|Moscow Mathematical Papyrus|
|Pushkin State Museum of Fine Arts in Moscow|
14th problem of the Moscow Mathematical Papyrus (V. Struve, 1930)
|Date||13th dynasty, Second Intermediate Period of Egypt|
|Place of origin||Thebes|
|Size||Length: 5.5 metres (18 ft)|
Width: 3.8 to 7.6 cm (1.5 to 3 in)
The Moscow Mathematical Papyrus is an ancient Egyptian mathematical papyrus, also called the Golenishchev Mathematical Papyrus, after its first owner outside of Egypt, Egyptologist Vladimir Golenishchev. Golenishchev bought the papyrus in 1892 or 1893 in Thebes. It later entered the collection of the Pushkin State Museum of Fine Arts in Moscow, where it remains today.
Based on the palaeography and orthography of the hieratic text, the text was most likely written down in the 13th dynasty and based on older material probably dating to the Twelfth dynasty of Egypt, roughly 1850 BC. Approximately 5½ m (18 ft) long and varying between 3.8 and 7.6 cm (1.5 and 3 in) wide, its format was divided into 25 problems with solutions by the Soviet Orientalist Vasily Vasilievich Struve in 1930. It is a well-known mathematical papyrus along with the Rhind Mathematical Papyrus. The Moscow Mathematical Papyrus is older than the Rhind Mathematical Papyrus, while the latter is the larger of the two.
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.
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 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 ½ times and added to 4 to make 10. In other words, in modern mathematical notation one is asked to solve
A higher pefsu number means weaker bread or beer. The pefsu number is mentioned in many offering lists. For example problem 8 translates as:
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.
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).
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."
The solution amounts to computing the area as
This means the scribe of the Moscow Papyrus used to approximate π.
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" 
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.
Other mathematical texts from Ancient Egypt include:
For the 2/n tables see:
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.