Subject: Mathematics — Circles
Question: In a non-right angled triangle $\bigtriangleup PQR$,
let $p,q,r$ denote the lengths of the sides opposite to the angles at
$P,Q,R$ respectively. The median from $R$ meets the side $PQ$ at $S$,
the perpendicular from $P$ meets the side $QR$ at $E$, $RS$ and $PE$
intersect at $O$. If $p = \sqrt{3}$, $q = 1$ and the radius of the
circumcircle of $\bigtriangleup PQR$ equals $1$, then which of the
following options is/are correct?
\(a\) Find the Radius of incircle of $\bigtriangleup PQR$
\(b\) Length of $RS$
Answer: (a) $\frac{\sqrt{3}}{2}\left( 2 - \sqrt{3} \right)$
\(b\) $\frac{\sqrt{7}}{2}$
Solution: Given: $RS$ is median of $\bigtriangleup PQR$ and
$PE\bot QR$,
$$p = \sqrt{3}\quad\text{and}\quad q = 1.$$
Radius $(R)$ of circumcircle of $\bigtriangleup PQR = 1$.
[IMAGE] Figure 1: Diagram for Question 39.
Using sine law in $\bigtriangleup PQR$,
$$\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = 2 \times \text{radius of circumcircle}.$$
$$\Rightarrow \frac{\sqrt{3}}{\sin P} = \frac{1}{\sin Q} = \frac{r}{\sin R} = 2 \times 1 = 2.$$
$$\Rightarrow sinP = \frac{\sqrt{3}}{2}\quad\text{and}\quad\sin Q = \frac{1}{2}.$$
$$\Rightarrow \angle P = 60^{\circ}\ \text{or}\ 120^{\circ}\quad\text{and}\quad\angle Q = 30^{\circ}\ \text{or}\ 150^{\circ}.$$
Since any triangle cannot have two obtuse angles, therefore
$\angle P = 120^{\circ}$ and $\angle Q = 150^{\circ}$ is rejected.
Since sum of three angles of a triangle can not be greater than
$180^{\circ}$, therefore $\angle P = 60^{\circ}$ and
$\angle Q = 150^{\circ}$ is rejected.
Since $\bigtriangleup PQR$ is non right triangle, therefore
$\angle P = 60^{\circ}$ and $\angle Q = 30^{\circ}$ is rejected.
Hence, $\angle P = 120^{\circ}$ and $\angle Q = 30^{\circ}$ is the
only option.
$$\Rightarrow \angle QPE = \angle RPE = 60^{\circ}\quad\text{and}\quad\angle PRQ = 30^{\circ}.$$
$$\Rightarrow \frac{r}{\sin 30^{\circ}} = 2\quad \Rightarrow \quad r = 1.$$
Area of $\bigtriangleup PQR$
$$= \frac{pqr}{4\left( \text{radius of circumcircle} \right)}\quad\quad\left\lbrack \because\ \Delta = \frac{abc}{4R} \right\rbrack$$
$$= \frac{\sqrt{3} \times 1 \times 1}{4 \times 1} = \frac{\sqrt{3}}{4}.$$
$$\therefore\text{In } \bigtriangleup PQR,\ \text{radius of incircle} = \frac{area( \bigtriangleup PQR)}{\text{Semi perimeter}}\quad\quad\left\lbrack \because\ r = \frac{\Delta}{s} \right\rbrack$$
$$= \frac{\frac{\sqrt{3}}{4}}{\frac{\sqrt{3} + 1 + 1}{2}} = \frac{\sqrt{3}}{2}\left( 2 - \sqrt{3} \right),$$
Since, $RS$ and $PE$ are medians of $\bigtriangleup PQR$ intersecting
at $O$, therefore $O$ is centroid of $\bigtriangleup PQR$.
$$\therefore\ area( \bigtriangleup OQR) = \frac{1}{3}area( \bigtriangleup PQR) = \frac{1}{3} \times \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{12}.$$
$$\Rightarrow \frac{1}{2} \times \sqrt{3} \times OE = \frac{\sqrt{3}}{12} \Rightarrow OE = \frac{1}{6}\text{.}$$
Now,
$$area( \bigtriangleup SOE) = \frac{1}{2}OE \times SF = \frac{1}{2} \times \frac{1}{6} \times \frac{r}{2}\sin 60^{\circ}$$
$$= \frac{1}{12} \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{48}\quad\quad\lbrack\because\ r = 1\rbrack,$$
Also, in $\bigtriangleup RQS$, by cosine law,
$$\cos 30^{\circ} = \frac{QS^{2} + QR^{2} - RS^{2}}{2QS \times QR}.$$
$$\Rightarrow \frac{\sqrt{3}}{2} = \frac{\left( \frac{1}{2} \right)^{2} + \left( \sqrt{3} \right)^{2} - RS^{2}}{2 \times \frac{1}{2} \times \sqrt{3}} \Rightarrow \frac{3}{2} = \frac{13}{4} - RS^{2}.$$
$$\Rightarrow RS = \frac{\sqrt{7}}{2}\text{.}$$