JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Circle

A straight line through the vertex

P

of a triangle

P QR

intersects the side

QR

at the point

S

and the circumcircle of the triangle

P QR

at the point

T

. If

S

is not the centre of the circumcircle, then

Answer: (B), (D)

Solution

By the power of point

S

with respect to the circumcircle:

P S \times S T = QS \times S R

Applying AM-GM inequality:

\frac{1}{P S} + \frac{1}{S T} \geq 2 \frac{1}{P S \cdot S T} = \frac{2}{P S \cdot S T} = \frac{2}{QS \times S R}

Equality holds only when

P S = S T

, which doesn't occur since

S

is not the center. Therefore:

\frac{1}{P S} + \frac{1}{S T} > \frac{2}{QS \times S R}

Also, by AM-GM on

QS

and

S R

\frac{QS + S R}{2} > QS \times S R

\frac{QR}{2} > QS \times S R

\frac{2}{QS \times S R} > \frac{4}{QR}

Combining:

\frac{1}{P S} + \frac{1}{S T} > \frac{2}{QS \times S R} > \frac{4}{QR}

Therefore, options (B) and (D) are correct.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · One or more correct options