# Best way to split up a circle into subsections

I'm not really sure how to word this question so feel free to edit it to make more sense:

I have a big circle of radius 32, and I want to divide it into smaller segments by both drawing smaller circle(s) with the same midpoint and drawing sectors (?) in such a way that each segment will fit into a circle of radius 9.5

I brought pictures!:

Imagine that if I were to divide the big circle into 8 slices and make 2 sub-circles of 20 and 10 radii, then you can see that it wont fit (the biege circle is the 9.5 radius one):

But if I make 16 slices, it will fit and I'll only need one smaller circle with a raidus of 18:

So my question is: is there any optimal way to figure out the least amount of cuts and sub-circles that are needed to make this work?

The cuts don't need to go all the way through the from the edge of the big circle to the center, it can start or stop at any smaller circle.

By the way, I used Blender to draw the pictures so it may not be exact.

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Interesting problem! If we draw a regular $n$-gon on the periphery, the main problem comes from the distance between adjacent vertices. A bit of trig (the angle whose sine is $9.5/32$) shows $n=10$ is not good enough, while $n=11$ looks OK. Perhaps, partly for simplicity, we might go to $n=12$. The rest should not be hard, but some experimentation s needed. – André Nicolas Jul 9 '12 at 18:41
@AndréNicolas Can you explain how you got to that number. Also What is the radius of the inner circle and how many segments does the inner circle need? – qwertymk Jul 9 '12 at 19:10
Have not done the necessary calculations, so cannot tell you radius of inner circle. Draw the $n$-gon, and join its vertices to the centre. Then we get angles of $360/n$. bisect one of these angles, extend bisector to the side of the $n$-gon. Have angle of $180/n$. Call this $\theta$. We need $32\sin\theta\le 9.5$. That gives $\theta$ about $17.27$ degrees. So $180/n \le 17.27$, giving $n$ at least $11$. – André Nicolas Jul 9 '12 at 19:21

Let $R$ (=32) be the radius of the large circle and $r$ (=9.5) be the radius of the small disk.

We start with the following simpler problem: Given an annular sector $S$ with inner radius $a$, outer radius $b$ and central angle $\phi$, what is the radius $\rho$ of the smallest circle that covers $S$?

Using the above figure one gets $$\rho^2={a^2+b^2-2 a b\cos\phi\over2(1+\cos\phi)}\qquad(0<\phi<\pi)\ .$$

As suggested by Ross Millikan we now aim at minimizing the number of pieces. To this end I propose the following greedy algorithm:

1. Put $r_0:=R$.
2. Given $r_k$ for a $k\geq0$ put $b=r_k$, choose a suitable $n$ and put $\phi:={2\pi\over n}$. Then solve the equation $$r^2={a^2+b^2-2 a b\cos\phi\over2(1+\cos\phi)}$$ for $a$ (choose the smaller of the two solutions).
3. Repeat step 2. for various values of $n$ until you have found the $n$ maximizing the quantity $$q_n:={b^2-a^2\over n}\ .$$ (Comment: We want to maximize the area gain per piece.)
4. If $a<1$ goto 5., else put $r_{k+1}:=a$ and goto 2.
5. Cover the center with a last disk.
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I don't know how to do any of that. Is there a wikipedia page on annular sector? Also what's the f function? – qwertymk Jul 9 '12 at 19:25
@qwertymk: It is natural to make things symmetric, so $\phi =\frac {2\pi}n$ for $n$ the number of radial cuts in the annulus. So you can start with $r=9.5, n=11, b=32$ and calculate $a$. This is the radius of your outermost circular cut. Put that in for $b$ and you can calculate the radius of the next circular cut. Keep going until you cover the center. If you increase $n$, you can decrease $a$, so try again with $n=12$. Keep going until you are happy. You haven't given a criterion for what is optimal, but maybe total number of pieces is a good one. – Ross Millikan Jul 9 '12 at 20:41
@qwertymk: You might also want to play with changing $n$, the number of radial cuts, as you move inward. It seems like it should decrease. – Ross Millikan Jul 9 '12 at 20:42