Proportion of the area covered by circles

[Dhamnekar Win,odd]

For the two attached pictures below, let [math]{P_n}[/math] denote the proportion of the big circle covered by the small circles as function of the number of the small circle in the outer layer. Find the expression for [math]P_n[/math] and compute [math]\lim_{n\to \infty} P_n [/math]. Would you answer this question in details?

 

**Author's answer to this question as follows:** Let R be the radius of the big circle, [math]r_n[/math] be the radius of the small circles in the outer layer, and \(R_n\) be the radius of the circle encompassing all small circles except those in the outer layer. Then in both cases \(P_n R^2\pi = n r^2_n \pi + P_n R^2_n \pi (\star)\) and \(\sin{\displaystyle\frac{\pi}{n} }= \displaystyle\frac{r_n}{R - r_n}\) In the first case, \(R_n= R- 2r_n\), giving $$ P_n = \displaystyle\frac{nr_n}{4(R-r_n)} = \displaystyle\frac{n \sin{\displaystyle\frac{\pi}{n}}}{4}, \lim_{n \to \infty} =\displaystyle\frac{\pi}{4}$$ In the second case  we have \((R-r_n) \cos{\displaystyle\frac{r}{n}}- \sqrt{(r_n + \hat{r}_n )^2 - r^2_n} + \hat{r}_n = R_n \) where \(\hat{r}_n \) is the radius of the circle in the second outer layer \(\sin{\displaystyle\frac{\pi}{n}} = \displaystyle\frac{\hat{r}_n}{R_n -\hat{r}_n}\). The last two relations yield \(R_2\) to be substituted in \((\star)\) whence we get an expression for \(P_n\) and $$\lim_{n \to \infty} P_n = \displaystyle\frac{\pi}{2\sqrt{3}}$$ Now I agreed with author's answer in the first case. But in the second case, I got [math] \lim_{n\to\infty} P_n = \displaystyle\frac{\pi\cdot r_n}{2R- r_n}[/math]. How is [math]P_n = \displaystyle\frac{\pi}{2\sqrt{3}}[/math] ? 

Edited January 5 by Dhamnekar Win,odd

[Genady]

1 hour ago, Dhamnekar Win,odd said:

Then in both cases PnR2π=nr2nπ+PnR2nπ

I see how \( P_n R^2\pi = n r^2_n \pi + P_n R^2_n \pi \) is correct in the first case, but not in the second one because in the second case the radius \( R_n \) partially covers the circles in the outer layer.

[Dhamnekar Win,odd]

@Genady, Would you explain your answer in details?

 

There are two relations we know in the second case. \(1) (R -r_n) \cos{\displaystyle\frac{\pi}{n}} - \sqrt{(r_n +\hat{r}_n)^2 -\hat{r}_n^2} + \hat{r}_n = R_n,  2) \sin{\displaystyle\frac{\pi}{n}}=\displaystyle\frac{\hat{r}_n}{R_n - \hat{r}_n}\) The last relation said \(\hat{r}_n=\displaystyle\frac{R_n \cdot \sin{\displaystyle\frac{\pi}{n}}}{1 +\sin{\displaystyle\frac{\pi}{n}}}\). Replacing n in \(R_n \)by 2, and replacing \(\cos^2{\displaystyle\frac{\pi}{n}}\) by \(1- \sin^2{\displaystyle\frac{\pi}{n}}\) and squaring and expanding the brackets in the first relation we get \(R_2 = (R- r_n)\). Replacing the \(R_2\)'s value in the \(P_nR^2\pi = nr^2_n \pi + P_n\pi R^2_2\) we get \(P_n = \displaystyle\frac{\pi r_n}{2R-r_n} \) But author said \(P_n=\displaystyle\frac{\pi}{2\sqrt{3}}\)

 

How is that?   

Edited January 6 by Dhamnekar Win,odd

[Genady]

1 hour ago, Dhamnekar Win,odd said:

Would you explain your answer in details?

Never mind, I got it.

1 hour ago, Dhamnekar Win,odd said:

How is that?

Looking at it ...

