Question

Prove that : $\frac{1}{r^2}+\frac{1}{{r_1}^2}+\frac{1}{{r_2}^2}+\frac{1}{{r_3}^2}=\frac{a^2+b^2+c^2}{S^2}$

Where in $△ABC,$ $r$ and $R$ are inradius and circumradius and $r_1,r_2,r_3$ are exradius respectively.

Also, $a,b,c$ are the corresponding sides and $S$ is the semiperimeter.

Answer 1

To prove: $\frac{1}{r^2}+\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}=\frac{a^2+b^2+c^2}{s^2}$

Proof: Formula to be used:

$r=\frac{\Delta }{s};r_1=\frac{\Delta }{s-a};r_2=\frac{\Delta }{s-b};r_3=\frac{\Delta }{s-c}$

$\Delta =\sqrt{s(s-a)(s-b)(s-c)};2s=a+b+c$

Now;

$LHS=\frac{1}{r^2}+\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}$

$=\frac{s^2}{{\Delta }^2}+\frac{{(s-a)}^2}{{\Delta }^2}+\frac{{(s-b)}^2}{{\Delta }^2}+\frac{{(s-c)}^2}{{\Delta }^2}$

$=\frac{1}{{\Delta }^2}[s^2+s^2+a^2-2as+s^2+b^2-2bs+s^2+c^2-2sc]$

$=\frac{1}{{\Delta }^2}[{4s}^2-2s(a+b+c)+a^2+b^2+c^2]$

$=\frac{a^2+b^2+c^2}{{\Delta }^2}$

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