# Difference between revisions of "The Mathematical Papyri"

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## The Rhind Mathematical Papyrus

This papyrus is sometimes called the Ahmes papyrys. Written by a scribe named Ahmes (Ahmose) in ca. 1550 BCE and bought in modern times by Alexander Henry Rhind. The papyrus lists many problems covering topics in algebra, geometry and other areas in mathematics.

### Problems from the Ahmes (or Rhind) Papyrus

The Rhind mathematical papyrus

Some of the following examples of the problems presented come from a website by G. Donald Allen, Professor at Texas A&M University, College Station. [Egyptian Mathematics]

Others come from Corinna Rossi's book [1] and an article by R.C. Archibald [2].

In the Rhind papyrus there are nineteen problems of a geometric nature: nos. 41-46 and 48-60. The first six problems show how to calculate volume. The problems include finding volumes of right circular cylinders (granaries) and finding volumes and dimensions of paralellopipedal granaries.

Several problems from nos. 48-60 show how to compute area. Formulas are given for the area of a triangle and several four sided figures. The remaining problems "deal with the relation of the lengths of two sides of a right triangle which corresponds to the cotangent of e angle which a face of a regular pyramid makes with its base. This is called the seked of the pyramid." [2]

Definition: The unknown, x , is called the heep.

Problem 24. Find the heep if the heep and a seventh of the heep is 19.

Problems 44-46 Discuss the volumes and dimensions of Parallelopipedal granaries.[2]

Problem 48 This problem shows how the formula from problem 50 is arrived at.

Trisect each side. Remove the corner triangles. The resulting octagonal figure approximates the circle. The area of the octagonal figure is:

$9^2 -4 \frac{1}{2} (3) (3) = 63$

Next we approximate 63 to be 64 and note that $64=8^2$

Thus the number $4(\frac{8}{9})^2 = 3 \frac{13}{81}$ plays the role of π. That this octagonal figure, whose area is easily calculated, so accurately approximates the area of the circle is just plain good luck. Obtaining a better approximation to the area using finer divisions of a square and a similar argument is not simple. [3]

Problem 50. A circular field of diameter 9 has the same area as a square of side 8.

Problem 51 Find the area of a triangle given its base and altitude.[2]

Problem 52 Find the area of a trapezium with (apparently) eqully slanting sides. The lengths of the parallel sides and the distance between them being the given numbers.[2]

Problem 56 This problem indicates an understanding of the idea of geometric similarity. This problem discusses the ratio $\frac{rise}{run}$ The problem essentially asks to compute the cotangent for some angle . Such a formula would be need for building pyramids.

Example of reckoning a pyramid 360 in its "ukha-thebet" (lenght of a side) and 250 in its "peremus" (height). Cause thou that I know the seked of it. You are to take half of 360; It becomes 180. You are to reckon with 250 to find 180. Result: 1/2 + 1/5 + 1/50.

A cubit being 7 palms, you are to multiply by 7.

1 ---- 7

1/2 --- 3 + 1/2

1/5 --- 1 + 1/3 + 1/15

1/50 --- 1/10 + 1/25

Its seked is 5 1/25 palms. [4]

Problem 57. The height of a pyramid is calculated from the base length and the seked (egyptian for slope).

A pyramid 140 in its ukha-thebt (length of a side), and 5 palms, 1 finger in its seked. What is the peremus thereof?

You are to divide 1 cubit by the seked doubled, which amounts to 10 1/2.

You are to reckon with 10 1/2 to find 7 for this one cubit.

Reckon with 10 1/2.

Two-thirds of 10 1/2 is 7.

You are to reckon with 140, for this is the ukha thebt.

Make two-thirds of 140, namely 93 1/3.

This is the peremus thereof. [4]

Problem 58. Determine the seked, given the length of one side and the height. The seked is again five palms, one finger (per cubit). [4] In Clagett the exact problem is given as:

In a pyramid whose altitude is 93 1/3 make known the seqed of it when its base side is 140 [cubits]." (The answer is 5 palms 1 finger)[5]

Problem 59. This exercise contains two parts.

59A In a pyramid whose base side is 12 [cubits] and whose altitude is 8 [cubits]; what is its seqed? (Answer: 5 palms 1 finger)

59B may be a computation to check the answer: If you construct a pyramid with base side 12 [cubits] and with a seqed of 5 palms 1 finger; what is its altitude? (Answer: 8 cubits)[5]

Problem 60. This exercise seems to concern a pillar or a cone instead of a pyramid. It is rather small and steep, with a seked of four palms (per cubit). [4] The problem and its solution as decribed in the papyrus are given in Clagett. [5]

[In] a pillar (iwn) [or perhaps a cone?] with a base side (sntt) [or perhaps diameter] of 15 cubits and a height of 30 [ubits]; what is its seqed?

Take 1/2 of 15; it is 7 1/2. Operate on 30 so as to get 7 1/2. The result is 1/4; which is the seqed.

