Question

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

• A. $\frac{1}{PS}+\frac{1}{ST}<\frac{2}{\sqrt{QS\times SR}}$
• B. $\frac{1}{PS}+\frac{1}{ST}>\frac{2}{\sqrt{QS\times SR}}$
• C. $\frac{1}{PS}+\frac{1}{ST}<\frac{4}{QR}$
• D. $\frac{1}{PS}+\frac{1}{ST}>\frac{4}{QR}$

Answer 1

Answer: (B), (D)

$PS\times ST=QS\times SR$
$\frac{\frac{1}{PS}+\frac{1}{ST}}{2}>\sqrt{\frac{1}{PS}\times \frac{1}{ST}}$
$\Rightarrow \frac{1}{PS}+\frac{1}{ST}>\frac{2}{\sqrt{QS\times SR}}$
$\frac{QS+SR}{2}>\sqrt{QS\times SR}$
$\frac{QR}{2}>\sqrt{QS\times SR}$
$\Rightarrow \frac{1}{\sqrt{QS\times SR}}>\frac{2}{QR}$
$\Rightarrow \frac{1}{PS}+\frac{1}{ST}>\frac{4}{QR}$