Question

In any $\Delta ABC,\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}$ is equal to

• A. $\frac{a^2+b^2+c^2}{2{\Delta }^2}$
• B. $\frac{a^2+b^2+c^2}{{\Delta }^2}$
• C. $\frac{a^2+b^2+c^2}{3{\Delta }^2}$
• D. None of these

Answer 1

The correct option is B $\frac{a^2+b^2+c^2}{{\Delta }^2}$

For the exterior circle of the triangle,

$\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{r}$

$r=\frac{\Delta }{s}$

$r_1=\frac{\Delta }{s-a}$

$r_2=\frac{\Delta }{s-b}$

$r_3=\frac{\Delta }{s-c}$

So according to this ,

$\frac{(s-a)+(s-b)+(s-c)}{\Delta }=\frac{s}{\Delta }$

$3s-(a+b+c)=s$

$s=\frac{a+b+c}{2}$

According to the question,

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

$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}=\frac{4s^2+(a^2+b^2+c^2)-2s(a+b+c)}{{\Delta }^2}$

$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}=\frac{4s^2+(a^2+b^2+c^2)-2s\times 2s}{{\Delta }^2}$

$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}=\frac{4s^2+(a^2+b^2+c^2)-4s^2}{{\Delta }^2}$

$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}=\frac{(a^2+b^2+c^2)}{{\Delta }^2}$