A straight line through the vertex $P$ of a $\triangle P Q R$ intersects the side $Q R$ at the point $S$ and the circumcircle of the $\triangle P Q R$ at the point $T$. If $S$ is not the centre of the circumcircle, then

• A

  $\frac{1}{P S}+\frac{1}{S T} < \frac{2}{\sqrt{Q S \cdot S R}}$
• Correct answer
  B

  $\frac{1}{P S}+\frac{1}{S T}>\frac{2}{\sqrt{Q S \cdot S R}}$
• C

  $\frac{1}{P S}+\frac{1}{S T} < \frac{4}{Q R}$
• Correct answer
  D

  $\frac{1}{P S}+\frac{1}{S T}>\frac{4}{Q R}$

Solution

Let a straight line through the vertex $P$ of a given $\triangle P Q R$ intersects the side $Q R$ at the point $S$ and the circumcircle of $\triangle P Q R$ at the point $T$. Points $P, Q, R, T$ are concyclic, hence $$ \begin{array}{rlrl} P S \cdot S T & =Q S \cdot S R & & \\ \Rightarrow \quad P_1 & \geq 2 \sqrt{P S \cdot S T} & \\ & \text { Now, } \frac{P S+S T}{2} & \geq \sqrt{P S \cdot S T} & {[\because \mathrm{AM} \geq \mathrm{GM}]} \end{array} $$
and $\quad \frac{1}{P S}+\frac{1}{S T} \geq \frac{2}{\sqrt{P S \cdot S T}}=\frac{2}{\sqrt{Q S \cdot S R}}$ Also, $\quad \frac{S Q+Q R}{2} \geq \sqrt{S Q \cdot S R}$ $\Rightarrow \quad \frac{Q R}{2} \geq \sqrt{S Q \cdot S R}$ $$ \begin{array}{ll} \Rightarrow & \frac{1}{\sqrt{S Q \cdot S R}} \geq \frac{2}{Q R} \Rightarrow \frac{2}{\sqrt{S Q \cdot S R}} \geq \frac{4}{Q R} \\ \text { Hence, } & \frac{1}{P S}+\frac{1}{S T} \geq \frac{2}{\sqrt{Q S \cdot S R}} \geq \frac{4}{Q R} . \end{array} $$

Hence, the correct answer is options: 'B', 'D'.