1 hour ago, Dhamnekar Win,odd said:

Replacing n in Rn by 2

How can you replace \(n\) by \(2\) in \(R_n\), but leave it \(n\) everywhere else in the expression?

[Genady]

7 hours ago, Dhamnekar Win,odd said:

Replacing n ... by 2

Anyway, n cannot be 2 in these configurations if n is:

On 1/5/2024 at 6:35 AM, Dhamnekar Win,odd said:

the number of the small circle in the outer layer.

 

[Dhamnekar Win,odd]

I am sorry. I wrongly computed \(P_n\) My new rectified \(P_n=\displaystyle\frac{n\cdot r_n}{2R-r_n}\)

Edited January 7 by Dhamnekar Win,odd

[Genady]

10 minutes ago, Dhamnekar Win,odd said:

I am sorry. I wrongly computed Pn My new rectified Pn=n⋅rn2R−rn

You need to express it as a function of \(n\).

[Dhamnekar Win,odd]

But in this case [math]\lim_{n \to \infty} P_n = \infty [/math] Isn't it?

Edited January 7 by Dhamnekar Win,odd

[Genady]

9 minutes ago, Dhamnekar Win,odd said:

But in this case limn→∞Pn=∞ Isn't it?

Not necessarily. As \(n \rightarrow \infty\), \(r_n \rightarrow 0\).

[Dhamnekar Win,odd]

In that case \(P_n\) would be  $$\lim_{n \to \infty, r_n \to 0} P_n = \displaystyle\frac{n \cdot r_n}{2R- r_n}=0 $$

 

But author said \( \lim_{n \to \infty} P_n =\displaystyle\frac{\pi}{2\sqrt{3}}\)

Edited January 7 by Dhamnekar Win,odd

[Genady]

2 hours ago, Dhamnekar Win,odd said:

In that case Pn would be 

limn→∞,rn→0Pn=n⋅rn2R−rn=0

 

 

 

No, not necessarily.

For example, if \(r_n= \frac c n\) then \[\lim_{n \to \infty} n \cdot r_n=c\]

Edited January 7 by Genady

[phyti]

Dehamnekar:

The author made the problem more complicated than necessary. No need to count the circles in the outer band.

Let B=area of outer circle and A= area of inner circle.

The ratio P =(B-A)/B. The limit of A=B, thus P=0/B.

 

[Dhamnekar Win,odd]

If [math]\lim_{n \to \infty} n\cdot r_n \to 2 \cdot \pi \cdot R_2[/math]

 

We got \(R-r_n = R_2 \Rightarrow R= R_2 +  R_2= 2R_2\) 

 

Putting these values in our final expression for \(P_n\) we get \(\displaystyle\frac{2 \cdot \pi \cdot R_2}{4 \cdot  R_2- R_2}=\frac23 \cdot \pi = P_n=2.0944 \) approx.

 

But author's answer is \(P_n=0.9069\) approx. 

How is that?

 

 

 

Edited January 8 by Dhamnekar Win,odd

[Genady]

42 minutes ago, Dhamnekar Win,odd said:

If limn→∞n⋅rn→2⋅π⋅R2

 

We got R−rn=R2⇒R=R2+R2=2R2  

 

Putting these values in our final expression for Pn we get 2⋅π⋅R24⋅R2−R2=23⋅π=Pn=2.0944 approx.

 

But author's answer is Pn=0.9069 approx. 

How is that?

 

 

 

I'd need to see your step-by-step derivation.

[Genady]

1 hour ago, Genady said:

I'd need to see your step-by-step derivation.

Surely your result, \(P_n=2.0944\) is wrong because \(P_n\) has to be \(\lt 1\) by its definition.

[Dhamnekar Win,odd]

Do you agree with my derivation for the expression of \(P_n = \displaystyle\frac{n \cdot r_n}{2\cdot R -r_n}, R_2= R-r_n\). If not, I shall show you my working over these derivation.

Edited January 8 by Dhamnekar Win,odd

[Genady]

3 minutes ago, Dhamnekar Win,odd said:

Do you agree with my derivation for the expression of Pn=n⋅rn2⋅R−rn,R2=R−rn . If not, I shall show you my working over these derivation.

