Rhind Mathematical Papyrus (7 of 8) / © 1979-2001 by Franz Gnaedinger, Zurich, fg@seshat.ch / www.seshat.ch

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RMP 53 (highly demanding)

The drawing of RMP 53 shows an isosceles triangle with a pair of additional lines parallel to the base. The numbers are (P = peak, AL = additional line, B = base):

7 (P)  7 '2 '4 '8   2 '4 (AL)  3 '4   6 (AL)  5   6 (B)

The additional lines mark a trapezoid with these measurements:

top       2 '4 khet (225 royal cubits)

height    3 '4 khet (325 royal cubits)

base      6    khet (600 royal cubits)

Advanced learners may answer the following questions: a) how long are the oblique sides? b) how long are the sides of the regular octagon of the same area? Let us multiply the above numbers by a factor of 8:

top       18          18

height    26

base      48       15+18+15

The trapezoid is composed of a rectangle of 26 x 18 units and a pair of two rectangular triangles measuring 15 units (base) and 26 units (height). By removing the rectangle and joining the pair of triangles we obtain a triangle. Its base measures 30 units while the height measures 26 units. How long are its slopes? They measure practically 26 units again, according to the following number pattern, which can be used both for approximating the diagonal of a cube and the side and height measurements of an equilateral triangle:

1       1       3

2       4       6

1       2       3

3       5       9

8      14      24

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

These numbers approximate the equilateral triangle:

half side   4    7   15   26    56    97   ...

height      7   12   26   45    97   168   ...

side        8   14   30   52   112   194   ...

Now the trapezoid of RMP 53 can be defined as follows:

top        18

slope      30

height     26

base       48    (angles 60 and 120 degrees)

We answered the first question. Now for the second task. I multiply the above numbers by a factor of 3 and obtain:

top 54   height 78   slope 90   base 144

The area of the trapezoid is found as follows:

(base + upper side) x height x '2

(144 + 54)         x   78   x '2  =  198 x 39

The regular octagon of about the same area can be approximated by means of these numbers:

1       1       2       2

2       3       4       6

5       7      10      14

12      17      24      34

29      41      58      82

70      99     140     198  and so on

side x side     12  17   29   41  ...

area octagon    58  82  140  198  ...

The area of the trapezoid is given by the product 39 x 198, while the area of the octagon may be defined thus:

side x side x 198 x '41

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

side x side x 198 x '41  =  198 x 39

side x side x '41 = 39   side x side = 39 x 41 = 1599

side of octagon  =  practically 40

Now let me divide all numbers by 3 x 8 = 24 in order to obtain measurements in khet:

top trapezoid        2 '4

height trapezoid     3 '4

slope trapezoid      3 '2 '4

base trapezoid       6

side octagon         1 "3

The other numbers of the drawing may represent isosceles triangles of the following measurements:

height   5    7 '2 '4 '8    7              7

base     6    2 '4          7 '2 '4 '8     6

I transform the triangles into octagons of about the same area:

base          height         factor    octagon side

6             5              21        37 or 1 '2 '4 '84

6             7              12        25 or 2 '12

2 '4          7 '2 '4 '8     12        23 or 1 '2 '3 '12

7 '2 '4 '8    7              72       172 or 2 '3 '18

In problem no. 53 of the Rhind Mathematical Papyrus we may assemble 8 or more similar problems, which can hopefully be reconstructed in the context of the previous calculations. The drawing of RMP 53 led me to a trapezoid and four triangles whose areas are quite easily transformed into regular octagons. However, I overlooked one possibility. The numbers

7     7 '2 '4 '8     2 '4

define a triangle with a base of 2 '4 khet and a height of 7 khet with an area measuring 7 '2 '4 '8 square khet or setat (or aroure, Greek name for setat). This area is seen in the last part of the written calculation:

2 '4 x 7 setat x '2  =  7 '2 '4 '8 setat

Now let me transform this area, 7 '2 '4 '8 setat or 78,500 square cubits, into a regular octagon:

side x side     12  17   24   29  ...

area octagon    58  82  116  140  ...

