Question

In a triangle, if $r_1=2r_2=3r_3,$then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ is equal to:

• A. $\frac{75}{60}$
• B. $\frac{155}{60}$
• C. $\frac{176}{60}$
• D. $\frac{191}{60}$

Answer 1

The correct option is D

$\frac{191}{60}$

Solving for $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ in a triangle where $r_1=2r_2=3r_3$:

Step 1: Given information

$r_1=2r_2=3r_{3}where{r}_1,r_{2}and{r}_3$are ex-circles.

$a,b,c$ are sides of the triangle.

Let the area of the triangle be $∆$ and semiperimeter be $s.$

Now, $r_1=\frac{∆}{s-a},r_2=\frac{∆}{s-b},r_3=\frac{∆}{s-c}$

Also, $\frac{r_1}{r_2}=2or\frac{s-b}{s-a}$.

Step 2: Finding $\frac{a}{b}:$

Equating we get:

$$\begin{gathered} \Rightarrow \frac{r_1}{r_2}=2 \\ \Rightarrow \frac{s-b}{s-a}=2 \\ \Rightarrow 2s-2a=s-b \\ \Rightarrow s-2a+b=0 \\ \Rightarrow \frac{a+b+c}{2}-2a+b=0 \\ \Rightarrow c-3a+3b=0 \\ \Rightarrow c=3a-3b \end{gathered}$$

Also, $\frac{r_1}{r_3}=3\mathrm{or}\frac{\mathrm{s}-\mathrm{c}}{\mathrm{s}-\mathrm{a}}$

Equating we get:

$$\begin{gathered} \Rightarrow \frac{r_1}{r_3}=3 \\ \Rightarrow \frac{s-c}{s-a}=3 \\ \Rightarrow 3s-3a=s-c \\ \Rightarrow 2s-3a+c=0 \\ \Rightarrow \frac{2a+2b+2c}{2}-3a+c=0 \\ \Rightarrow b-2a+2c=0 \\ \Rightarrow b-2a+2(3a-3b)=0 \\ \Rightarrow b-2a+6a-6b=0 \\ \Rightarrow 4a-5b=0 \\ \Rightarrow 4a=5b \\ \Rightarrow \frac{a}{b}=\frac{5}{4} \end{gathered}$$

Step 3: Solving for $\frac{b}{c}and\frac{c}{a}:$

$$\begin{gathered} b-2a+2c=0\mathrm{and}c=3a-3b \\ \Rightarrow b-\frac{2(c+3b)}{3}+2c=0 \\ \Rightarrow 3b-2c-6b+6c=0 \\ \Rightarrow 4c-3b=0 \\ \Rightarrow \frac{b}{c}=\frac{4}{3} \end{gathered}$$

Again, we have

$$\begin{gathered} \frac{(3a-c)}{3}-2a+2c=0 \\ \Rightarrow 3a-c-6a+6c=0 \\ \Rightarrow 5c-3a=0 \\ \Rightarrow \frac{c}{a}=\frac{3}{5} \end{gathered}$$

Step 4: Calculating $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$

$$\begin{gathered} \frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{5}{4}+\frac{4}{3}+\frac{3}{5} \\ \Rightarrow \frac{75+80+36}{60} \\ \Rightarrow \frac{191}{60} \\ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{191}{60} \end{gathered}$$

Hence, Option (D) is correct.