Please show it.

[Genady]

2 hours ago, Dhamnekar Win,odd said:

Do you agree with my derivation for the expression of Pn=n⋅rn2⋅R−rn,R2=R−rn . If not, I shall show you my working over these derivation.

Look at the picture for case 2:

You can count, \(n=13\). You can measure, \(R/r_n \approx 5\). If your formula, \(P_n = \displaystyle\frac{n \cdot r_n}{2\cdot R -r_n}\) were correct, then \(P_{13}=\displaystyle\frac{13 \cdot 1}{2\cdot 5 -1} \gt 1\), which cannot be correct because it has to be \(\lt 1\) for any \(n\).

So, your result is wrong.

 

Edited January 8 by Genady

[phyti]

Dehamnekar:

 

Let B=outer circle area πR2 and A= inner circle area πr2 .

P=ratio of outer band to B, = (B-A)/B.

Case 1.

P=1-A/B=(1- r2/R2) .

The graph plots P as r varies.

Case 2.

The inner circle overlaps the outer band.

When A=B, there are small portions of the band remaining.

 

[Dhamnekar Win,odd]

Author's answer to first case is \(\displaystyle\frac{\pi}{4}\) Do you disagree with that? Author assumed R be the radius of big circle, \(r_n\) be the radius in the outer layer of the big circle and \(\hat{r}_n\) be the radius of next inner layer circle of the big circle and \(R_n\) be the radius of the concentric circles inside the big circle. Following are the concentric circles.

 Do you agree with this author's assumption? Author computed the first case answer considering all these assumption.

I think the following question with answer will be considered for more reference. 

 

Win_odd Dhamnekar (https://math.stackexchange.com/users/153118/win-odd-dhamnekar), Inscribed circles inside an equilateral triangle, URL (version: 2023-07-14): https://math.stackexchange.com/q/4735973

[Genady]

5 hours ago, Dhamnekar Win,odd said:

Rn be the radius of the concentric circles inside the big circle.

No, this is not so. Here it is:

 

On 1/5/2024 at 6:35 AM, Dhamnekar Win,odd said:

Rn be the radius of the circle encompassing all small circles except those in the outer layer.

 

[phyti]

Dehamnekar:

Was uncertain about case-2, didn't make any calculations.

Because of overlap, (A+B)>1 initially.

"When A=B, there are small portions of the band remaining."

This needs clarification.

As A expands toward B, the band shrinks to 0 at A=B.

My response is based on the case-2 picture.

 

[KJW]

The following is a solution to the second problem:

Let [math]R[/math] be the radius of the large circle.
Let [math]r[/math] be the radius of the small circles in the outermost layer.
Let [math]\alpha r[/math] be the radius of the small circles in the second outermost layer.
Let [math]n[/math] be the number of small circles around the circumference of the large circle.
Let [math]P[/math] be the proportion of the area of the large circle covered by small circles for [math]n[/math] small circles around the circumference of the large circle.

Partitioning the large circle into [math]n[/math] sectors, the area of each sector is:

[math]= \dfrac{\pi R^2}{n}[/math]

The total area of small circles covering a sector:

[math]= \pi r^2 (1 + \alpha^2 + \alpha^4 + \alpha^6 + \cdots) = \dfrac{\pi r^2}{1 - \alpha^2}[/math]

Then the proportion of the area of a sector covered by small circles:

[math]P = \dfrac{n r^2}{(1 - \alpha^2) R^2}[/math]

[math]\dfrac{r}{R - r} = \sin\dfrac{\pi}{n}[/math]

[math]r = (R - r) \sin\dfrac{\pi}{n}[/math]

[math]r (1 + \sin\dfrac{\pi}{n}) = R \sin\dfrac{\pi}{n}[/math]

[math]\dfrac{r}{R} = \dfrac{\sin\dfrac{\pi}{n}}{1 + \sin\dfrac{\pi}{n}}[/math]

[math]P = \dfrac{n\>\sin^2\dfrac{\pi}{n}}{(1 - \alpha^2) (1 + \sin\dfrac{\pi}{n})^2}[/math]

