A straight line through the vertex P of a P Q R intersects the side Q R at the point S and the circumcircle of the P Q R at the point T . If S is not the centre of the circumcircle, then

Correct answer: B. (1)/(P S)+(1)/(S T)> 2 Q S · S R Let a straight line through the vertex P of a given P Q R intersects the side Q R at the point S and the circumcircle of P Q R at the point T . Points P, Q, R, T are concyclic, hence array rlrl P S · S T & =Q S · S R & & P_1 & ≥ 2 P S · S T & & Now, (P S+S T)/(2) & ≥ P S · S T & [ AM ≥ GM] array and (1)/(P S)+(1)/(S T) ≥ 2 P S · S T = 2 Q S · S R Also, (S Q+Q R)/(2) ≥ S Q · S R (Q R)/(2) ≥ S Q · S R array ll & 1 S Q · S R ≥ (2)/(Q R) 2 S Q · S R ≥ (4)/(Q R) Hence, & (1)/(P S)+(1)/(S T) ≥ 2 Q S · S R ≥ (4)/(Q R) . array

JEE Advanced 2008Paper 1

Exam: JEE Advanced
Year: 2008
Paper: Paper 1
Subject: Mathematics
Chapter: Straight Lines

JEE Advanced 2008 · Paper 1 · Mathematics · Straight Lines

A straight line through the vertex P of a P Q R intersects the side...

A straight line through the vertex

P

of a

\triangle P Q R

intersects the side

Q R

at the point

S

and the circumcircle of the

\triangle P Q R

at the point

T

. If

S

is not the centre of the circumcircle, then

Options

A $\frac{1}{P S}+\frac{1}{S T} < \frac{2}{\sqrt{Q S \cdot S R}}$
B $\frac{1}{P S}+\frac{1}{S T}>\frac{2}{\sqrt{Q S \cdot S R}}$
C $\frac{1}{P S}+\frac{1}{S T} < \frac{4}{Q R}$
D $\frac{1}{P S}+\frac{1}{S T}>\frac{4}{Q R}$

Correct answer

B. $\frac{1}{P S}+\frac{1}{S T}>\frac{2}{\sqrt{Q S \cdot S R}}$

Step-by-step solution

Let a straight line through the vertex

P

of a given

\triangle P Q R

intersects the side

Q R

at the point

S

and the circumcircle of

\triangle P Q R

at the point

T

. Points

P, Q, R, T

are concyclic, hence

\begin{array}{rlrl} P S \cdot S T & =Q S \cdot S R & & \\ \Rightarrow \quad P_1 & \geq 2 \sqrt{P S \cdot S T} & \\ & \text{ Now, } \frac{P S+S T}{2} & \geq \sqrt{P S \cdot S T} & {[\because \mathrm{AM} \geq \mathrm{GM}]} \end{array}

and

\quad \frac{1}{P S}+\frac{1}{S T} \geq \frac{2}{\sqrt{P S \cdot S T}}=\frac{2}{\sqrt{Q S \cdot S R}}

Also,

\quad \frac{S Q+Q R}{2} \geq \sqrt{S Q \cdot S R} \Rightarrow \quad \frac{Q R}{2} \geq \sqrt{S Q \cdot S R}

\begin{array}{ll} \Rightarrow & \frac{1}{\sqrt{S Q \cdot S R}} \geq \frac{2}{Q R} \Rightarrow \frac{2}{\sqrt{S Q \cdot S R}} \geq \frac{4}{Q R} \\ \text{ Hence, } & \frac{1}{P S}+\frac{1}{S T} \geq \frac{2}{\sqrt{Q S \cdot S R}} \geq \frac{4}{Q R} . \end{array}

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