A "Simple" Plane Geometry Question

I'm trying to divide a circle that's 32" in diameter into nine sections using 3/4" thick cross-lapped rails that divide it up. Here's the picture:
http://i156.photobucket.com/albums/t...leDividers.gif
My objective is to join the rails so that the nine partitioned areas are as close to being equal as possible. Two pairs of four areas (1) and (2) are identical, and the center (3) should be a square.

It's impossible to make the areas of all nine sections equal, but what should the dimension X be so that all nine sections are as close to each other in size as possible? (i.e., minimize the deviation between them)

First of all, how did you figure out it is impossible to make all areas equal? I certainly can't see that directly, it would require some math. If you did this math you probably also know how to make them as close as possible, and you wouldn't ask this question :)

On to your question...

I haven't thought about it for very long so there might be a better way, but here is what I thought of at first:

If you can express each area (1, 2 and 3) as a function of x alone, you end up with 3 equations.
To minimize the deviation between the area's you have to find this deviation, as a function of x again, and optimize the function.

Expressing the areas as a function of x can be a bit tricky, but with some calculus (integration in particular) I don't think it will be too hard.

So a question for you, do you know how to calculate the area under the graph of a function f(x)?

You need to use integration:

The area decided by the lines x = a, x = b and y = f(x) can be calculated as follows:

A = int[a to b]( f(x) dx ) = F(b) - F(a)
where F(x) denotes the antiderivative of f(x).


If you know how to do this, also try to figure out from where to where you should calculate the area each time.

For example for area 2, you need a as the right side of the rail:
a = (1/2)d + 3/4
(d is what you called x, to avoid confusion I have called it d)

you need b as the x-intercept of the circle, which is simply the radius:
b = 32

Then you need to realise that you only need the part of the area beneath the line y = (1/2)d - 3/4 (which is the horizontal rail) AND that you have only counted one half of the area (only the part above the x-axis).

The function for a circle if you didn't know is:
x² + y² = r², or solved for y and plugging in r = 32:
y = sqrt( 32² - x² )

(Note, you only have the upper half of the circle here because of the square root, but since the circle is symmetric you can simply multiply the appropriate areas by 2)


I hope you understand any of this... I will try to help you more tomorrow, too late now...

Alright I did some more thinking about this and I came up with a very simple solution, although I'm not sure if it actually works...

I reasoned as follows:

You know the total area of the circle, which is A = pi * r² = pi * 32².
Dividing this area into 9 equal areas will result in a 'sub'area of:
A_sub = (1/9) * pi * 32²

You know the area of the square (number 3), it's simply A₃ = d²

Conclusion:
d² = (1/9) * pi * 32²

(So d is the square root of this).


Note, I don't know if this works! I'm going to try and put in some values here to see if it's even remotely close to being the optimal value... I will also try to do it in the way I though of before.. It might even come up with the same answer.

Alright I tried computing the values with the value for d I thought of in my last post.

If my calculations of the areas are correct, it's wrong.
A_1 = 293.07
A_2 = 213,14
A_3 = 357,44

I am still working on the other solution, it's not really working out lol...


EDIT
I am also posting this on another forum asking for help, I'm quite interested in this aswell, it is somewhat related to what I'm have been doing recently... I hope you are ok with this?

Nick, the problem I see is that the area partitions can never all be equal. Suppose you started with a square rather than a circle and all partitions defined by the four rails equal the area of the center square. Then you draw the large circle and slice off the corners and sides of the large square.

Note that if you next adjust the rails to make the center square smaller to equal the outer triangular-shaped partitions, the size of the other four partitions also increase, but not at the same rate. When the outer triangles eventually equal the area of the square, the other four partitions will be larger than any of the rest.

Therefore, there is no way that all nine partitions can equal each other. At best, you can adjust the rails to get the square to match the outer four triangular-shaped partitions, or the other four. Regardless, all 9 will never be equal, but there must be an optimal location for the rails so that the sum of the deviations from each other is minimized.

That's what I am trying to determine. My gut feeling is that when the square is shrunk to about the size of the triangular partitions, the sum of the deviations in area partitions will be minimized, but I hate to rely on gut feelings. LOL!

Hmm your theory does make sense... You could be completely right that there is no solution to this problem that makes all the areas equal.

However, even so, my first method should be able to find the best value for d.
I think I have done something wrong though since when calculating two different areas for area2 (the rightmost and the topmost for example) using integrals, I didn't come up with the same value... I can't see why, but I'll figure it out, (possibly with the help of the other forum)..