Let [math](1 - \alpha^2) = \beta \dfrac{\pi}{n}[/math]

[math]\displaystyle \lim_{n\to\infty} P = \dfrac{n\>\left( \dfrac{\pi}{n} \right)^2}{\beta\>\dfrac{\pi}{n}} = \dfrac{\pi}{\beta}[/math]

Applying the law of cosines:

Note that [math](1 + \alpha) r[/math] is the distance between the centres of the outermost and nearest second outermost circles, and that [math](R - r)[/math] and [math]\alpha (R - r)[/math] are the distances from the centre of the large circle to the centres of the outermost circle and second outermost circle, respectively.

[math](1 + \alpha)^2 r^2 = (1 + \alpha^2) (R - r)^2 - 2 \alpha (R - r)^2 \cos\dfrac{\pi}{n}[/math]

[math](1 + \alpha)^2 \dfrac{r^2}{(R - r)^2} = 1 + \alpha^2 - 2 \alpha \cos\dfrac{\pi}{n}[/math]

[math]\dfrac{r}{R - r} = \sin\dfrac{\pi}{n}[/math]

[math](1 + \alpha)^2 \sin^2\dfrac{\pi}{n} = 1 + \alpha^2 - 2 \alpha \cos\dfrac{\pi}{n}[/math]

[math](1 + \alpha)^2 \sin^2\dfrac{\pi}{n} = (1 + \alpha)^2 - 2 \alpha (1 + \cos\dfrac{\pi}{n})[/math]

[math](1 + \alpha)^2 \cos^2\dfrac{\pi}{n} - 2 \alpha (1 + \cos\dfrac{\pi}{n}) = 0[/math]

[math](1 + \alpha)^2 \cos^2\dfrac{\pi}{n} - 2 (1 + \alpha) (1 + \cos\dfrac{\pi}{n}) + 2 (1 + \cos\dfrac{\pi}{n}) = 0[/math]

Let [math]\phi = \dfrac{\pi}{n}[/math]

[math](1 + \alpha)^2 \cos^2\phi - 2 (1 + \alpha) (1 + \cos\phi) + 2 (1 + \cos\phi) = 0[/math]

Also let [math]\phi \approx 0[/math]

[math]\cos\phi \approx 1 - \dfrac{1}{2} \phi^2[/math]

[math](1 + \alpha)^2 (1 - \phi^2) - (1 + \alpha) (4 - \phi^2) + (4 - \phi^2) = 0[/math]

This is a quadratic equation to be solved for [math](1 + \alpha)[/math]. However, only the lesser of the two solutions is of interest.

[math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{(4 - \phi^2)^2 - 4 (1 - \phi^2)(4 - \phi^2)}}{2 (1 - \phi^2)}[/math]

[math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{(16 - 8 \phi^2 + \phi^4) - (16 - 20 \phi^2 + 4 \phi^4)}}{2 (1 - \phi^2)}[/math]

[math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{12 \phi^2 - 3 \phi^4}}{2 (1 - \phi^2)}[/math]

[math]1 + \alpha = 2 - \sqrt{3} \phi[/math]

[math]\alpha = 1 - \sqrt{3} \phi[/math]

[math]\alpha^2 = 1 - 2 \sqrt{3} \phi + 3 \phi^2[/math]

[math]1 - \alpha^2 = \beta \dfrac{\pi}{n} = 2 \sqrt{3} \phi = 2 \sqrt{3} \dfrac{\pi}{n}[/math]

[math]\beta = 2 \sqrt{3}[/math]

Therefore:

[math]\displaystyle \lim_{n\to\infty} P = \dfrac{\pi}{2 \sqrt{3}}[/math]

 

 

Edited January 12 by KJW

[KJW]

With the solution to the second problem in place, I decided to provide a solution to the generalised intermediate version of this problem, where the first and second problems are the two extremes. Because much of the solution to the second problem is unchanged in the solution to the generalised problem, I shall only include the parts of the solution that have changed.

