How To Divide A Circle Into 9 Equal Parts

Nick89

Hi,

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

Consider a circle with a radius of 32 units. We desire to split up the area of the circumvolve into 9 areas that take, if possible, exactly the same expanse. Come across the post-obit paradigm:

The ruby lines are the 'dividing lines', spaced by a distance [itex]d[/itex] (in the x also every bit the y management).

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

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

The get-go thing I thought about (simply which doesn't seem to be working, see later) is simply to practice the following:

Nosotros know the area of the complete circle: [tex]A_{tot} = \pi 32^2[/tex]
Therefore, if the nine areas are to exist divided in equal areas, the area of ane the subareas will be: [tex]A_{sub} = \frac{ \pi 32^2}{9}[/tex]
Nosotros also know the area [itex]A_3[/itex] since it'south only a square: [tex]A_3 = d^ii[/tex]
Therefore: [tex]d = \sqrt{ \frac{ \pi 32^ii}{ix}}[/tex].

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

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

I came up with the post-obit expanse's; [itex]A_1[/itex] is calculated from the summit-right area1 and [itex]A_2[/itex] is calculated from the rightmost area2.

[tex]A_1 = \int_\frac{d}{2}^b \left( \sqrt{ 1024-10^two} - \frac{d}{2} \right) \, dx[/tex]
[tex]A_2 = two \times \left( \int_b^{32} \sqrt{1024-10^2} \, 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^two}{4}}[/tex]

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

Where have I gone wrong:
1) Assuming at that place is a solution where all areas are equal;
2) Bold this solution was merely to separate the total surface area by ix and equaling this to d^ii;
iii) Calculating the areas using integrals?

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

Could anyone help me out here?

mathwonk

i would exist merely amazed if this is possible along the lines suggested by your moving picture. given the expanse of the circle, the centre square is known and then it would have to be true that the other regions all have the aforementioned area every bit the foursquare, which seems highly unlikely. calculus will tell you.

Nick89

Well that'south what I'g request... I also suspect information technology'due south impossible to make the areas exactly equal, but there is obviously 1 distance d for which the departure betwixt each area is at a minimum...

Tedjn

I oasis't taken a await at your integrals in detail, just my calculations also show that this is impossible. In essence, it is the aforementioned every bit what you did, just with geometry:

Let r stand for the radius of the circumvolve and ten = d/ii be the length we want to find. For the aforementioned reason as yours, [itex]d^2 = 4x^two = \pi r^2 / ix \implies x^ii = \pi r^2 / 36[/itex]

To notice the surface area of region 2, sectionalisation it into pieces. Outset, fill in the chord connecting the two intersections betwixt the boundaries of region 2 and the circle. And then, depict a radius down the middle of region two to dissever information technology in half. We'll notice the area of one of those halves, call information technology region R.

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

[tex]\begin{tabular}{rcl}\vspace{0.15in}A(2) & = & 2\,\left[A(\,\text{side triangle}\,)\,+\,A(\,\text{sector}\,)\,-\,A(\,\text{piddling triangle}\,)\right] \\ \vspace{0.25in} & = & \displaystyle\frac{x(\sqrt{r^2-ten^2}-10)^2}{\sqrt{r^ii-x^2}} + r^two\sin^{-ane}\left(\frac{10}{r}\right) - \frac{ten^3}{\sqrt{r^two-x^2}} \\ & = & \displaystyle ten\sqrt{r^two - ten^ii} - 2x^ii + r^2\sin^{-1}\left(\frac{x}{r}\right) \\ \end{tabular}[/tex]

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

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

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

If you want to minimize deviation, i 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 exist [itex]M = \pi r^2/9[/itex]. We desire to minimize:

[tex]E = (A_1-Thou)^2 + (A_2-K)^2 + (A_3-Chiliad)^ii[/tex]

For a radius of 32, I found 10 = 8.9162, so d = 17.8324. The squared mistake was a whopping 5098.15.

Information technology's belatedly, however, and so I could take made a lot of errors.

Nick89

Thanks.

I can't really follow your adding of Area2, only the final expanse looks similar what I got... but not exactly.

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

I believe csgn(d) is always i since d > 0, right? And then then I go:
[tex]-\frac{ane}{iv} \,\sqrt {4096-{d}^{2}} \, d-1024\,\arcsin
\left( {\frac {ane}{64}}\,\sqrt {4096-{d}^{2}} \correct) +512\,\pi +\frac{i}{2}\,
d\sqrt {4096-{d}^{2}}[/tex]

I got this in the following way:

If you lot wait at the rightmost area 2. I split up it up into two areas:
1). The rectangle from x = 1/second to ten = the intersection of circle with y = 1/2d
ii). The remaining part nether the circle, from x = the intersection of circle with y = ane/second to ten = 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 phone call it b):
[tex]ten = \frac{1}{2} \sqrt{ 4096 - d^two} = \sqrt{ 1024- \frac{d^2}{4}} = b[/tex]

The areas are and then:
ane).: Just the rectangle: [tex]A = \frac{d}{2} b[/tex]
ii).: 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 = ii \times \left( \int_{b}^{32} \sqrt{1024-x^ii} \, dx + \frac{d}{two} b \right)[/tex]
This yields, according to maple, what I said to a higher place...

