 4,800 quadruple hekat = 19,200 hekat.

Now let us calculate the volume of the granary more exactly:

diameter   9 royal cubits  or  63 palms  or  252 fingers

height    10 royal cubits  or  70 palms  or  280 fingers

By using the value '7 of 22 for re we obtain:

diameter         63 palms  or   252 fingers

circumference   198 palms  or  1386 fingers

area wall 13,860 square palms  area floor 49,896 square fingers

volume  13,970,880 cubic fingers or about 636 cubic cubits

A better result than the first one. Now, as a game for advanced learners, we may exchange the numbers:

diameter  10 royal cubits  or  70 palms  or  280 fingers

height     9 royal cubits  or  63 palms  or  252 fingers

circumference                 220 palms  or  880 fingers

area wall  13,860 square palms

area floor  3,850 square palms  or  61,600 square fingers

volume  15,523,200 cubic fingers  or  about 707 cubic cubits

The walls of the two cylinders have the same area while their volumes maintain the ratio   9 to 10.

The diameter of another cylinder measures 15 royal cubits and the height 6 royal cubits. The wall again has the same area while the volume increases to 23,284,800 cubic fingers or about 1061 cubic cubits.

A further cylinder may have a diameter of 90 royal cubits and a height of 1 royal cubit. The area of the wall would again be the same while the volume would increase to about 6364 cubic cubits.

Now please consider a sequence of cylinders:

diameter  1   2   3   5   6   9  10  15  18  30  45  90  rc

height   90  45  30  18  15  10   9   6   5   3   2   1  rc

The walls have the same area while the volume increases in a peculiar way: divide the height by any number and the volume will increase by the same factor ... The lower the cylinder the bigger the volume. And if the wall has no height at all? In that case the volume would be infinitely huge ... A pretty paradox to be discussed in the seminary of professor Ahmes.

RMP 42

The inner diameter of another cylindrical granary measures 10 royal cubits and the inner height again 10 royal cubits. Using the same formula (diameter 9, side 8), Ahmes obtains a capacity of 790 '18 '27 '54 or 790 '9 cube cubits instead of 790 '9 '81 cube cubits. Using the value '7 of 22 for re we would obtain 785 '2 '7 '14 cubic cubits. Using the still better value '120 of 377 we obtain 785 '3 '12 cube cubits.

diameter    1    2    4    5   10   20   25   50  100

height    100   50   25   20   10    5    4    2    1

The walls of these cylinders have the same area, while the volume increases from about 78 '2 to about 7854 cube cubits. Let us consider the following pair of examples:

diameter  2   height 50   volume  157   re   '50 of 157

diameter 50   height  2   volume 3927   re '1250 of 3927

The second value for re is generated by a number sequence:

33  (plus 66)  99  165  ...  3795  3861  3927

11  (plus 21)  32   53  ...  1208  1229  1250

The same value can be found by means of two multiplication processes:

1250 x 3 '9 '33   equals  3750  139  38  or  3927 (rounded)

1250 x 3 '10 '24  equals  3750  125  52  or  3927 (rounded)

RMP 43

Ahmes calculates the capacity of a cylindrical granary whose inner diameter measures 6 royal cubits while the inner height measures 9 royal cubits. However, his result is wrong, because in the middle of his calculation he jumps from one formula to a different one. Perhaps an intentional mistake to challenge his pupils? Here I am not concerned with any mistake or possible intentions, but only with the numbers of the cylinder.

If a circle of the diameter 9 rc and a square of the side 8 rc have the same area, the volume of the above cylinder measures 256 cubic cubits. Now let me compare two cylinders:

diameter   6 or 9 royal cubits

height     9 or 6 royal cubits

If you remember my interpretation of RMP 41, you can easily calculate the measurements of the second cylinder: its round wall has the same area while its volume equals

'6 x 9 x 256 cubic cubits  =  384 cubic cubits

Now please imagine a granary in the shape of a hemi-ellipsoid within the frame of the second cylinder: what is the volume?

The circle of the second cylinder has a diameter of 9 royal cubits. Let us first imagine a sphere with the same diameter, 9 rc, and calculate its volume by means of the formula

volume sphere  =  '6 diameter x diameter x diameter x re

By using the value '81 of 256 or '81 x 256 for re we obtain:

'6 x 9 x 9 x 9 x '81 x 256 ccc  =  384 cubic cubits

What a tidy result: a cylinder with diameter 9 and height 6 and a sphere with diameter 9 have exactly the same volume.

Now for the ellipsoid. This geometric form is nothing other than a sphere lengthened in one dimension. The diameter of  the above sphere measures 9 cubits in every dimension while the vertical diameter of the ellipsoid measures 6 + 6 = 12 cubits. We obtain the volume of the ellipsoid by multiplying that of the sphere by a factor of '9 x 12 = '3 x 4 as follows:

sphere      diameter 9 x 9 x 9     volume  384 ccc

ellipsoid   diameter 9 x 9 x 12    volume  512 ccc

Now the volume of the hemi-ellipsoid equals '2 x 512 ccc = 256 cubic cubits: exactly the volume of the first cylinder.

A granary in the shape of a cylinder has a diameter of 6 royal cubits and a height of 9 royal cubits -- another granary in the shape of a hemi-ellipsoid has a diameter of 9 royal cubits and a height of 6 royal cubits -- the two granaries have exactly the same volume

RMP 44 and 45

These problems concern a container in the form of a cube measuring 10 by 10 by 10 royal cubits. If you wish to build such a granary you have to know the lengths of the floor diagonals, of the face diagonals, and of the cubic diagonals.

1       1       2

2       3       4

5       7      10

12      17      24

29      41      58

70      99     140   and so on

If the edge of a cube measures 10 royal cubits or 70 palms, the diagonal of a face measures practically 99 palms (mistake 0.4 millimeters).

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   and so on

If the edge of a cube measures 10 royal cubits or 70 palms or 280 fingers = 5 x 56 fingers, the cubic diagonal measures practically 5 x 97 = 485 fingers (mistake 0.5 millimeters).

Now we may go a step further and imagine a cone, a sphere, a hemi-ellipsoid and a cylinder in the frame of the cube. By using the simple value '50 of 157 for re we obtain:

CONE             diameter base   10 royal cubits

area base       78 '2 square cubits

height          10 royal cubits

volume         261 "3 cubic cubits

SPHERE           diameter        10 royal cubits

volume         523 '3 cubic cubits

HEMI-ELLIPSOID   diameter base   10 royal cubits

area base       78 '2 square cubits

height          10 royal cubits

volume         523 '3 cubic cubits

CYLINDER         diameter        10 royal cubits

area base       78 '2 square cubits

height          10 royal cubits

volume         785 cubic cubits

Volumes  CONE : HEMI-ELLIPSOID : CYLINDER  =  1 : 2 : 3

The granary in the form of a cube has a volume of 1,000 cubic cubits, while a granary in the shape of a hemi-ellipsoid in the frame of the same cube has a volume of about 523 cubic cubits. When we transform the cube into a right parallelepiped, the ratio of the volume of the parallelepiped to the volume of the inscribed hemi-ellipsoid remains the same, about 2. This generates a very simple practical formula:

If you see a granary in the shape of a hemi-ellipsoid and wish to estimate its volume, carry out the following calculation: '2 x diameter base x diameter base x height and you obtain the approximate volume of the granary.

Rhind 1 / Rhind 2 / Rhind 3 / Rhind 4 / Rhind 5 / Rhind 6 / Rhind 7 / Rhind 8