JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 14)
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 :
$${1 \over {PS}} + {1 \over {ST}} < {2 \over {\sqrt {QS \times SR} }}$$
$${1 \over {PS}} + {1 \over {ST}} > {2 \over {\sqrt {QS \times SR} }}$$
$${1 \over {PS}} + {1 \over {ST}} < {4 \over {QR}}$$
$${1 \over {PS}} + {1 \over {ST}} > {4 \over {QR}}$$
Explanation
As S is not the centre of the circumcircle, there for PS $$\ne$$ ST and QS $$\ne$$ SR
Also, (PS)(ST) = (QS)(SR)
[$$\therefore$$ from the properties of two intersecting chords in a circle]
$$\Rightarrow$$ A.M $$\ge$$ G.M
We get $${{{1 \over {PS}} + {1 \over {ST}}} \over 2} \ge \sqrt {{1 \over {PS}} \times {1 \over {ST}}} $$
or
$${1 \over {PS}} + {1 \over {ST}} \ge {2 \over {\sqrt {QS \times SR} }}$$
$${{QS + SR} \over 2} \ge \sqrt {QS \times SR} $$ ..... (i)
$${1 \over {\sqrt {QS \times SR} }} \ge {2 \over {QR}}$$ .... (ii)
$$ \Rightarrow {1 \over {PS}} + {1 \over {ST}} \ge {4 \over {QR}}$$ From (i) and (ii)
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