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We divide the unit circle into equal sectors and inscribe circles within each sector. Assuming we make smaller and smaller sectors, what does the total area of the inscribed circles converges to?

I think I have solved it, but since I'm not a mathematician, I wonder if this is correct.

I began by asking: What is the total area of all the circles in a single sector?

We will start with the unit circle as the outer perimeter, with $\theta$ being half of the subtended angle, and $r_0$ being the radius of the first ring of circles, $r_1$ being the radius of the second ring of circles, and so on.

Since the unit circle has a radius of 1, we can immediately deduce the following: $$r_{0} = ({\sin{\theta} \over 1+\sin{\theta}})$$ and we also know that: $$r_{n+1} = r_{n} ({1-\sin{\theta} \over 1+\sin{\theta}})$$ that means that: $$r_{1} = ({\sin{\theta} \over 1+\sin{\theta}}) ({1-\sin{\theta} \over 1+\sin{\theta}}) = {(\sin{\theta})({1-\sin{\theta})} \over (1+\sin{\theta})^2}$$ and: $$r_{2} = {(\sin{\theta})({1-\sin{\theta})^2} \over (1+\sin{\theta})^3}$$ so generally we can say that: $$r_{n} = {(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}}$$ the area of a single circle in a sector is: $$A_{n} = \pi r_{n}^2 = \pi ({(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}})^2$$ so the area of a single sector of circles will be: $$A_{s} = \pi \sum_{n=0}^{\infty} ({(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}})^2 $$ and since we're dealing with two or more sectors, we can say that: $$ 0 < \theta \leq {\pi \over 2} $$ so this means the sum of areas of all the circles inscribed in a sector converges to: $$A_{s} = \pi \sum_{n=0}^{\infty} ({(\sin{\theta})({1-\sin{\theta})^n} \over (1+\sin{\theta})^{n+1}})^2 = {\pi \sin{\theta} \over 4}$$

I then continued with: What is the total area of all the circles in all the sectors?

Assuming we have $k$ sectors, then $\theta = {\pi \over k}$ and our sum would be: $$A_{k} = {k A_{s}} = {\pi k \sin{\pi \over k} \over 4}$$ and if we take the limit of $k$, we will get: $$\lim_{k \to \infty} {\pi k \sin{\pi \over k} \over 4} = {\pi^2 \over 4} \tag*{$\blacksquare$}$$

Is this valid?

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The first question is "What is the total area of all the circles in a single sector?".

If "the total area of all the circles in a single sector" means $\displaystyle\lim_{n\to\infty}\sum_{i=0}^{n} \pi r_i^2$, then I think that you've correctly got $\displaystyle\lim_{n\to\infty}\sum_{i=0}^{n} \pi r_i^2=\dfrac{\pi \sin{\theta}}4$.


The second question is "What is the total area of all the circles in all the sectors?".

I think that the meaning of "the total area of all the circles in all the sectors" is ambiguous since here we have two variables.

Let $$f(n,k):=k\cdot \frac{\pi\sin\frac{\pi}{k}}{4}\bigg(1-\bigg(\frac{1-\sin\frac{\pi}{k}}{1+\sin\frac{\pi}{k}}\bigg)^{2n}\bigg)$$ be the sum of the areas of the $nk$ circles where we have $k$ sectors each of which has $n$ circles.

  • If "the total area of all the circles in all the sectors" means $\displaystyle\lim_{k\to\infty}\lim_{n\to\infty}f(n,k)$ (first take $n\to\infty$ and then take $k\to\infty$), then your argument looks valid to me, and I think that you've correctly got $\displaystyle\lim_{k\to\infty}\lim_{n\to\infty}f(n,k)=\dfrac{\pi^2}{4}$.

  • If "the total area of all the circles in all the sectors" means $\displaystyle\lim_{n\to\infty}\lim_{k\to\infty}f(n,k)$ (first take $k\to\infty$ and then take $n\to\infty$), then your argument is not valid since $\displaystyle\lim_{n\to\infty}\lim_{k\to\infty}f(n,k)=0$.

  • If "the total area of all the circles in all the sectors" means $\displaystyle\lim_{(n,k)\to (\infty,\infty)}f(n,k)$, then your argument is not valid since $\displaystyle\lim_{(n,k)\to (\infty,\infty)}f(n,k)$ does not exist. To see this, we set $n=ak$ to have $$\lim_{(n,k)\to (\infty,\infty)}f(n,k)=\lim_{k\to\infty}\frac{\pi^2}{4}\cdot \frac{\sin\frac{\pi}{k}}{\frac{\pi}{k}}\bigg(1-\bigg(\bigg(1+\frac{1}{\frac{1-\sin\frac{\pi}{k}}{2\sin\frac{\pi}{k}}}\bigg)^{\frac{1-\sin\frac{\pi}{k}}{2\sin\frac{\pi}{k}}}\bigg)^{\frac{2}{1-\sin\frac{\pi}{k}}\cdot\frac{\sin\frac{\pi}{k}}{\frac{\pi}{k}}\cdot(-2a\pi)}\bigg)$$This equals $\dfrac{\pi^2}{4}(1-e^{-4a\pi})$ which depends on $a$. So, $\displaystyle\lim_{(n,k)\to (\infty,\infty)}f(n,k)$ does not exist.


Some comments :

  • If we assume that we have $n$ sectors instead of $k$ sectors, then "the total area of all the circles in all the sectors" should be $\displaystyle\lim_{n\to\infty}f(n,n)=\dfrac{\pi^2}{4}(1-e^{-4\pi})$.

  • If we assume that we have $2n$ sectors instead of $k$ sectors, then "the total area of all the circles in all the sectors" should be $\displaystyle\lim_{n\to\infty}f(n,2n)=\dfrac{\pi^2}{4}(1-e^{-2\pi})$.

  • If we assume that we have $mn$ sectors instead of $k$ sectors where $m$ is a fixed positive integer which does not depend on $n$, then "the total area of all the circles in all the sectors" should be $\displaystyle\lim_{n\to\infty}f(n,mn)=\dfrac{\pi^2}{4}(1-e^{-\frac{4\pi}{m}})$.

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    $\begingroup$ Thank you for answering this, the comments section is also very interesting, I love how playing around with the variables you get beautiful results. In response to @yves-daoust idea, it seems possible to remove $n$ and keep the ratio solely dependent on the number of $k$ sectors. Anyways, great answer, thanks again! $\endgroup$ – Amos Haviv Dec 8 '20 at 13:42
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The ratio of the total area of the disks over the outer circle is the same as the ratio of the area of a single disk over the tight annular sector that bounds it.

If there are $k$ circles of radius $r$ in the outer ring, the subtended angle is $\theta=\dfrac{2\pi}k$.

The area of a sector is $$\theta(1-(1-2r)^2)$$

and the requested ratio

$$\frac{kr}{4(1+2r)}.$$

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  • $\begingroup$ Thanks for your answer! If we go by this ratio, we say that: $$ \textrm{Area of annulus} = {π (1 - (1 - 2 r)^2)}$$ and: $$ \textrm{Area of circles} = k \pi r^2 $$ then we get a ratio of: $$ kr \over {4(1-r)} $$ if we take it further the and say that: $$ r = {\sin({\theta \over 2}) \over (1 + \sin({\theta \over 2}))} $$ then the ratio is: $$ {{1 \over 4} k {\sin({\pi \over k})}} $$ and so $$ \lim_{k \to \infty} {{1 \over 4} k {\sin({\pi \over k})}} = {π \over 4}$$ $\endgroup$ – Amos Haviv Dec 1 '20 at 22:16

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