78,750 x 29 x '140  =  16,312 '2

127 x 127 = 16,129     128 x 128 = 16,384

No simple numbers. Let me begin anew and try it with palms instead of royal cubits:

700 x 7 x 225 x 7 x '2  =  3,858,750 (square palms)

3,858,750 x 29 x '140  =  799,312 '2

893 x 893  =  797,449  (1863 '2 less)

894 x 894  =  799,236  (  76 '2 less)

895 x 895  =  801,025  (1712 '2 more)

A triangle with a base of 225 royal cubits and a height of 700 royal cubits and a regular octagon whose side measures 894 palms have roughly the same area. Now let me look for a suitable grid. The number 894 does not appear in the above number pattern, but we may proceed like this:

6 x 140  plus  41  plus  10  plus  3  =  894

6 x  99  plus  29  plus   7  plus  2  =  632

9 x 99  plus  3  =  894

9 x 70  plus  2  =  632

Hence the grid of the octagon of the same area measures roughly

632+894+632 by 632+894+632 palms

while the octagon's area measures 3,858,116 square palms or practically 78,737 square cubits, only 13 square cubits less than 78,750 square cubits.

One of RMP 53's calculations seems to concern a triangle with the following measurements:

base  2 '4 khet  (225 rc)   height  4 '2 khet (450 rc)

area    '2 x 225 cubits x 450 cubits  =  50,625 setat

(1 setat being 1 square khet or 10,000 square cubits)

Transforming this area into a regular octagon:

side x side     5  12   29  ...

area octagon   24  58  140  ...

50,625 x 29 x '140  =  10,487 (rounded)

101 x 101  =  10,201  ---  286 less

102 x 102  =  10,404  ---   83 less

103 x 103  =  10,609  ---  122 more

The side length of the octagon lies between 102 and 103 royal cubits, nearer to 102 cubits. By multiplying these numbers by a factor of 5 we obtain:

'2 x 225 x 5 x 450 x 5  =  1,265,625

1,265,625 x 29 x '140  =  262,165 (rounded)

511 x 511  =  261,121  ---  1044 less

512 x 512  =  262,144  ---    21 less

513 x 513  =  263,169  ---  1004 more

We have found a good value: the side of the octagon measures practically 512 x '5 cubits. Now for the grid:

3 x 140  plus  58  plus  17  plus  17  =  512

3 x  99  plus  41  plus  12  plus  12  =  362

5 x 99  plus  17  =  512

5 x 70  plus  12  =  362

Hence the grid measures

362+512+362 by 362+512+362  fifth royal cubits

while the area of the octagon in the grid is found as follows:

362+512+362  =  1236 fifth royal cubits

1236 x 1236  -  2 x 362 x 362  =  1,265,608

1,265,608 x '5 x '5  =  50,624 '5 '10 '50 square cubits

area of triangle  =  50,625 square cubits

mistake only  '2 '6 '75 square cubit

We have been lucky: we found a result with a tiny margin of error of less than 1 square cubit in an area of over 50,000 square cubits. Or the other way round: the numbers of the triangle were chosen because they would yield a fine result.

Now for the remaining calculation of RMP 53 (translation by Thomas Eric Peet, measurement transformed by myself):

1/10 of an area measures 14,750 square cubits

1/10 of the area subtracted, then this is the area

There are two areas, one of them '10 smaller than the other - perhaps an octagon in a circle? Let us consider an octagon in a simple grid:

side octagon  10   square  24 by 24

partition  7+10+7 = 24   grid  7+10+7 by 7+10+7

diameter of the inscribed circle  24

diameter of the circumscribed circle  26

according to the triple  2 x 5-12-13

diameter 26 / side 10  =  13 to 5

The area of the octagon measures

24x24 - 2x7x7  =  478 square units

The area of the circle is given by the following formula:

13 x 13 x re

Is there a good value for re? Yes, 531 above 169:

3  (plus 22)  22  47  69  91  113  ...  531

1  (plus  7)   8  15  22  29   36  ...  169 = 13x13

13 x 13 x 531 x '13 x '13  =  531

Area of the circle                531 square units

area of the inscribed octagon     478 square units

The ratio is practically 10 to 9, according to a crosswise multiplication:

531 x 9 = 4779    478 x 10 = 4780

Hence the two areas mentioned in RMP 53 may really be a circle and an octagon.

