Question

The incircle of $△ABC$ having integral side lengths and least perimeter, touches $AB,BC,CA$ at $R,P,Q$ respectively. If $\frac{2}{AR}+\frac{5}{BP}+\frac{5}{CQ}=\frac{6}{r}$, where $r$ is the radius of the incircle, then

• A. $\Delta ABC\text{is isosceles}$
• B. $\Delta ABC\text{is equilateral}$
• C. $\text{Area of}△ABC\text{is 27 sq. units}$
• D. $\text{Inradius of}△ABC\text{is 4 units}$

Answer 1

Answer: (A), (D)

The correct options are
A $\Delta ABC\text{is isosceles}$
D $\text{Inradius of}△ABC\text{is 4 units}$

Let $\tan \frac{A}{2}=x,\tan \frac{B}{2}=y,\tan \frac{C}{2}=z$


$x=\frac{r}{AR},y=\frac{r}{BP},z=\frac{r}{CQ}$
Putting the value of $AR,BP$ and $CQ$ in given relation, we get
$\frac{2x}{r}+\frac{5y}{r}+\frac{5z}{r}=\frac{6}{r}\Rightarrow 2x+5y+5z=6\cdots \left(1\right)$

Using $\sum \tan \frac{A}{2}\tan \frac{B}{2}=1,$ we get
$xy+yz+zx=1\cdots \left(2\right)$
If we interchange $y$ and $z$ in both equations $\left(1\right)$ and $\left(2\right)$, then they remain unchanged.
$\Rightarrow △ABC$ is isosceles with $\angle B=\angle C\Rightarrow y=z$

Now, both the equation reduce to
$x=3–5y2xy+y^2=1$
Solving both, we get
$9y^2-6y+1=0\Rightarrow y=z=\frac{1}{3}\Rightarrow x=\frac{4}{3}$
Therefore,
$AR=\frac{3r}{4},BP=3r,CQ=3r$
Now, $AB=AR+BR$
$\Rightarrow AB=\frac{3r}{4}+3r$
$\Rightarrow AB=\frac{15r}{4}=AC,BC=6r$
Now, the perimeter of the triangle
$=2(AR+BP+CQ)=\frac{27r}{2}$
For perimeter to be smallest integer and sides also integer, $r$ must be $4.$
$\therefore △=rs=4\times \frac{27r}{4}=108\text{sq. unit}$