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Hard geometry Problem

C

Chu

Banned
Ok, we take a circle, and we draw 10 circles inside it s.t. each circle shares the same center, and the distance between each circle is equal.

Now, we want to subdivide this creation into 100 parts. That's pretty easy - just draw 9 lines that intersect the center that do not overlap.

The question is, what if we want to divide the circle into 100 sections with equal area? What would our lines have to look like?

And no, this is not a homework problem, just something I am messing around with for a program I want to write.

EDIT : Sorry I wasn't specific enough, equal area, not equal shapes (which would make the problem impossible). Also fixed some obvious errors in the setup.
 
how about dividing it into a pie? so each line starts from the middle and goes out? actual thats wrong, nevermind
 
100 equal area parts? or 100 equal shape & area parts? and are we restricted to the GENERAL idea of 10 circles and 10 lines to make 100 parts?
 
Originally posted by: Chu
Ok, we take a circle, and we draw 9 circles inside it s.t. each circle shares the same center, and the distance between each circle is equal.

The question is, what if we want to divide the circle into 100 EQUAL parts? What would our lines have to look like?
Click to expand...
If you have 10 circles, which of the 10 is the circle? Or do you want to divide all circles into parts?

 
What on earth are you talking about? If all the circles share the same center and have the same area, then what is the deal about distance between the circles? It would be better if you rephrased your question or represented it with some sort of a diagram.
 
Actually, I think Kevman is correct, he just didn't realize it.

I started with the assumption of each portion being 1/100 of the original area.
I also like cutting it like a pie, every 36 degrees to make 10 equal areas...

Now, you just have to calculate the radii of the other 9 circles, such that the area between two consecutive circles is the same for each 2 consecutive circles.
pi*r(10)^2 - pi*r(9)^2 =
pi*r(9)^2 - pi*r(8)^2 =
pi*r(8)^2 - pi*r(7)^2 =
pi*r(7)^2 - pi*r(6)^2 =
pi*r(6)^2 - pi*r(5)^2 =
. . .
=1/10 pi r(10)^2

(I called the radius of the outermost circe (the original circle) r(10), then the next largest radius = r(9) and so on.)
 
Wow, I just realized how much either I suck at using microsoft paint, or how much microsoft paint sucks...
Attempting to provide the solution graphically...
 
Originally posted by: DrPizza
I started with the assumption of each portion being 1/100 of the original area.
Click to expand...
Unfortunately that assumption is incorrect. The original poster said the difference in diameters must be equal.

As far as I can tell, this problem is solvable in a few limited special cases. But the general case is not possible. For example if the area of the innermost circle is 1 m^2 and the area of the next innermost circle is 1.0001 m^2. How can you subdivide those such that all parts are equal area? To do so would take far more than 100 sub-parts.
 
Equal area... well.. the problem becomes easy. So you draw ten lines that are separated by 36 degrees each. Assuming R is the radius of the largest circle, the innermost circle would have a radius of R/sqrt(10). The next would be R*sqrt(2)/sqrt(10). The next would have a radius of R*sqrt(3)/sqrt(10) and so on...

Edited for Clarity
r1 (the innermost) = R*sqrt(1/10)
r2 = R*sqrt(2/10)
r3 = R*sqrt(3/10)

etc.....

The derivation comes from this.

The innermost circle has to have an area that's 1/10 of the overall area.

pi*r1^2 = 1/10*pi*R^2
r1 = R*sqrt(1/10)

The next circle has to have an ENTIRE area of 2/10 of the overall area. r2 = R*sqrt(2/10)... etc...
 
Is this graphic an accurate representation of your original description?

geometry.gif

And now the question is how do we modify the above diagram so that each small section of the large circle has an equal area?
 
Originally posted by: TuxDave
Equal area... well.. the problem becomes easy. So you draw ten lines that are separated by 36 degrees each. Assuming R is the radius of the largest circle, the innermost circle would have a radius of R/sqrt(10). The next would be R*sqrt(2)/sqrt(10). The next would have a radius of R*sqrt(3)/sqrt(10) and so on...
Click to expand...
This is your math:
R1 = R*(1/10)^0.5 = 0.316 R
R2 = R*(2/10)^0.5 = 0.477 R
R3 = R*(3/10)^0.5 = 0.548 R

R2 - R1 = 0.131 R
R3 - R2 = 0.101 R

But that violates the original condition that
Originally posted by: Chu
and the distance between each circle is equal.
Click to expand...
The distance (R2 - R1) does not equal the distance (R3 - R2).

 
Originally posted by: dullard
Originally posted by: TuxDave
Equal area... well.. the problem becomes easy. So you draw ten lines that are separated by 36 degrees each. Assuming R is the radius of the largest circle, the innermost circle would have a radius of R/sqrt(10). The next would be R*sqrt(2)/sqrt(10). The next would have a radius of R*sqrt(3)/sqrt(10) and so on...
Click to expand...
This is your math:
R1 = R*(1/10)^0.5 = 0.316 R
R2 = R*(2/10)^0.5 = 0.477 R
R3 = R*(3/10)^0.5 = 0.548 R

R2 - R1 = 0.131 R
R3 - R2 = 0.101 R

But that violates the original condition that
Originally posted by: Chu
and the distance between each circle is equal.
Click to expand...
Click to expand...

