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Divide circle into 9 areas

  1. May 23, 2008 #1

    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:
    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...

    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


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    i would be just amazed if this is possible along the lines suggested by your picture. given the area of the circle, the middle square is known and then it would have to be true that the other regions all have the same area as the square, which seems highly unlikely. calculus will tell you.
  4. May 23, 2008 #3
    Well that's what I'm asking... I also suspect it's impossible to make the areas exactly equal, but there is obviously one distance d for which the deviation between each area is at a minimum...
  5. May 23, 2008 #4
    I haven't taken a look at your integrals in detail, but my calculations also show that this is impossible. In essence, it is the same as what you did, but with geometry:

    Let r represent the radius of the circle and x = d/2 be the length we want to find. For the same reason as yours, [itex]d^2 = 4x^2 = \pi r^2 / 9 \implies x^2 = \pi r^2 / 36[/itex]

    To find the area of region 2, partition it into pieces. First, fill in the chord connecting the two intersections between the boundaries of region 2 and the circle. Then, draw a radius down the middle of region 2 to split it in half. We'll find the area of one of those halves, call it region R.

    Draw a radius going through R to the point of intersection between its boundaries and the circle. Finally, complete the rectangle by drawing a line through the middle of region 3 parallel to the chord we drew earlier. The area of R is the the area of the side triangle plus the area of the sector minus the little triangular area inside region 3. The area of region 2 is twice that. From similar triangles, we find that:

    [tex]\begin{tabular}{rcl}\vspace{0.15in}A(2) & = & 2\,\left[A(\,\text{side triangle}\,)\,+\,A(\,\text{sector}\,)\,-\,A(\,\text{little triangle}\,)\right] \\ \vspace{0.25in} & = & \displaystyle\frac{x(\sqrt{r^2-x^2}-x)^2}{\sqrt{r^2-x^2}} + r^2\sin^{-1}\left(\frac{x}{r}\right) - \frac{x^3}{\sqrt{r^2-x^2}} \\ & = & \displaystyle x\sqrt{r^2 - x^2} - 2x^2 + r^2\sin^{-1}\left(\frac{x}{r}\right) \\ \end{tabular}[/tex]​

    When [itex]x = \pi r^2/36[/itex] is plugged in, it simplifies to

    [tex]\sin^{-1}\left(\frac{\sqrt{\pi}}{6}\right) = \frac{\pi}{6} - \frac{\sqrt{\pi(36-\pi)}}{36}[/tex]​

    which isn't true, so there isn't d for any r if I didn't make any algebra errors.

    If you want to minimize deviation, one possible way is to minimize the squared differences from what would be equal area. The area for each region if they are all equal would be [itex]M = \pi r^2/9[/itex]. We want to minimize:

    [tex]E = (A_1-M)^2 + (A_2-M)^2 + (A_3-M)^2[/tex]​

    For a radius of 32, I found x = 8.9162, so d = 17.8324. The squared error was a whopping 5098.15.

    It's late, however, so I could have made a lot of errors.
  6. May 24, 2008 #5

    I can't really follow your calculation of Area2, but the final area looks like what I got... but not exactly.

    I got (using Maple):
    [tex]-\frac{1}{4} \,\sqrt {4096-{d}^{2}} \, \text{csgn} \left( d \right) d-1024\,\arcsin
    \left( {\frac {1}{64}}\,\sqrt {4096-{d}^{2}} \right) +512\,\pi +\frac{1}{2}\,
    d\sqrt {4096-{d}^{2}}[/tex]

    I believe csgn(d) is always 1 since d > 0, right? So then I get:
    [tex]-\frac{1}{4} \,\sqrt {4096-{d}^{2}} \, d-1024\,\arcsin
    \left( {\frac {1}{64}}\,\sqrt {4096-{d}^{2}} \right) +512\,\pi +\frac{1}{2}\,
    d\sqrt {4096-{d}^{2}}[/tex]

    I got this in the following way:

    If you look at the rightmost area 2. I divide it up into two areas:
    1). The rectangle from x = 1/2d to x = the intersection of circle with y = 1/2d
    2). The remaining part under the circle, from x = the intersection of circle with y = 1/2d to x = 32.
    Since this is only the part above the x-axis, I multiply by 2.