Is there any reason why you couldnt find the areas of 1,2 and 3 (Which I will represent with A,B and C) and use the formula

pi*16^2 = C + 4B + 4A


If we work out the areas of A,B and C using x as the only variable then we would be left with an equation?

Edit: Using my above idea I got an equation and then optimised it, getting the optimum value to be 5.8934577654678 for x but that assumes that the dividers are just lines and don't have a thickness.

What did you get for the areas?

When I try to solve the equation you posted, the only answer I find is " 0." (Including the dot, which I THINK means that there is no solution..)


I tried it from scratch, assured myself that the areas I calculated were correct (they were not before) and I have now find a value of d = 16.559

I drew an image and this is what I got:
http://i30.tinypic.com/142u8nr.jpg

Doesn't look too good does it..? At the naked eye I think I can do better by hand lol...

Although it looks better than d = 5.893 which yields:
http://i30.tinypic.com/vg1njn.jpg


I don't know if this is really the best solution, according to my maths it should be, but somehow it doesn't look right...

I binned all the working I had done as I didnt think I would use it. If each section had the same area and there are 9 sections then the total area according to mine is 312.595599900174 square inches. But if I use the circle formula then the area is 804.247719318987.

Im not sure how you have done it. You seem to have just calculated x for the centre square. I did it for each line. Which means that if you cut of the edge for the 2 sections then they would be squares.

I will write up a word file on how exactly I calculated my value, it's way to difficult to recognize the formulas without latex or equation editor :p

That's a clever way to check if I'm correct though, totalling up the areas! Didn't think of that before.
When I enter my value for d (16.559) and calculate all my areas and summing them, I get a total area for the circle of 3205.04. The actual area is pi*1024 = 3216.99, pretty close!

Hold on for the word file, ill post it in a minute.

Here's some of my work on the problem without using calculus. I found values of X that equated pairwise the areas of each set (1), (2), and (3):
http://i156.photobucket.com/albums/t...ividers2-2.gif

Then I plotted the sums of the deviations and connected the points with a polynomial regression line:
http://i156.photobucket.com/albums/t...eDividers3.gif

Interpolating the line, I get X = 8.1" as close to optimal. Perhaps this is rather crude but please look it over. WDYT?

Quote:

---

Originally Posted by NickThissen

I will write up a word file on how exactly I calculated my value, it's way to difficult to recognize the formulas without latex or equation editor :p

That's a clever way to check if I'm correct though, totalling up the areas! Didn't think of that before.
When I enter my value for d (16.559) and calculate all my areas and summing them, I get a total area for the circle of 3205.04. The actual area is pi*1024 = 3216.99, pretty close!

Hold on for the word file, ill post it in a minute.

---

Nick, the area of a circle that is 32" in diameter is 804.248 sq in. That's 256(pi).

And how did you calculate the areas for any value of x? Without calculus this is nearly impossible to do accurately...


Attached is the word file which describes in detail how I came up with d = 16.559. I hope I didn't make any mistakes, tell me what you think!


EDIT

OOH! The diameter is 32". I thought it was the radius, sorry! :) In that case, I think my value for d simply needs to be halved, which brings it close to 8.1! I will calculate it exactly again using a radius of 16, the calculations don't change, merely the numbers, hold on!

Also, I didn't take into account the thickness of the beams. I am assuming the calculation does not change if you take d as the distance between the center of each beam, but I'm not 100% sure on that...

Thanks, Nick. I was able to read this Word file and save it to my permanent library. You did a marvelous job!

In answer to your question concerning how I found the areas without using integral calculus, the answer is a "a pretty good CAD program." I have one that will approximate areas of irregular shapes that you draw. Microsoft also has a couple of these available. You can also overlay small square grids onto your areas and count the squares, thus getting pretty close.

This "simple" problem turned out to be a dandy, no less. It appears that the length of the side of the ideal center square will always be a little over half the radius of the circle, which is not exactly obvious to anyone. Maple is a rather amazing math software package, to say the least.

I'm just glad that my integral calculus teacher back in my college days did not include this problem on the final exam. :afrog:

No problem, it was fun trying to figure it out and it helped me brush up on Maple skills... (Yes it's a great program but it can also be a great pain in the ass sometimes lol!)

I'm not surprised you don't get questions like this in an exam since there is no way you are going to solve this by hand in the limited time you have... Maybe a similar problem with much easier equations could be done but still...