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RMP 53, 54, 55
Posted:
May 27, 2009 8:14 AM


Dear Forum members:
Translating RMP 53 as three separate problems, two triangle areas and an area of undefined shape, consider
http://planetmath.org/encyclopedia/CubitsEgyptianGeometryAreasCalculatedIn.html by:
A scribal area of triangles and other shapes area calculation method is reported by three Rhind Mathematical Papyrus problems RMP 5355. The scribal geometry utilized quotients and remainders in an arithmetic context that was looked like scribal weights and measures, algebra, and/or 2/n tables calculations.
RMP 53 calculated the area of two triangles, of 45/8 setat and 63/8 setat, and a third area of an undefined shape by the note, 1/10 of 1 3/8 mh added to 10 cubits of land (COL) is the desired area. A setat was 100 cubit by 100 cubit, or 10,000 square cubits. A cubit of land (COL), or mh, was one cubit wide by 100 cubits long, or 1/100 setat.
The first triangle had an altitude of 5 khet and a base of 9/4 khet. Using the area of a triangle formula, 1/2 the base times the altitude, 5*(9/4)*(1/2)= (45/8) = 5 5/8 setat.
The second triangle had an altitude of 7 khet and a base of 9/4 khet. Using the area of a triangle formula, 1/2 the base times the altitude, Ahmes calculated 7*(9/4)*(1/2) = 63/8 = 7 7/8 setat
The third calculation found the area of undefined shape discussed by:
11/8 mh = 110/8 mh + 10 mh = 23 3/4 mh = 1/8 setat + 11 1/4 mh
since 12 1/2 mh = 1/8 setat.
Scholars have suggested that a truncated pyramid or a triangle defined the third shape.
To assist the decoding of the third RMP 53 area RMP 54, and RMP 55 setat and mh data have been consulted.
RMP 54 partitioned 7/10 setat by 10, 5, 2 1/2 and 1 1/4 segments. Proof was provided by multiplying one setat by 7/10, 14/10, 28/10 and 56/10 within a quotient and remainder context. A quotient setat and a scaled remainder mh were scaled as the 2/n table and a ro unit in hekat (volume unit) were scaled, by writing:
a. (7/10)*(4/4) = 28/40 = (24 + 3)/40 = 3/8 setat + 300/40 mh = 5/8 setat + 7 1/2 mh
b. (14/10)*(4/4) = 56/10 = (55 + 1)/40 = 11/8 setat + 100/4 mh = 1 3/8 setat + 2 1/2 mh
c. (28/10)*(2/2) = 56/20 = (55 + 1)/20 = 11/4 setat + 100/20 mh = 2 3/4 setat + 5 mh
d. (56/10) = (55 + 1)/10 = 11/2 setat + 100/10 COL = 5 1/2 setat + 10 mh
Ahmes may have also made calculations thinking in mh unuts. For example,
Ahmes shorthand partition of 7/10 setat, (1/2 + 1/5) setat, may have focused upon 1/5 setat written as 20 mh. Knowing 12 1/2 mh was 1/8 setat, an answer may have been recorded by:
(1/2 + 1/5)setat = (1/2 + 1/8 + (20  12 1/2 mh) = 5/8 setat + 7 1/2 mh.
RMP 55 takes 3/5 of 5 setat to obtain 3 setat by three steps:
a. 1/2 setat + 10 mh
b. 1 1/8 setat + 7 1/2 mh
c. 1 3/8 setat + 2 1/2 setat
d. adding steps a. and c, knowing that 12 1/2 mh = 1/8 setat
(1/2 setat + 10 mb) + (1 3/8 setat + 2 1/2 mh) = 2 7/8 setat + 12 1/2 mh = 3 setat
Comments would be apppreciated.
Best Regards to all,
Milo Gardner



