Question

$A$ circle $C$ of radius $1$ is inscribed in an equilateral triangle $PQR$. The points of contact of
$C$ with the sides $PQ,QR,RP$ are $D,E,F$ respectively. The line $PQ$ is given by the equation
$\sqrt{3}x+y-6=0$ and the point $D$ is $(\frac{\sqrt{3}}{2},\frac{3}{2})$. Further it is given that the origin
and the centre of $C$ are on the same side of $PQ.$ Points $E$ and $F$ are given by

• A. $(\frac{\sqrt{3}}{2},\frac{3}{2}),(\sqrt{3},0)$
• B. $(\frac{\sqrt{3}}{2},\frac{1}{2}),(\sqrt{3},0)$
• C. $(\frac{\sqrt{3}}{2},\frac{3}{2}),(\frac{\sqrt{3}}{2},\frac{1}{2})$
• D. $(\frac{3}{2},\frac{\sqrt{3}}{2}),(\frac{\sqrt{3}}{2},\frac{1}{2})$

Answer 1

Answer: (A)

$(\frac{\sqrt{3}}{2},\frac{3}{2}),(\sqrt{3},0)$

Given $PQ:\sqrt{3}x+y–6=0$

$D=(\frac{\sqrt{3}}{2},\frac{3}{2})$

$r=1$

Let $C(x_1,y_1)$ be the incenter.

$⟹CD=1$ = perpendicular ditstance from $C$ to $PQ$

$⟹\sqrt{(x_1-\frac{\sqrt{3}}{2}{)}^2+(y_1-\frac{3}{2}{)}^2}=1–\left(1\right)$

$\frac{|\sqrt{3}{x}_1+y_1-6|}{\sqrt{3+1}}=1$

$⟹|\sqrt{3}{x}_1+y_1-6|=2\left(2\right)$

Solving (1) and (2)

We get $x=\sqrt{3},y=1$

$C=(\sqrt{3},1)$

Given $PQR$ is equilateral

Let $QR$ be $y=m_{1}x+c_1$

$RP$ be $y_2=m_{2}x+c_2$

All the sides are at angle ${60}^{\circ }$ to each other.

$PQ=\sqrt{3}x+y–6,m=-\sqrt{3}$

$\tan 60=\frac{|m–m^{\prime }|}{1+m{m}^{\prime }}$

$⟹\sqrt{3}=\frac{|m^{\prime }+\sqrt{3}|}{1-\sqrt{3}{m}^{\prime }}$

$⟹(\sqrt{3}–3m^{\prime }{)}^2=(m^{\prime }+\sqrt{3}{)}^2$

$⟹8m^{\prime 2}=8\sqrt{3}{m}^{\prime }$

$m^{\prime }=0\sqrt{3}$

Let $m_2=0,m_2=\sqrt{3}$

Distance from $C$ to $QR$ and $RP=r=1$

From $PR$

$\frac{|1–0-c_1|}{\sqrt{1^2+0^2}}=1$

$⟹c_1=0,PR$ is $y=0$

From $QR$

$\frac{|1–3-c_2|}{\sqrt{1^2+3^2}}=1$

$⟹c_2=0,QR$ is $y=\sqrt{3}x$