Question

If $r_1,r_2,r_3$ are the radii of the escribed circles of a triangle $ABC$ and if $r$ is the radius of its incircle,then $r_1{r}_2{r}_3-r(r_1{r}_2+r_2{r}_3+r_3{r}_1)$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$r_1{r}_2{r}_3-r[r_1{r}_2+r_2{r}_3+r_3{r}_1]$
$=\frac{△}{s-a}\frac{△}{s-b}\frac{△}{s-c}-\frac{△}{s}[\frac{{△}^2}{(s-a)(s-b)}+\frac{{△}^2}{(s-b)(s-b)}+\frac{{△}^2}{(s-c)(s-a)}]$
$=\frac{{△}^3}{(s-a)(s-b)(s-c)}-\frac{△}{s}\frac{{△}^2[s-c+s-a+s-b]}{(s-a)(s-b)(s-c)}$
$=\frac{{△}^3}{(s-a)(s-b)(s-c)}-\frac{{△}^3}{s}\frac{3s-(a+b+c)}{\frac{{△}^2}{s}}$
$=\frac{{△}^3}{(s-a)(s-b)(s-c)}-\frac{{△}^3}{s}\frac{(3s-2s)}{\frac{{△}^2}{s}}$
$\frac{{△}^3}{\frac{{△}^2}{s}}-\frac{{△}^3}{s}\frac{(3s-2s)}{\frac{{△}^2}{s}}$
On simplifying,we get
$△s-△s=0$