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Thread: Rhind papyrus problem 41 (pi)

  1. #1
    Apr 2013

    Rhind papyrus problem 41 (pi)

    Okay, i got a little problem. I have to do a research for school about the history of pi. Not that hard you'd think... But then i found that the Egyptians found 3,1605 for pi in 1650 BC. So now i got to find out how he found it out.
    he used the formula: Volume of a cylinder = ((1-1/9)d)h if you calculate it you get 256/81 r h so he found 256/81 for pi. But i don't get the start. where did that ((1-1/9)d)h came from? i know it replaces pi r in our modern formula but that's about it...
    I thought it came from the relation between a square and circle, but after looking into that for the last 2 days, i still haven't figured it out.
    If someone could help me with this i'd be very grateful.
    with regards, danielrowling.

    P.S. I hope i posted it in the correct section.
    Last edited by danielrowling; Apr 25th 2013 at 07:32 AM.
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  2. #2
    MHF Contributor
    Oct 2009

    Re: Rhind papyrus problem 41 (pi)

    I am not a specialist in the history of mathematics, but according to the Wikipedia article about Rhind papyrus (sections Volumes and Areas), the idea is as follows.

    Rhind papyrus problem 41 (pi)-pi.png

    We take a circle (blue) and its circumscribing square, divide each side of the square in three equal parts and remove the corners. The resulting octagon (red) approximates the circle. If the side of the square is 1, then the area of the octagon is $\displaystyle 1 - 4\left(\frac{1}{2}\cdot\frac13\cdot\frac13\right) = 1-\frac29$. So, for a circle of some diameter d, the corresponding octagon's area is $\displaystyle \left(1-\frac29\right)d^2$. However, for some reason the Egyptian mathematicians wanted to express this as $\displaystyle (a d)^2$ for some number $\displaystyle a$. I am guessing that since $\displaystyle \sqrt{1-\frac29}$ is irrational, they decided to approximate it as $\displaystyle 1-\frac19$. Indeed, $\displaystyle \left(1-\frac19\right)^2=\left(\frac89\right)^2 =\frac{64}{81}\approx\frac{63}{81} =\frac79=1-\frac29$. That's how they got the formula $\displaystyle ((1-1/9)d)^2$ for the area of the circle.

    Note that the octagon's approximation of the circle is not too bad because it both adds to and subtracts some areas from the circle. However, according to Wikipedia, "That this octagonal figure, whose area is easily calculated, so accurately approximates the area of the circle is just plain good luck".

    If you need a more reliable source than Wikipedia, you should probably look for a book about the history of $\displaystyle \pi$ or ancient Egyptian mathematics.
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