Problem 63. 700 loaves are to be divided among recipients where the amounts they are to receive are in the continued proportion $\frac{2}{3} : \frac{1}{2} : \frac{1}{3} : \frac{1}{4}$

Definition: The strength is given by the following formula $strength = \frac{1}{grain \ density}$

Problem 72. How many loaves of strength 45 are equivalent to 100 loaves of strength 10?

Problem 79. This problem cites only seven houses, 49 cats, 343 mice, 2401 ears of spelt, 16,807 hekats."

## The Moscow Papyrus

The Moscow mathematical papyrus

The Moscow Mathematical Papyrus is also called the Golenischev Mathematical Papyrus, after its first owner, Egyptologist Vladimir Goleniščev. The Papyrus dates to the Middle Kingdom, ca. 1700 BCE. The Moscow papyrus is smaller than the Rhind papyrus. The most often quoted problem is problem 14. In this problem the volume of a truncated pyramid is computed.

### Problems form the Moscow Papyrus

Problem 4. Identical to problem 51 of the Rhind Mathematical Papyrus. [2]

Problem 6. Given a rectangular enclosure of 12 units area and the ratio of the sides as 1 : 3/4, find the lengths of the sides. This problem seems to be identical to one of the Kahun Papyri in London.

The steps of the solution given in both papyri is the same:

$1 : 3/4 = 4/3; \ 4/3 * 12 =16; \ \sqrt{16} = 4 \ (=x);\ 4 * 3/4 =3 \ (=y)$ [2]

Problem 7. "A triangle of given area is such that it altitude is 2.5 times its base; find both."[2]

Problem 10. Computes the area of a hemisphere. [2]

Problem 14. The base is a square of side 4 cubits, the top is a square of side 2 cubits and the height of the truncated pyramid is 6 cubits.

The text reads: " Example of calculating a truncated pyramid. If you are told: a truncated pyramid of 6 for the vertical height by 4 on the base by 2 on top:

You are to square this 4; result 16. You are to double 4; result 8. You are to square this 2; result 4. You are to add the 16 and the 8 an the 4; result 28. You are to take 1/3 of 6; result 2. You are to take 28 twice; result 56. See it is of 56. You will find (it) right." [2]

## Other Mathematical sources

### Berlin Papyrus

Also known as Berlin P. 6619, this papyrus dates to the Middle Kingdom. It dates to roughly the same historical period as the Moscow papyrus mentioned above. The papyrus contains two problems that solve simultaneous equations. Another papyrus now in Berlin (P. 11529) contains exercises that compute the volume of a tapered column. P. 11529 dates to the second century AD and appears to contain several mistakes. [6]

### Egyptian Mathematical Leather Roll

The leather roll was bought in Egypt by Rhind at approximately the same time that the Rhind Mathematical Papyrus was purchased (mid 19th century). The problems are mainly algebraic in nature. "It consists of 26 unit fraction series each of which is an expression of a rational number of the form or into an Egyptian fraction.". [7]

### Lahun Papyri

The Lahun papyri consist of several fragments. They are sometime referred to as the Kahun papyri. The papyri date to the Middle Kingdom which is the same time period that gave us the Moscow papyrus. The papyri were discovered in the late 19th centuy in the city of Lahun in Egypt.

A problem appearing in IV.3 computes the volume of a granary with a circular base of diameter 12 "mH" and height 8 "mH". A similar problem and procedure can be found in the Rhind papyrus (problem 43).

Problem 1 in LV.4 is given as: "An area of 40 "mH" by 3 "mH" shall be divided in 10 areas, each of which shall have a width that is 1/2 1/4 of their length." [8]

A translation of the problem and its solution as it appears on the fragment is given on the website maintained by University College London. [9]

The other mathematical problems in these papyrus fragments are algebraic in nature.

## References

1. C. Rossi Architecture and Mathematics in Ancient Egypt Cambridge University Press 2004
2. R.C. Archibald Mathematics before the Greeks Science, New Series, Vol.71, No. 1831, (Jan. 31, 1930), pp.109-121
3. Don Allen Egyptian and Babylonian Mathematics
4. R.Greenberg The Rhind Mathematical Papyrus [Rhind Papyrus]
5. M. Clagett Ancient Egyptian Science: Ancient Egyptian Mathematics 1999, retrieved through Google Books
6. K Vogel The truncated Pyramid in Egyptian Mathematics The Journal of Egyptian Archaeology, Vol. 16 No. 3/4 (Nov. 1930), pp. 242-249
7. Gardner, Milo. Egyptian Mathematical Leather Roll. From MathWorld--A Wolfram Web Resource, created by Eric W. Weisstein. [Egyptian Mathematical Leather Roll]
8. Anette Imhausen Digitalegypt website: Lahun Papyrus [IV.3 ]
9. Anette Imhausen Digitalegypt website: Lahun Papyrus [LV.4 ]