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
1PS+1ST<1QS×SR
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B
1PS+1ST>2QS×SR
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C
1PS+1ST<4QR
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D
1PS+1ST>4QR
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Solution

## The correct options are B 1PS+1ST>2√QS×SR D 1PS+1ST>4QR We know by geometry PS×ST=QS×SR ......(1) ∴ S is not the centre of circulm circle, PS≠ST And we know that for two unequal real numbers. H.M < G.M. ⇒21PS+1ST<√PS×ST⇒1PS+1ST>2√PS×ST ⇒1PS+1ST>2√QS×SR [using eqn (1)] ....(2) ∴ (b) is the correct option. Also √QS×SR<QS+SR2(∵GM<AM) ⇒1√QS×SR>2QR⇒2√QS×SR>4QR....(3) From equations (2) and (3) we get 1PS+1ST>4QR ∴ (d) is also the correct option.

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