IIT-JEE 2008 · Paper 1 · Mathematics Q7

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MathematicsMultiple correct+4 / -1Question 7 of 69

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

\dfrac1{PS}+\dfrac1{ST}<\dfrac2{\sqrt{QS\times SR}}

\dfrac1{PS}+\dfrac1{ST}>\dfrac2{\sqrt{QS\times SR}}

\dfrac1{PS}+\dfrac1{ST}<\dfrac4{QR}

\dfrac1{PS}+\dfrac1{ST}>\dfrac4{QR}

Answer and worked solution

Official answer: B, D

Intersecting chords. $PS\cdot ST=QS\cdot SR$ .

AM-HM. For positive $PS\ne ST$ or $QS\ne SR$ :
$\frac1{PS}+\frac1{ST}\ge\frac2{\sqrt{PS\cdot ST}}=\frac2{\sqrt{QS\cdot SR}}\ge\frac2{(QS+SR)/2}=\frac4{QR}.$
Equality throughout would need $PS=ST$ and $QS=SR$ : two chords bisecting each other, which happens only at the centre. Since S is not the centre, the inequalities are strict.

(B) True, and (A) false.
(D) True, and (C) false.

(B), (D)

Topics:Equation of a circle AM-GM-HM inequalities

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IIT-JEE 2008 · Paper 1, Mathematics question 7