I am doing some research on the egyptian mathematics. The so called Moscow papyrus, from 1850 bc, the famous problem 14 shows how to calculate the volume of a frustum. The most authoritive translation i have been able to find on the internet so far, describes it 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"
All sources i have found derives this formula from the text:
V = h/3 (a^2 + a*b +b^2)
which is, to everyones amazement, the correct formula.
I am on track with the "h/3", the "a^2" and the "b^2", but if the translation above is correct, the "a*b" is not correct, but should rather be "2*b" according to the text which says ".. you are to double 4" not "multiply 2 and 4".
Could it be that it has been mistaken since a=2 in the example? Or is the translation above flawed?
Thanks for any comments and help
You are to square the 4;
You are to double 4;
You are to square this 2;
(interesting that he chooses to write "this" here instead of "the" like with the other numbers. Either he wants to emphasize that "this 2" is not the same as the "double" before but actually a. Or "this" could mean "the 2" we just used for "double 4". Much depends on the accuracy of the translation)
You are to add the 16 and the 8 and the 4;
b^2 + 2*b + a^2
You are to take 1/3 of 6;
You are to take 28 twice;
h/3 (b^2 + 2*b + a^2) = 56
(here he uses the word "twice" since it is not one of the given numbers but a (calculated) factor. This is what makes me think that the "double" in "double 4" means a factor of 2 not the a = 2)