Nick89

Alright I tried again, and got a slightly dissimilar area this time:

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

I calculated areas A2 and A1 two times both, one time for the topmost and one time for the rightmost areas. They should ofcourse be equal simply they were not in the last mail... They are equal now so I think my areas are right.

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

I couldn't minimize this however...

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

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

Hmm??

Tedjn

Hm... I reread CRGreathouse's mail on the other thread, and I remember he's right about this. Since there are a dissimilar number of regions, it is better to minimize:

[tex]4(A_1-M)^2+4(A_2-M)^2+(A_3-Yard)^two[/tex]

The TI-89 gives me different results for my areas and your areas, so I figure I must take made a error with my geometry.

Nick89

Yep sorry nigh the other thread, I got a database error when I posted the outset fourth dimension so I retried and apparently it posted it twice...

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

Drawing an epitome with this value gives me:

Although it looks equally though it's not the all-time approach it might exist... But I think if d is a little bigger it would be better... Although I don't sympathize why my results don't yield this... Hmff..

Tedjn

I tried plugging your values into Mathematica, and it simply kept on running forever. I probably did something wrong (not too bang-up with the program however), but I'll try again later on. 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 dorsum after dejeuner and maybe take another wait.

CRGreathouse · May 24, 2008

Since the functions hither are smooth, a 'skillful' fashion 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 grade this requires that nosotros agree on the integrals first. :)

(Is it likewise much to enquire for a moderator to merge the two threads?)

Nick89 · May 24, 2008

That's usually a good way to cheque, aye, but in this case you go three values for the 3 areas and it'southward not existent like shooting fish in a barrel to directly encounter which values are the best.

A amend way is probably to but check if the integrals are correct. Is in that location whatever concrete manner to check them (their values)?

I tried computing the area2 2 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 reply (numerically) to about 5 decimals. Unless I made the aforementioned error twice (which is unlikely because the calculations are different) I call back this is a expert pointer that the surface area is correct.

I accept one question though, why are 'we' minimizing the squares of the deviation between each area and the optimal expanse?

Why are we minimizing:
[tex]f(d) = (A_3 - 1000)^ii + 4(A_2-M)^2 + 4(A_1-Thousand)^2[/tex]

When graphing these four functions, their minimum values all occur around sixteen, but not quite at the same signal exactly... Why is one method better than another, and which of these 4 minimums should exist called?

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

When I sum all my areas I go a total of 3205.04.
The bodily area is pi*1024 = 3216.99.
Pretty shut!

Tedjn

Yeah, I don't see anything wrong with your integrals, and so I'll accept to take a await at my geometry afterwards (or peradventure different results is just due to roundoff error), because I calculated the area of region i using my values for region two and 3.

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

As for the absolute values functions, they would work too, especially since we are using Mathematica, Maple, etc. In theory, it is easier to work out the squared deviations since that role is differentiable wherever A_i, A_two, and A_three are differentiable, while the absolute values make information technology difficult to work out analytically. With a solver, information technology's a whole different story.

Nick89

I have institute one affair I don't like most my integral, in the adding of area 1. I'1000 non certain if this is a fault I made or that it is right both ways...

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

I tried to compute the expanse over again using this intersection as the upper limit, and I suddenly got an imaginery area... Something gone wrong lol...

epenguin

I tried a uncomplicated geometrical approach to this, and I ended that you cannot have 9 equal areas in Nick'due south motion-picture show, just please check. :uhh:

Let the circle take unit of measurement radius, then its surface area is [tex]\pi[/tex]

If all nine areas in Nick'south moving-picture show are equal then the area of the central sqare (3) is [tex]d^2 = \frac{\pi}{ix}[/tex] and [tex]d = \sqrt\pi/3 \simeq 0.59088[/tex]

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

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

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

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

Thus BC > d. So independent inside the area two 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'due south 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 elementary. The ugliness :yuck: may exist because essentially we are squaring the circumvolve which we do not expect to give anything nice.)

Nick89

Yes, I know nosotros can't get all values equal, it'due south pretty easy to ostend that once you lot know all the areas in terms of d (try d to be [itex]\sqrt{\pi} / three[/itex] and y'all will run across 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 = sixteen.559
Graphing it with this distance does wait kind of ok, but information technology'southward hard to see with the naked center...

I am still non sure virtually the calculation of area one though. While I recollect 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 thing, right??

epenguin · May 29, 2008

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

My showtime reaction now I take seen the rest was

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

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

Whatever - I call back the areas you want will all the same be calculable geometrically without integrations.