Question

In $\Delta ABC,x,y$ and $z$ are the distances of incentre from angular points $A,B,$ and $C$ respectively. If $\frac{xyz}{abc}=\frac{\lambda r}{s}$, then $\lambda =$

• A. $1$
• B. $2$
• C. $3$
• D. none of these

Answer 1

Answer: (A)

$1$

From the figure:
In $\Delta AIC$
$\angle AIC=\pi -\left(\frac{A+C}{2}\right)$
From sine rule:
$\frac{AI}{\sin \frac{C}{2}}=\frac{b}{\sin \angle AIC}$
$\Rightarrow AI=\frac{b\sin \frac{C}{2}}{\sin (\pi -\frac{A+C}{2})}$
$\Rightarrow AI=\frac{2R\sin B\sin \frac{C}{2}}{\sin \frac{B}{2}}=\frac{2R\left(2\sin \frac{B}{2}\cos \frac{B}{2}\right)\sin \frac{C}{2}}{\sin \frac{B}{2}}$ [ using sine rule in $\Delta ABC$ ]
$\Rightarrow x=AI=4R\sin \frac{B}{2}\sin \frac{C}{2}$
Similarly for $\Delta AIB$ and $\Delta BIC$, we get
$\Rightarrow y=BI=4R\sin \frac{A}{2}\sin \frac{C}{2}$
and $z=CI=4R\sin \frac{B}{2}\sin \frac{A}{2}$
$\frac{xyz}{abc}=\frac{\lambda r}{s}$
$\Rightarrow \frac{64R^3\left(\sin \frac{B}{2}\sin \frac{C}{2}\right)\left(\sin \frac{A}{2}\sin \frac{C}{2}\right)\left(\sin \frac{B}{2}\sin \frac{A}{2}\right)}{{8R}^3\sin A\sin B\sin C}=\frac{\lambda r}{s}$
$\Rightarrow \frac{8{\left(\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\right)}^2}{\left(2\sin \frac{A}{2}\cos \frac{A}{2}\right)\left(2\sin \frac{B}{2}\cos \frac{B}{2}\right)\left(2\sin \frac{C}{2}\cos \frac{C}{2}\right)}=\frac{\lambda r}{s}$
$\Rightarrow \tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}=\frac{\lambda r}{s}$
$\Rightarrow \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=\frac{\lambda r}{s}$
$\Rightarrow \sqrt{\frac{s(s-a)(s-b)(s-c)}{s^4}}=\frac{\lambda △}{s^2}$
$\Rightarrow \lambda =1$
Ans: A