Question

If $r_1,r_2,r_3$ be the radii of excircles of the triangle $ABC$ then $\sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}$ is equal to.

• A. $\sum _{}cot\frac{A}{2}$
• B. $\sum _{}cot\frac{A}{2}cot\frac{B}{2}$
• C. $\sum _{}\tan \frac{A}{2}$
• D. $\prod _{}\tan \frac{A}{2}$

Answer 1

The correct option is C

$\sum _{}\tan \frac{A}{2}$

Explanation for the correct option:

Step $1:$ Concept and formula to be used.

According to the Herons formula, Radii of excircles of a triangle can be given by

$$\begin{gathered} r_1=\frac{∆}{s-a} \\ {r}_2=\frac{∆}{s-b} \\ {r}_3=\frac{∆}{s-c} \end{gathered}$$

Where, $∆$ is the area of the triangle given by $∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.

$r_1,r_2,r_3$ are the radii of excircles.

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

$s$ is the semi-perimeter of the triangle given by $s=\frac{1}{2}\left(a+b+c\right)$.

$$\begin{gathered} \tan \frac{A}{2}=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}} \\ \tan \frac{B}{2}=\sqrt{\frac{\left(s-a\right)\left(s-c\right)}{s\left(s-b\right)}} \\ \tan \frac{C}{2}=\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{s\left(s-c\right)}} \end{gathered}$$

Step $2:$ Compute the required value.

Expand the given expression as follows:

$\sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\frac{r_1+r_2+r_3}{\sqrt{r_1{r}_2+r_2{r}_3+r_1{r}_3}}$

Substitute the values of radii of excircles.

$$\begin{gathered} \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\frac{\frac{∆}{s-a}+\frac{∆}{s-b}+\frac{∆}{s-c}}{\sqrt{\frac{∆}{s-a}\frac{∆}{s-b}+\frac{∆}{s-b}\frac{∆}{s-c}+\frac{∆}{s-c}\frac{∆}{s-a}}} \\ \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\frac{\frac{∆}{s-a}+\frac{∆}{s-b}+\frac{∆}{s-c}}{\sqrt{\frac{{∆}^2}{\left(s-a\right)\left(s-b\right)}+\frac{{∆}^2}{\left(s-b\right)\left(s-c\right)}+\frac{{∆}^2}{\left(s-c\right)\left(s-a\right)}}} \\ \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\frac{\frac{∆}{s-a}+\frac{∆}{s-b}+\frac{∆}{s-c}}{∆\sqrt{\frac{1}{\left(s-a\right)\left(s-b\right)}+\frac{1}{\left(s-b\right)\left(s-c\right)}+\frac{1}{\left(s-c\right)\left(s-a\right)}}} \\ \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\frac{\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}}{\sqrt{\frac{1}{\left(s-a\right)\left(s-b\right)}+\frac{1}{\left(s-b\right)\left(s-c\right)}+\frac{1}{\left(s-c\right)\left(s-a\right)}}} \\ \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\frac{\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}}{\sqrt{\frac{s-a+s-b+s-c}{\left(s-a\right)\left(s-b\right)\left(s-c\right)}}} \\ \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\frac{\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}}{\sqrt{\frac{3s-\left(a+b+c\right)}{\left(s-a\right)\left(s-b\right)\left(s-c\right)}}} \\ \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\frac{\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}}{\sqrt{\frac{3s-2s}{\left(s-a\right)\left(s-b\right)\left(s-c\right)}}} \\ \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\frac{\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}}{\sqrt{\frac{s}{\left(s-a\right)\left(s-b\right)\left(s-c\right)}}} \end{gathered}$$

Multiply and divide by $s$ inside the square root.

$$\begin{gathered} \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\frac{\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}}{\sqrt{\frac{s^2}{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}} \\ \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\frac{\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}}{{\displaystyle \frac{s}{∆}}} \\ \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\frac{∆}{s\left(s-a\right)}+\frac{∆}{s\left(s-b\right)}+\frac{∆}{s\left(s-c\right)} \end{gathered}$$ [Since, $∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$]

]Which can be written as follows:

$$\begin{gathered} \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}}+\sqrt{\frac{\left(s-a\right)\left(s-c\right)}{s\left(s-b\right)}}+\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{s\left(s-c\right)}} \\ \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2} \\ \Rightarrow \sum _{}\frac{r_1}{\sqrt{r_1{r}_2}}=\sum \tan \frac{A}{2} \end{gathered}$$

Hence, option $C$ is the correct answer.