Question

In $△ABC,$ the incircle touches the sides $BC,CA$ and $AB$, respectively, at $D,E$ and $F$.

If the radius of the incircle is $4$ units and $BD,CE$ and $AF$ are consecutive integers,

then the value of $\frac{s}{3}$, where $s$ is a semi-perimeter of the triangle, is?

Answer 1

Given in radius $r=4$

$BD,CE,AF$ are in A.P of

common difference $1$

$s=\frac{△}{r}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{r}$

$=2y=x+z$

$z-y=y-x=1$

$a=x+z,b=y+z,c=x+y$

$S=\left(\frac{a+b+c}{2}\right)=\left(3y\right)$

$△=\sqrt{\left(3y\right)[3y-(x+z\left)\right][3y-(y+z\left)\right][3y-(x+4\left)\right]}$

$=\sqrt{\left(3y\right)\left(y\right)+(2y-z)(2y-x)}$

$=\sqrt{\left(3y\right)y(y-1)(y+1)}$

$=\frac{\sqrt{3y^2(y^2-1)}}{r}$

$s=\frac{△}{r}=s^2=\frac{{△}^2}{r^2}$

$9y^2=\frac{3y^2(y^2-1)}{16}\Rightarrow 3\times 16=(y^2-1)$

$y^2=49\Rightarrow y=7$

$\frac{s}{3}=\frac{3y}{3}=7$