Now let me invent a problem: The radius of a circle measure 2 '6 khet or '3 x 650 royal cubits - how long is the side of the inscribed octagon? I calculate the area of the circle by means of the formula:

'3 x 650 x '3 x 650 x '169 x 531  =  147,500 square cubits

area of the circle                   147,500 square cubits

minus one tenth (RMP 53)              14,750 square cubits

= area of the inscribed octagon      132,750 square cubits

Diameter / side  =  13 to 5

diameter  =  2 x 2 '6 khet  =  4 '3 khet  =  '3 x 13 khet

side  =  '3 x 13 x 5 x '13 khet  =  '3 x 5  or  1 '6 khet

If the radius of a circle measures 2 '6 khet, the side of the inscribed regular octagon measures about 1 '6 khet.

A more precise result is derived by means of another octagon. Draw the square 24 by 24: the diagonals measure practically  34 units. Prolong the axes by 5 units on every side: they will measure 34 units. Join the ends of the axes and the corners of the square, and you obtain an octagon: its side measures 13 units, according to the triple 5-12-13

Diameter / side  =  34 to 13

The diameter of my circle measures

4 '3 khet or '3 x 1300 royal cubits

while the side of the inscribed octagon measures

'3 x 1300 x 13 x '34  =  165 '2 '6 '51 royal cubits

margin of error: only 1 palm on 165 royal cubits

This is my interpretation of RMP 53 so far. It would no doubt have provided subject matter for a whole semester in the seminary of professor Ahmes.

RMP 54 and 55

7 setat of land are divided into 10 fields, each one measuring 7,000 square cubits.

3 setat of land are divided into 5 fields, each one measuring 6,000 square cubits.

You may say that these are very simple calculations compared to RMP 53. But Ahmes will smile, and propose that you transform these areas into regular octagons. You will try this, and find the following solutions:

OCTAGON A

side  38 royal cubits

square  92 by 92 royal cubits

partition  27 + 38 + 27  =  92 royal cubits

grid  27+38+27 by 27+38+27  royal cubits

area octagon (grid)  7,006 square cubits

OCTAGON B

side  35 royal cubits

square  85 by 85 royal cubits

grid  25+35+25 by 25+35+25 royal cubits

area octagon (grid)  5,975 square cubits

Ahmes will be pleased. Then he may propose that you multiply the numbers by a factor of 20, refine the grids and look out for a pair of new octagons. With your previous learning you will find the following solutions quite easily:

OCTAGON A

side  704   (20 units = 1 royal cubit)

square  1700 x 1700  (85 by 85 royal cubits)

partition  498 + 704 + 498  =  1700  (85 royal cubits)

grid  498+704+498 x 498+704+498

OCTAGON B

side  762   (20 units = 1 royal cubit)

square  1840 x 1840  (92 by 92 royal cubits)

partition  539 + 762 + 539  =  1840  (92 royal cubits)

grid  539+762+539 x 539+762+539

Then Ahmes will ask you to draw the grids and octagons, using Maantef marks instead of royal cubits (20 Maantef marks equal 10 Nut marks or 1 finger or about 1.87 centimeters):

OCTAGON A and B

partitions  539+762+539  and  498+704+498 Maantef marks

Next, Ahmes will ask you to draw a circle around the smaller octagon and another one inside of the larger octagon. By doing so you will be surprised: the circles have the same diameter:

OCTAGON A    side  762 Maantef marks

square  1840 by 1840 Maantef marks (92 by 92 fingers)

grid  539+762+539 by 539+762+539 Maantef marks

diameter of the inscribed circle  1840 Mm (92 fingers)

OCTAGON B    side  704 Maantef marks

square  1700 by 1700 Maantef marks (85 by 85 fingers)

grid  498+704+498 by 498+704+498 Mantef marks

diameter of the circumscribed circle  1840 Mm (92 fingers)

according to the pseudo-triple  4 x 176-425-460

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