Question

$ABC$ is a triangle. $3$ Circles with radii $1,4$ and $9$ as shown are drawn inside the triangle each touching two sides and the incircle. Then the radius of the incircle of the $△ABC$ is

Answer 1

Let the radius of the incircle of the $△ABC$ is $r.$
$\therefore XF=\sqrt{(r+1{)}^2-(r-1{)}^2}=\sqrt{4r}=2\sqrt{r}$
Similarly,
$YF=\sqrt{(r+4{)}^2-(r-4{)}^2}=\sqrt{16r}=4\sqrt{r}$
$PQ=\sqrt{(r+9{)}^2-(r-9{)}^2}=6\sqrt{r}$
$\tan \frac{A}{2}=\frac{r-1}{2\sqrt{r}};\tan \frac{B}{2}=\frac{r-4}{4\sqrt{r}};\tan \frac{C}{2}=\frac{r-9}{6\sqrt{r}}$

Using $\sum \tan \frac{A}{2}\tan \frac{B}{2}=1$
$\Rightarrow \frac{r-1}{2\sqrt{r}}\cdot \frac{r-4}{4\sqrt{r}}+\frac{r-4}{4\sqrt{r}}\cdot \frac{r-9}{6\sqrt{r}}+\frac{r-9}{6\sqrt{r}}\cdot \frac{r-1}{2\sqrt{r}}=1$
Multiplying it throughout by 24r,
$\Rightarrow 3(r-1)(r-4)+(r-4)(r-9)+2(r-9)(r-1)=24r$
$\Rightarrow 3(r^2-5r+4)+(r^2-13r+36)+2(r^2-10r+9)=24r$
$\Rightarrow 6r^2-48r+66=24r$
$\Rightarrow 6r^2-72r+66=0$
$\Rightarrow r^2-12r+11=0$
$\Rightarrow (r-11)(r-1)=0$
$\therefore r=11(\text{as}r\ne 1)$