Chapter 8.2, Problem 57E

To determine

To find: an expression for the diameter of $C_n$ .

Answer to Problem 57E

  $r_c=\frac{1}{2n(n+1)}.$

Explanation of Solution

Given information: Given figure shows two circle C and D of radius 1 that touch at P . T

is a common tangent line; $C_1$ is the circle that touches $C,D$ , and T ; $C_2$ is the circle that touches $C,D\text{and}{C}_1:C_3$ is the circle that touches C, D and $C_2$ . This procedure can be continued indefinitely and produces an infinite sequence of circles $\left\{C_n\right\}.$

  

Calculation:

  

To solve this problem find enough values of $r_c$ to find a pattern looking at the circles and using Pythagoras Theorem for the first circle.

  $\begin{array}{c}$ {(r_0+r_1)}^2={(r_0-r_1)}^2+r_0{}^2.$r_0{}^2+2r_0{r}_1+r_1{}^2=r_0{}^2-2r_0{r}_1+r_1{}^2+r_0{}^2.\\ r_0{}^2-4r_0{r}_1=0.\\ \text{Substitute}\text{and}\text{solve:}\\ 1-4r_1=0\\ r_1=\frac{1}{4}.\end{array}$

Looking at the circles and using Pythagoras Theorem for the second circle:

  $\begin{array}{c}$ {(r_0+r_2)}^2={(r_0-2r_1-r_2)}^2+r_0{}^2.$\text{Substitute}\text{and}\text{solve}:\\ $ {(1+r_2)}^2={(\frac{1}{2}-r_2)}^2+r_0{}^2$1+2r_2+r_2{}^2=\frac{1}{4}-r_2+r_2{}^2+1\\ 2r_2=\frac{1}{4}-r_2\\ 3r_2=\frac{1}{4}.\\ r_2=\frac{1}{12}.\end{array}$

Looking at the circles and using Pythagoras Theorem for the third circle:

  $\begin{array}{c}$ {(r_0+r_3)}^2={(r_0-2r_1-2r_2-r_3)}^2+r_0{}^2.$$ \text{Substitute}\text{and}\text{solve}:$$ {(1+r_3)}^2={(1-\frac{1}{2}-\frac{1}{6}-r_3)}^2+1$$ {(1+r_3)}^2={(\frac{1}{3}-r_3)}^2+1$1+2r_3+r_3{}^2=\frac{1}{9}-\frac{2}{2}{r}_3+r_3{}^2+1\\ 2r_3+\frac{2}{3}{r}_3=\frac{1}{9}\\ \frac{8}{3}{r}_3=\frac{1}{9}.\\ r_3=\frac{1}{24}.\end{array}$

So, the pattern for the radius is given by:

  $\begin{array}{l}{r}_c=\frac{1}{2n(n+1)}.\\ a=r_0-2r_1-2r_2-2r_3-\mathrm{......}-r_n.\\ a=r_0-2{\displaystyle \sum _1^{n-1}{r}_n}-r_n.\end{array}$