Question

If $r_1<r_2<r_3$ are the ex-radii of a right angled triangle and $r_1=1,r_2=2$, then $r_3=....$

Answer 1

$r_1=\frac{S}{s-a}=1,r_2=\frac{S}{s-b}=2,r_3=\frac{S}{s-c}=?$
$s-a=k.1,s-b=k.\frac{1}{2},s-c=k.\frac{1}{r_3}$
Adding pairwise and putting $2s=a+b+c$
$c=k(1+\frac{1}{2}),a=k(\frac{1}{2}+\frac{1}{r_3}),b=k(1+\frac{1}{r_3})$
Since triangle is right angled, $c^2=a^2+b^2$
$\therefore k^2.{\left(\frac{3}{2}\right)}^2=\frac{k^2{(r_3+2)}^2}{4{\left(r_3\right)}^2}+k^2\frac{{(r_3+1)}^2}{{\left(r_3\right)}^2}$
$\therefore 9r_3^2={(r_3+2)}^2+4{(r_3+1)}^2$
$4r_3^2-12r_3-8=0$
$\therefore r_3=\frac{3\pm \sqrt{17}}{2}=\frac{3+\sqrt{17}}{2}$ as $r_3$ is +ive