Let [math]\sigma[/math] be a parameter, [math]0 \leqslant \sigma \leqslant 1[/math], such that [math]\dfrac{\sigma \pi}{n}[/math] is the angle between the line joining the centre of the large circle to the centre of the outermost circle and the line joining the centre of the large circle to the centre of the nearest second outermost circles.

Applying the law of cosines and continuing as shown in solution to the second problem:

[math](1 + \alpha)^2 \cos^2\dfrac{\pi}{n} - 2 (1 + \alpha) (1 + \cos\dfrac{\sigma\pi}{n}) + 2 (1 + \cos\dfrac{\sigma\pi}{n}) = 0[/math]

Note that the angle associated with [math]\cos^2\dfrac{\pi}{n}[/math] has a different origin to the angle associated with [math]1 + \cos\dfrac{\sigma\pi}{n}[/math].

[math]\phi = \dfrac{\pi}{n}[/math]

[math](1 + \alpha)^2 \cos^2\phi - 2 (1 + \alpha) (1 + \cos\sigma\phi) + 2 (1 + \cos\sigma\phi) = 0[/math]

[math]\phi \approx 0[/math] ,     [math]\cos\phi \approx 1 - \dfrac{1}{2} \phi^2[/math] :

[math](1 + \alpha)^2 (1 - \phi^2) - (1 + \alpha) (4 - \sigma^2 \phi^2) + (4 - \sigma^2 \phi^2) = 0[/math]

Solving the quadratic equation for [math](1 + \alpha)[/math] :

[math]1 + \alpha = \dfrac{(4 - \sigma^2 \phi^2) - \sqrt{(4 - \sigma^2 \phi^2)^2 - 4 (1 - \phi^2)(4 - \sigma^2 \phi^2)}}{2 (1 - \phi^2)}[/math]

[math]1 + \alpha = \dfrac{(4 - \sigma^2 \phi^2) - \sqrt{(16 - 8 \sigma^2 \phi^2 + \sigma^4 \phi^4) - (16 - (16 + 4 \sigma^2) \phi^2 + 4 \sigma^2 \phi^4)}}{2 (1 - \phi^2)}[/math]

[math]1 + \alpha = \dfrac{(4 - \sigma^2 \phi^2) - \sqrt{(16 - 4 \sigma^2) \phi^2 - (4 \sigma^2 - \sigma^4) \phi^4}}{2 (1 - \phi^2)}[/math]

[math]1 + \alpha = 2 - \sqrt{4 - \sigma^2}\>\phi[/math]

[math]\alpha = 1 - \sqrt{4 - \sigma^2}\>\dfrac{\pi}{n}[/math]

[math]1 - \alpha^2 = 2 \sqrt{4 - \sigma^2}\>\dfrac{\pi}{n}[/math]

 

[math]\displaystyle \lim_{n\to\infty} P = \dfrac{\pi}{2 \sqrt{4 - \sigma^2}}[/math]

 

For problem #1,  [math]\sigma = 0:[/math]    [math]\displaystyle \lim_{n\to\infty} P = \dfrac{\pi}{4}[/math]

For problem #2,  [math]\sigma = 1:[/math]    [math]\displaystyle \lim_{n\to\infty} P = \dfrac{\pi}{2 \sqrt{3}}[/math]

 

 

Edited January 13 by KJW

[phyti]

On 1/9/2024 at 12:14 AM, Dhamnekar Win,odd said:

Author's answer to first case is π4 Do you disagree with that? Author assumed R be the radius of big circle, rn be the radius in the outer layer of the big circle and r^n be the radius of next inner layer circle of the big circle and Rn be the radius of the concentric circles inside the big circle. Following are the concentric circles.

 Do you agree with this author's assumption? Author computed the first case answer considering all these assumption.

I think the following question with answer will be considered for more reference. 

 

Win_odd Dhamnekar (https://math.stackexchange.com/users/153118/win-odd-dhamnekar), Inscribed circles inside an equilateral triangle, URL (version: 2023-07-14): https://math.stackexchange.com/q/4735973

As Rn approaches R, rn approaches 0.

When Rn equals R, rn equals 0.

Why doesn't P=outer band/area of circle=0?