Question

A straight line through the vertex P of a traingle 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. $\frac{1}{PS}+\frac{1}{ST}<\frac{1}{\sqrt{QS\times SR}}$
• 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)

$\frac{1}{PS}+\frac{1}{ST}>\frac{4}{QR}$


We know by geometry $PS\times ST=QS\times SR$ ......(1)
$\therefore$ S is not the centre of circulm circle,
$PS\ne ST$
And we know that for two unequal real numbers.
H.M < G.M.
$\Rightarrow \frac{2}{\frac{1}{PS}+\frac{1}{ST}}<\sqrt{PS\times ST}\Rightarrow \frac{1}{PS}+\frac{1}{ST}>\frac{2}{\sqrt{PS\times ST}}$
$\Rightarrow \frac{1}{PS}+\frac{1}{ST}>\frac{2}{\sqrt{QS\times SR}}$ [using eqn (1)] ....(2)
$\therefore$ (b) is the correct option.
Also $\sqrt{QS\times SR}<\frac{QS+SR}{2}(\because GM<AM)$
$\Rightarrow \frac{1}{\sqrt{QS\times SR}}>\frac{2}{QR}\Rightarrow \frac{2}{\sqrt{QS\times SR}}>\frac{4}{QR}....\left(3\right)$
From equations (2) and (3) we get $\frac{1}{PS}+\frac{1}{ST}>\frac{4}{QR}$
$\therefore$ (d) is also the correct option.