    The intersection of the circle with y = 1/2d is, (lets call it b):
    [tex]x = \frac{1}{2} \sqrt{ 4096 - d^2} = \sqrt{ 1024- \frac{d^2}{4}} = b[/tex]

    The areas are then:
    1).: Just the rectangle: [tex]A = \frac{d}{2} b[/tex]
    2).: Using an integral: [tex]A = \int_{b}^{32} \sqrt{1024-x^2} \, dx[/tex]

    The total integral is now twice the sum of these:
    [tex]A = 2 \times \left( \int_{b}^{32} \sqrt{1024-x^2} \, dx + \frac{d}{2} b \right)[/tex]
    This yields, according to maple, what I said above...
  7. May 24, 2008 #6
    Alright I tried again, and got a slightly different area this time:

    [tex]A_1 = \int_{\frac{d}{2}}^{32} \left( \sqrt{1024-x^2}-\frac{d}{2} \right) \, dx[/tex]
    [tex]A_2 = \int_{\frac{-d}{2}}^{\frac{d}{2}} \left{ \sqrt{1024-x^2} - \frac{d}{2} \right) dx[/tex]
    [tex]A_3 = d^2[/tex]

    I calculated areas A2 and A1 two times both, once for the topmost and once for the rightmost areas. They should ofcourse be equal but they were not in the last post... They are equal now so I think my areas are correct.

    Then I tried to minimize:
    [tex]f(d) = (A_3 - M)^2 + 4(A_2-M)^2 + 4(A_1-M)^2[/tex]
    [tex]= \left( {d}^{2}-{\frac {1024}{9}}\,\pi \right) ^{2}+4\, \left( 1/4\,d
    \sqrt {4096-{d}^{2}}+1024\,\arcsin \left( {\frac {1}{64}}\,d \right) -
    1/2\,{d}^{2}-{\frac {1024}{9}}\,\pi \right) ^{2}+4\, \left( -1/8\,d
    \sqrt {4096-{d}^{2}}-512\,\arcsin \left( {\frac {1}{64}}\,d \right) +1
    /4\,{d}^{2}-16\,d+{\frac {1280}{9}}\,\pi \right) ^{2}[/tex]

    I couldn't minimize this however...

    I tried solving [tex]\frac{df}{dd} = 0[/tex] with Maple but it gave no answer...

    I made a plot of f(d) and it clearly had a minimum, somewhere around 15/16... But for some reason I can't calculate it...

  8. May 24, 2008 #7
    Hm... I reread CRGreathouse's post on the other thread, and I think he's right about this. Since there are a different number of regions, it is better to minimize:


    The TI-89 gives me different results for my areas and your areas, so I figure I must have made a mistake with my geometry.
  9. May 24, 2008 #8
    Yeah sorry about the other thread, I got a database error when I posted the first time so I retried and apparently it posted it twice...

    Anyhow... I finally figured out what I was doing wrong trying to solve df/dd=0 (just some maple detail) and now I got a value for d:
    d = 16.559

    Drawing an image with this value gives me:

    Although it looks as though it's not the best approach it might be... But I think if d is a little bigger it would be better... Although I don't understand why my results don't yield this... Hmff..
  10. May 24, 2008 #9
    I tried plugging your values into Mathematica, and it just kept on running forever. I probably did something wrong (not too great with the program yet), but I'll try again later. I plugged in my areas, now minimizing the weighted differences, and I got similar answers to what you have. My d is 16.6402. Well, I'll come back after lunch and maybe take another look.
  11. May 24, 2008 #10


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    Since the functions here are smooth, a 'good' way to test putative solutions is to jitter the value slightly, once to the right and once to the left, and verify that the areas are less close to each other than in the solution.

    Of course this requires that we agree on the integrals first. :)

    (Is it too much to ask for a moderator to merge the two threads?)
    Last edited: May 24, 2008
  12. May 24, 2008 #11
    That's usually a good way to check, yes, but in this case you get 3 values for the 3 areas and it's not real easy to directly see which values are the best.

    A better way is probably to simply check if the integrals are correct. Is there any concrete way to check them (their values)?

    I tried computing the area2 two times, once for the topmost area2 and once for the rightmost. Since their orientation is different, the calculation was quite different aswell, but it yielded the same answer (numerically) to about 5 decimals. Unless I made the same mistake twice (which is unlikely because the calculations are different) I think this is a good pointer that the area is correct.

    I have one question though, why are 'we' minimizing the squares of the difference between each area and the optimal area?

    Why are we minimizing:
    [tex]f(d) = (A_3 - M)^2 + 4(A_2-M)^2 + 4(A_1-M)^2[/tex]

    Why not one of these?:
    [tex]f(d) = (A_3 - M)^2 + (A_2-M)^2 + (A_1-M)^2[/tex]
    [tex]f(d) = \left| A_3 - M \right| + \left| A_2 - M \right| + \left| A_1 - M \right|[/tex]
    [tex]f(d) = \left| A_3 - M \right| + 4\left| A_2 - M \right| + 4\left| A_1 - M \right|[/tex]

    When graphing these 4 functions, their minimum values all occur around 16, but not quite at the same point exactly... Why is one method better than another, and which of these 4 minimums should be chosen?

