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 [2008]

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 [2008]

Smart Achievers · Accepted Answer

(2, 4)

$We know by geometry P S \times S T = Q S \times S R ...(i)$

$∵$ $S is not the centre of circumcircle, P S \neq S T$

And we know that for two unequal real numbers, H.M. < G.M.

$\Rightarrow \frac{2}{\frac{1}{P S} + \frac{1}{S T}} < \sqrt{P S \times S T} \Rightarrow \frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{P S \times S T}}$

$\Rightarrow \frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{Q S \times S R}} [using eqn (i)] ...(ii)$

$∴$ $(2) is the correct option.$

$Also \sqrt{Q S \times S R} < \frac{Q S + S R}{2} (∵ GM < AM)$

$\Rightarrow \frac{1}{\sqrt{Q S \times S R}} > \frac{2}{Q R} \Rightarrow \frac{2}{\sqrt{Q S \times S R}} > \frac{4}{Q R} ...(iii)$

$From equations (ii) and (iii), \frac{1}{P S} + \frac{1}{S T} > \frac{4}{Q R}$

$∴$ $(4) is also the correct option.$

Solution

Q.
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 [2008]

1 $\frac{1}{P S} + \frac{1}{S T} < \frac{2}{\sqrt{Q S \times S R}}$

2 $\frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{Q S \times S R}}$

3 $\frac{1}{P S} + \frac{1}{S T} < \frac{4}{Q R}$

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

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Solution

Q. 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 [2008]

1 1PS+1ST<2QS×SR

2 1PS+1ST>2QS×SR

3 1PS+1ST<4QR