Chapter 8, Problem 9P

To determine

To show: That $\underset{n\to \infty }{\lim }\frac{A_n}{A}=\frac{\pi }{2\sqrt{3}}$ .

Answer to Problem 9P

It is proved that $\underset{n\to \infty }{\lim }\frac{A_n}{A}=\frac{\pi }{2\sqrt{3}}$ .

Explanation of Solution

Given:

It is given that the circle of equal diameter is packed tightly in $n$ rows inside an equilateral triangle.

Calculation:

Draw the schematic diagram of the triangle.

  

Find the area of the triangle.

  $\begin{array}{c}A=\frac{1}{2}\times \text{base}\times \text{height}\\ \text{=}\frac{1}{2}\left(L\right)\left(\frac{\sqrt{3}}{2}L\right)\\ =\frac{\sqrt{3}}{4}{L}^2\\ L^2=\frac{4}{\sqrt{3}}A\end{array}$

Calculate the total length of the triangle from the figure shown.

  $\begin{array}{c}L=\sqrt{3}r+r+\left(n-2\right)\left(2r\right)+r+\sqrt{3}r\\ =r\left(2n-2+2\sqrt{3}\right)\\ r=\frac{L}{\left(2n-2+2\sqrt{3}\right)}\end{array}$

The circle are arranged in the triangle starting from $1,2,3,\mathrm{...}n$ in the form of natural number series, since there are $n$ number of circles. So, use the equation of natural number series.

Write the equation of the natural number series for $n$ number of circles.

  $1+2+\cdots +n=\frac{n\left(n+1\right)}{2}$

Calculate the total area of the $n$ number of circles using the formula,

  $A_n=\frac{n\left(n+1\right)}{2}\pi r^2$

Substitute $\frac{L}{\left(2n-2+2\sqrt{3}\right)}$ for $r$ in the formula $A_n=\frac{n\left(n+1\right)}{2}\pi r^2$ .

  $\begin{array}{c}{A}_n=\frac{n\left(n+1\right)}{2}\pi {\left(\frac{L}{\left(2n-2+2\sqrt{3}\right)}\right)}^2\\ =\frac{n\left(n+1\right)}{2}\pi \frac{L^2}{4{\left(n+\sqrt{3}-1\right)}^2}\end{array}$

Substitute $\frac{4}{\sqrt{3}}A$ for $L^2$ in the above equation.

  $\begin{array}{c}{A}_n=\frac{n\left(n+1\right)}{2}\pi \frac{\frac{4}{\sqrt{3}}A}{4{\left(n+\sqrt{3}-1\right)}^2}\\ =\frac{n\left(n+1\right)}{{\left(n+\sqrt{3}-1\right)}^2}\frac{\pi A}{2\sqrt{3}}\\ \frac{A_n}{A}=\frac{n\left(n+1\right)}{{\left(n+\sqrt{3}-1\right)}^2}\frac{\pi }{2\sqrt{3}}\\ =\frac{1+\frac{1}{n}}{{\left(1+\left(\frac{\sqrt{3}-1}{n}\right)\right)}^2}\frac{\pi }{2\sqrt{3}}\end{array}$

Apply the limits on both sides where $n\to \infty$ .

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{A_n}{A}=\underset{n\to \infty }{\lim }\frac{1+\frac{1}{n}}{{\left(1+\left(\frac{\sqrt{3}-1}{n}\right)\right)}^2}\frac{\pi }{2\sqrt{3}}\\ =\frac{1+0}{{\left(1+0\right)}^2}\frac{\pi }{2\sqrt{3}}\\ =\frac{\pi }{2\sqrt{3}}\end{array}$

Therefore, it is proved that $\underset{n\to \infty }{\lim }\frac{A_n}{A}=\frac{\pi }{2\sqrt{3}}$ .