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Moscow Mathematical Papyrus
please note:
- the content below is remote from Wikipedia
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- the content below is remote from Wikipedia
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Exercises contained in the Moscow Papyrus
The problems in the Moscow Papyrus follow no particular order, and the solutions of the problems provide much less detail than those in the Rhind Mathematical Papyrus. The papyrus is well known for some of its geometry problems. Problems 10 and 14 compute a surface area and the volume of a frustum respectively. The remaining problems are more common in nature.Ship's part problems
Problems 2 and 3 are ship's part problems. One of the problems calculates the length of a ship's rudder and the other computes the length of a ship's mast given that it is 1/3 + 1/5 of the length of a cedar log originally 30 cubits long.Aha problems
{{Hiero|Aha|P6-a:M35|align=left|era=nk}}Aha problems involve finding unknown quantities (referred to as Aha) if the sum of the quantity and part(s) of it are given. The Rhind Mathematical Papyrus also contains four of these type of problems. Problems 1, 19, and 25 of the Moscow Papyrus are Aha problems. For instance problem 19 asks one to calculate a quantity taken 1 and Â½ times and added to 4 to make 10. In other words, in modern mathematical notation one is asked to solve 3/2 times x + 4 = 10Pefsu problems
Most of the problems are pefsu problems (see: Egyptian algebra): 10 of the 25 problems. A pefsu measures the strength of the beer made from a hekat of grain
mbox{pefsu} = frac{mbox{number loaves of bread (or jugs of beer)}}{mbox{number of heqats of grain}}
A higher pefsu number means weaker bread or beer. The pefsu number is mentioned in many offering lists. For example problem 8 translates as:
(1) Example of calculating 100 loaves of bread of pefsu 20
(2) If someone says to you: "You have 100 loaves of bread of pefsu 20
(3) to be exchanged for beer of pefsu 4
(4) like 1/2 1/4 malt-date beer"
(5) First calculate the grain required for the 100 loaves of the bread of pefsu 20
(6) The result is 5 heqat. Then reckon what you need for a des-jug of beer like the beer called 1/2 1/4 malt-date beer
(7) The result is 1/2 of the heqat measure needed for des-jug of beer made from Upper-Egyptian grain.
(8) Calculate 1/2 of 5 heqat, the result will be 2 1/2
(9) Take this 2 1/2 four times
(10) The result is 10. Then you say to him:
(11) "Behold! The beer quantity is found to be correct."
Baku problems
Problems 11 and 23 are Baku problems. These calculate the output of workers. Problem 11 asks if someone brings in 100 logs measuring 5 by 5, then how many logs measuring 4 by 4 does this correspond to? Problem 23 finds the output of a shoemaker given that he has to cut and decorate sandals.Geometry problems
Seven of the twenty-five problems are geometry problems and range from computing areas of triangles, to finding the surface area of a hemisphere (problem 10) and finding the volume of a frustum (a truncated pyramid).Two Interesting Geometry Problems
Problem 10
The 10th problem of the Moscow Mathematical Papyrus asks for a calculation of the surface area of a hemisphere (Struve, Gillings) or possibly the area of a semi-cylinder (Peet). Below we assume that the problem refers to the area of a hemisphere.The text of problem 10 runs like this: "Example of calculating a basket. You are given a basket with a mouth of 4 1/2. What is its surface? Take 1/9 of 9 (since) the basket is half an egg-shell. You get 1. Calculate the remainder which is 8. Calculate 1/9 of 8. You get 2/3 + 1/6 + 1/18. Find the remainder of this 8 after subtracting 2/3 + 1/6 + 1/18. You get 7 + 1/9. Multiply 7 + 1/9 by 4 + 1/2. You get 32. Behold this is its area. You have found it correctly."Williams, Scott W. Egyptian Mathematical PapyriThe solution amounts to computing the area as
text{Area} = left (frac{2 times 8}{9}right)^2 times (text{diameter})^2 = frac{256}{81} (text{diameter})^2
This means the scribe of the Moscow Papyrus used frac{256}{81} approx 3.16049 to approximate pi.Problem 14: Volume of frustum of square pyramid
missing image!
- Pyramide-tronquÃ©e-papyrus-Moscou 14.jpg -
left
The 14th problem of the Moscow Mathematical calculates the volume of a frustum. Problem 14 states that a pyramid has been truncated in such a way that the top area is a square of length 2 units, the bottom a square of length 4 units, and the height 6 units, as shown. The volume is found to be 56 cubic units, which is correct.The text of the example runs like this: "If you are told: a truncated pyramid of 6 for the vertical height by 4 on the base by 2 on the top: You are to square the 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 and 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" as given in Gunn & Peet, Journal of Egyptian Archaeology, 1929, 15: 176. See also, Van der Waerden, 1961, Plate 5The solution to the problem indicates that the Egyptians knew the correct formula for obtaining the volume of a truncated pyramid:
- Pyramide-tronquÃ©e-papyrus-Moscou 14.jpg -
left
V = frac{1}{3} h(a^2 + a b +b^2)
where a and b are the base and top side lengths of the truncated pyramid and h is the height. Researchers have speculated how the Egyptians might have arrived at the formula for the volume of a frustum but the derivation of this formula is not given in the papyrus.{{citation
| last = Gillings | first = R. J.
| issue = 8
| journal = The Mathematics Teacher
| jstor = 27957144
| pages = 552â€“555
| quote = While it has been generally accepted that the Egyptians were well acquainted with the formula for the volume of the complete square pyramid, it has not been easy to establish how they were able to deduce the formula for the truncated pyramid, with the mathematics at their disposal, in its most elegant and far from obvious form
| title = The volume of a truncated pyramid in ancient Egyptian papyri
| volume = 57
| year = 1964}}.
| issue = 8
| journal = The Mathematics Teacher
| jstor = 27957144
| pages = 552â€“555
| quote = While it has been generally accepted that the Egyptians were well acquainted with the formula for the volume of the complete square pyramid, it has not been easy to establish how they were able to deduce the formula for the truncated pyramid, with the mathematics at their disposal, in its most elegant and far from obvious form
| title = The volume of a truncated pyramid in ancient Egyptian papyri
| volume = 57
| year = 1964}}.
Other papyri
Other mathematical texts from Ancient Egypt include:- Berlin Papyrus 6619
- Egyptian Mathematical Leather Roll
- Lahun Mathematical Papyri
- Rhind Mathematical Papyrus
References
Full Text of the Moscow Mathematical Papyrus
- Struve, Vasilij Vasil'eviÄ, and Boris Turaev. 1930. Mathematischer Papyrus des Staatlichen Museums der SchÃ¶nen KÃ¼nste in Moskau. Quellen und Studien zur Geschichte der Mathematik; Abteilung A: Quellen 1. Berlin: J. Springer
Other references
- Allen, Don. April 2001. The Moscow Papyrus and Summary of Egyptian Mathematics.
- Imhausen, A., Ã„gyptische Algorithmen. Eine Untersuchung zu den mittelÃ¤gyptischen mathematischen Aufgabentexten, Wiesbaden 2003.
- Mathpages.com. The Prismoidal Formula.
- O'Connor and Robertson, 2000. Mathematics in Egyptian Papyri.
- Truman State University, Math and Computer Science Division. Mathematics and the Liberal Arts: Ancient Egypt and The Moscow Mathematical Papyrus.
- Williams, Scott W. Mathematicians of the African Diaspora, containing a page on Egyptian Mathematics Papyri.
- Zahrt, Kim R. W. Thoughts on Ancient Egyptian Mathematics.
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