Properties Of Triangles Ques 44
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
$(2008,4 M)$
(a) $\frac{1}{P S}+\frac{1}{S T}<\frac{2}{\sqrt{Q S \times S R}}$
(b) $\frac{1}{P S}+\frac{1}{S T}>\frac{2}{\sqrt{Q S \times S R}}$
(c) $\frac{1}{P S}+\frac{1}{S T}<\frac{4}{Q R}$
(d) $\frac{1}{P S}+\frac{1}{S T}>\frac{4}{Q R}$
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Answer:
Correct Answer: 44.(b, d)
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, then $P S \cdot S T=Q S \cdot S R$
Now, $\quad \frac{P S+S T}{2}>\sqrt{P S \cdot S T} \quad[\because AM>GM]$
and $\quad \frac{1}{P S}+\frac{1}{S T}>\frac{2}{\sqrt{P S \cdot S T}}=\frac{2}{\sqrt{Q S \cdot S R}}$
Also, $\quad \frac{S Q+Q R}{2}>\sqrt{S Q \cdot S R}$
$\Rightarrow \quad \frac{Q R}{2}>\sqrt{S Q \cdot S R}$
$\Rightarrow \quad \frac{1}{\sqrt{S Q \cdot S R}}>\frac{2}{Q R}$
$\Rightarrow \quad \frac{2}{\sqrt{S Q \cdot S R}}>\frac{4}{Q R}$
$\therefore \quad \frac{1}{P S}+\frac{1}{S T}>\frac{2}{\sqrt{Q S \cdot S R}}>\frac{4}{Q R}$
BETA
