Question

The value of $\frac{1}{{r_1}^2}+\frac{1}{{r_2}^2}+\frac{1}{{r_3}^2}+\frac{1}{r^2}$ is

• A. $\frac{\left(a^2+b^2+c^2\right)}{{∆}^2}$
• B. $0$
• C. $1$
• D. $\frac{{∆}^2}{\left(a^2+b^2+c^2\right)}$

Answer 1

The correct option is A

$\frac{\left(a^2+b^2+c^2\right)}{{∆}^2}$

Explanation for the correct answer:

Consider the standard notation followed in triangle geometry.

Let $r$ be the radius of the incircle of the triangle.

Let $r_1,r_2,r_3$ be the radii of the excircles drawn opposite to the vertices $A,B,C$ respectively.

Let $s$ be the semi-perimeter of the triangle where $s=\frac{a+b+c}{2}$

Let $a,b,c$ be the lengths of the sides of the triangles opposite to the vertices $A,B,C$ respectively..

Let $∆$be the area of the triangle.

We know,

$∆=r.s=r_1\left(s-a\right)=r_2\left(s-b\right)=r_3\left(s-c\right)$

$$\begin{gathered} \therefore \frac{1}{r}=\frac{s}{∆} \\ \frac{1}{r_1}=\frac{s-a}{∆} \\ \frac{1}{r_2}=\frac{s-b}{∆} \\ \frac{1}{r_3}=\frac{s-c}{∆} \end{gathered}$$

$\Rightarrow$ $\frac{1}{{r_1}^2}+\frac{1}{{r_2}^2}+\frac{1}{{r_3}^2}+\frac{1}{r^2}=\frac{1}{{∆}^2}\left({\left(s-a\right)}^2+{\left(s-b\right)}^2+{\left(s-c\right)}^2+s^2\right)$

$=\frac{1}{{∆}^2}\left(s^2-2as+a^2+s^2-2bs+b^2+s^2-2cs+c^2+s^2\right)$

$=\frac{1}{{∆}^2}\left(4s^2-2s\left(a+b+c\right)+\left(a^2+b^2+c^2\right)\right)$

$=\frac{1}{{∆}^2}\left(4s^2-4s^2+\left(a^2+b^2+c^2\right)\right)$ $\left(\because a+b+c=2s\right)$

$\therefore$ $\frac{1}{{r_1}^2}+\frac{1}{{r_2}^2}+\frac{1}{{r_3}^2}+\frac{1}{r^2}=\frac{\left(a^2+b^2+c^2\right)}{{∆}^2}$

Hence, option(A) is the correct answer.