    Ofcourse, (why didn't I think of this?) we can easily check the areas by simply summing them and checking if they are equal to pi*r^2.

    When I sum all my areas I get a total of 3205.04.
    The actual area is pi*1024 = 3216.99.
    Pretty close!
    Last edited: May 24, 2008
  13. May 24, 2008 #12
    Yeah, I don't see anything wrong with your integrals, so I'll have to take a look at my geometry later (or maybe different results is just due to roundoff error), because I calculated the area of region 1 using my values for region 2 and 3.

    The first function you mentioned, without the 4's, I used in my first thread. I don't really know which is better, but there are probably arguments for both.

    As for the absolute values functions, they would work also, especially since we are using Mathematica, Maple, etc. In theory, it is easier to work out the squared deviations since that function is differentiable wherever A1, A2, and A3 are differentiable, while the absolute values make it difficult to work out analytically. With a solver, it's a whole different story.
  14. May 24, 2008 #13
    I have found one thing I don't like about my integral, in the calculation of area 1. I'm not sure if this is a mistake I made or that it is right both ways...

    The integral for area 1 is taken from d/2 to 32, the radius of the circle. But shouldn't the upper limit be the x value of the intersection of the circle with the line y = d/2 ?

    I tried to compute the area again using this intersection as the upper limit, and I suddenly got an imaginery area... Something gone wrong lol...
  15. May 29, 2008 #14


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    I tried a simple geometrical approach to this, and I concluded that you cannot have 9 equal areas in Nick's pic, but please check. :uhh:

    Let the circle have unit radius, then its area is [tex]\pi[/tex]

    If all 9 areas in Nick's pic are equal then the area of the central sqare (3) is [tex]d^2 = \frac{\pi}{9}[/tex] and [tex]d = \sqrt\pi/3 \simeq 0.59088[/tex]

    Now sorry, for the moment I will have to describe in words without fig. , the site and Latex are already slowing me enough.

    Draw the radius OC to C on the circle where area 1 meets 2 in Nik's pic. Height from midpoint of side of square (call A) to C, AC will be seen to be

    [tex]\sqrt{1-(d/2)^2} = \sqrt{1 - \pi/36} [/tex]

    Thus height from top (B) of central square, BC is [tex]\sqrt{1 - \pi/36} - \sqrt\pi/6 \simeq 0.66[/tex]

    Thus BC > d. So contained inside the area 2 is a rectangle that is already greater in area than area 3. So the circle cannot be divided into nine equal areas that fit Nik's pic.

    (The areas 1 must be less than that of 3. All the areas in the fig. can be calculated in terms of d. Although the formulae are quite simple they are not nice and simple. The ugliness :yuck: may be because essentially we are squaring the circle which we do not expect to give anything nice.)
  16. May 29, 2008 #15
    Yes, I know we can't get all values equal, it's pretty easy to confirm that once you know all the areas in terms of d (try d to be [itex]\sqrt{\pi} / 3[/itex] and you will see that the areas are not equal).

    I have now found a value for d which I believe (if my calculations are correct) to be the optimal value: d = 16.559
    Graphing it with this distance does look kind of ok, but it's hard to see with the naked eye...

    I am still not sure about the calculation of area 1 though. While I think I should only integrate up to the x-value of the intersection between circle and y=d/2, the graph is lying below the x-axis behind that value, so it shouldn't matter, right??
  17. May 29, 2008 #16


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    I thought the question was find what makes them equal if possible, and I think the answer is they cannot be all equal. Sorry if I missed the point as this developed and did not notice you had got there already.

    My first reaction now I have seen the rest was

    (1) You can decide to minimise anything, but if the question is to make the areas look as early equal as possible I would have thought [tex]f(d) = \left| A_3 - M \right| + \left| A_2 - M \right| + \left| A_1 - M \right|[/tex]
    was the thing to minimise. Not the squares of those because the areas are already squares or partly.

    (2) What you are going to get is your pic with the boundaries between are 1 and 2 all shifted inwards, right? You are going to get a larger area 1 with a corner nicked out. So you want to minimise as function of two parameters, the position of that boundary and d. But if I understand you have only varied one parameter in your calc? Perhaps the solution is not even a unique d.

    Whatever - I think the areas you want will still be calculable geometrically without integrations.
    Last edited: May 29, 2008
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