Uh.... if the distance between the circles are equal, and we're stuck with straight lines, then to achieve equal area would be impossible. I think... gimme a sec.
 
Originally posted by: TuxDave
Equal area... well.. the problem becomes easy. So you draw ten lines that are separated by 36 degrees each. Assuming R is the radius of the largest circle, the innermost circle would have a radius of R/sqrt(10). The next would be R*sqrt(2)/sqrt(10). The next would have a radius of R*sqrt(3)/sqrt(10) and so on...

Edited for Clarity
r1 (the innermost) = R*sqrt(1/10)
r2 = R*sqrt(2/10)
r3 = R*sqrt(3/10)

etc.....

The derivation comes from this.

The innermost circle has to have an area that's 1/10 of the overall area.

pi*r1^2 = 1/10*pi*R^2
r1 = R*sqrt(1/10)

The next circle has to have an ENTIRE area of 2/10 of the overall area. r2 = R*sqrt(2/10)... etc...
Click to expand...


except he said the circles have to be equidistant from each other.
 
Originally posted by: HomeBrewerDude
Originally posted by: TuxDave
Equal area... well.. the problem becomes easy. So you draw ten lines that are separated by 36 degrees each. Assuming R is the radius of the largest circle, the innermost circle would have a radius of R/sqrt(10). The next would be R*sqrt(2)/sqrt(10). The next would have a radius of R*sqrt(3)/sqrt(10) and so on...

Edited for Clarity
r1 (the innermost) = R*sqrt(1/10)
r2 = R*sqrt(2/10)
r3 = R*sqrt(3/10)

etc.....

The derivation comes from this.

The innermost circle has to have an area that's 1/10 of the overall area.

pi*r1^2 = 1/10*pi*R^2
r1 = R*sqrt(1/10)

The next circle has to have an ENTIRE area of 2/10 of the overall area. r2 = R*sqrt(2/10)... etc...
Click to expand...


except he said the circles have to be equidistant from each other.
Click to expand...

Oh... ok... try #2, given 10 circles that are equi-distant from each other.

The innermost circle would be left alone.
The next circle (not including any circle contained in it) would be divided into 3 equal parts
The next circle would be divided into 5 equal parts
The next circle would be divided into 7 equal parts
9
11
etc...

 
Here - I did it with 4 circles drawn inside. Note that as you work from the centers out, the circles need to get closer and closer together. I've created 50 (sort of) equal area sections. I'm on the kids computer, so I don't have any good CAD programs to use or else the picture would have been accurate.
circles.JPG

I highlighted 5 of the equal area sections in red, if it makes it any clearer. The difference in radius of the arclengths decreases.

Since I know your abilities, I'm pretty sure you'll have no problem from this point...
 
oh crap... I missed the circles equally spaced part of it..

So, given an area of A for the innermost circle, the area of the next circle will be 4A (twice the radius = 4 times the area), then 9A, then 16A, then... up to 100A

The areas of the rings will be A, 3A, 5A, 7A, ...

Divide each ring into areas = A. You'll have 100 areas when you're done.
(1 + 3 + 5 + ... + 19 = 100)
edit: And, I see TuxDave beat me to that solution too! I'm slow today!
 
Originally posted by: DrPizza
oh crap... I missed the circles equally spaced part of it..

So, given an area of A for the innermost circle, the area of the next circle will be 4A (twice the radius = 4 times the area), then 9A, then 16A, then... up to 100A

The areas of the rings will be A, 3A, 5A, 7A, ...

Divide each ring into areas = A. You'll have 100 areas when you're done.

edit: And, I see TuxDave beat me to that solution too! I'm slow today!
Click to expand...

*victory dance*
 
Originally posted by: TuxDave
Originally posted by: DrPizza
oh crap... I missed the circles equally spaced part of it..

So, given an area of A for the innermost circle, the area of the next circle will be 4A (twice the radius = 4 times the area), then 9A, then 16A, then... up to 100A

The areas of the rings will be A, 3A, 5A, 7A, ...

Divide each ring into areas = A. You'll have 100 areas when you're done.

edit: And, I see TuxDave beat me to that solution too! I'm slow today!
Click to expand...

*victory dance*
Click to expand...
Yes you both solved the specific case. But the general case, where there may be 101 equal parts or 99 equal parts is unsolvable.
 
I liked this problem... I just added it into my portfolio of extra credit problems for calculus and precalculus. 🙂

I hate to give them homework to start when they finish tests... it seems too much like busy work just to keep their mouths shut while others are still working on the test. Problems like these tend to challenge the faster kids a bit.
 
Originally posted by: dullard
Originally posted by: TuxDave
Originally posted by: DrPizza
oh crap... I missed the circles equally spaced part of it..

So, given an area of A for the innermost circle, the area of the next circle will be 4A (twice the radius = 4 times the area), then 9A, then 16A, then... up to 100A

The areas of the rings will be A, 3A, 5A, 7A, ...

Divide each ring into areas = A. You'll have 100 areas when you're done.

edit: And, I see TuxDave beat me to that solution too! I'm slow today!
Click to expand...

*victory dance*
Click to expand...
Yes you both solved the specific case. But the general case, where there may be 101 equal parts or 99 equal parts is unsolvable.
Click to expand...

Well... then I say this solution works when dividing it into X areas where X is a perfect square. :-D
 
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