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Divide circle into 9 areas

  1. May 23, 2008 #1
    Hi,

    I was asked this question on another forum and was interested in it... It's somewhat related to what I have been doing lately so I gave it a (few) tries, but I never really worked it out...


    Consider a circle with a radius of 32 units. We want to divide the area of the circle into 9 areas that have, if possible, exactly the same area. See the following image:
    ezgwig.jpg
    The red lines are the 'dividing lines', spaced by a distance [itex]d[/itex] (in the x aswell as the y direction).

    The areas 1 (blue) and 2 (green) and the area 3 (red) are marked with the colors. Note that there are four areas 1 and four areas 2, they should be equal in area.


    The question is how to find the distance [itex]d[/itex] that will yield the optimal result (if possible, that all areas are equal).



    The first thing I thought about (but which doesn't seem to be working, see later) is simply to do the following:

    We know the area of the complete circle: [tex]A_{tot} = \pi 32^2[/tex]
    Therefore, if the 9 areas are to be divided in equal areas, the area of one the subareas will be: [tex]A_{sub} = \frac{ \pi 32^2}{9}[/tex]
    We also know the area [itex]A_3[/itex] since it's just a square: [tex]A_3 = d^2[/tex]
    Therefore: [tex]d = \sqrt{ \frac{ \pi 32^2}{9}}[/tex].


    I tried to graph it and it seemed alright to the eye, but I wanted to be sure, so I went on...



    The following way I could think of was to calculate the subareas seperately using integrals and then looking for a [itex]d[/itex] that would minimize their deviation.

    I came up with the following area's; [itex]A_1[/itex] is calculated from the top-right area1 and [itex]A_2[/itex] is calculated from the rightmost area2.

    [tex]A_1 = \int_\frac{d}{2}^b \left( \sqrt{ 1024-x^2} - \frac{d}{2} \right) \, dx[/tex]
    [tex]A_2 = 2 \times \left( \int_b^{32} \sqrt{1024-x^2} \, dx \right) + d \sqrt{1024-\frac{d^2}{4}}[/tex]
    [tex]A_3 = d^2[/tex]
    where the limit b is the intersection of the circle with y = d/2:
    [tex]b = \sqrt{1024-\frac{d^2}{4}}[/tex]


    When I now plugged in the value for [itex]d[/itex] I found above I don't get the same result, I get a different result for each area...


    So, I thought, maybe my simple solution above wasn't right.
    But now I have found three areas each as a function of d. I should be able to minimize the deviation between the areas for one value of d, right? I can't see any way how to do that though... Maybe taking the absolute value of the deviation (A_1 - A_2 for example) and using solving it's derivative for 0? Even then I only minimized A_1 - A_2 and had nothing to do with A_3...


    Where have I gone wrong:
    1) Assuming there is a solution where all areas are equal;
    2) Assuming this solution was simply to divide the total area by 9 and equaling this to d^2;
    3) Calculating the areas using integrals?

    I can't see any other mistakes I may have made, so I assume it must be one of the three...

    Could anyone help me out here?
     
  2. jcsd
  3. May 23, 2008 #2

    CRGreathouse

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    Homework Helper

    I would calculate the areas (with integrals as needed) as a function of d, then try to minimize
    [tex]4(A_1-A)^2+4(A_2-A)^2+(A_3-A)^2[/tex] with [tex]A=1024\pi/9[/tex]
     
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