Question

A straight line through point \(P\) 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 which of the following is true?

• A. \(\dfrac{1}{PS} + \dfrac{1}{ST}<\dfrac{2}{\sqrt{(QS)(QR)}}\)
• B. \(\dfrac{1}{PS} + \dfrac{1}{ST}<\dfrac{4}{QR}\)
• C. \(\dfrac{1}{PS} + \dfrac{1}{ST}>\dfrac{1}{(QS)(QR)}\)
• D. \(\dfrac{1}{PS} + \dfrac{1}{ST}>\dfrac{4}{QR}\)
• E. None of the above

Hint:

When secant-tangent relations appear in geometry problems, always combine them with AM–GM or HM–GM inequalities to establish bounds on reciprocal sums.

Answer 1

The Correct Option is D

Step 1: Use Secant Property.
From point \(P\), the secant property gives: \[ PS \cdot ST = QS \cdot SR \quad \text{(1)} \]

Step 2: Harmonic Mean vs Geometric Mean.
For two positive numbers \(PS\) and \(ST\), we know: \[ \frac{2}{\frac{1}{PS} + \frac{1}{ST}}<\sqrt{PS \cdot ST} \] which implies \[ \frac{1}{PS} + \frac{1}{ST}>\frac{2}{\sqrt{PS \cdot ST}} \]

Step 3: Replace \(PS \cdot ST\).
Using equation (1): \[ PS \cdot ST = QS \cdot SR \] So, \[ \frac{1}{PS} + \frac{1}{ST}>\frac{2}{\sqrt{QS \cdot SR}} \quad \text{(2)} \]

Step 4: Apply AM ≥ GM for \(QS\) and \(SR\).
We know that: \[ \sqrt{QS \cdot SR} \leq \frac{QS + SR}{2} \] But \(QS + SR = QR\). So, \[ \sqrt{QS \cdot SR} \leq \frac{QR}{2} \]

Step 5: Use inequality.
From (2): \[ \frac{1}{PS} + \frac{1}{ST}>\frac{2}{\sqrt{QS \cdot SR}} \] and since \(\sqrt{QS \cdot SR} \leq \frac{QR}{2}\), \[ \frac{2}{\sqrt{QS \cdot SR}} \geq \frac{2}{QR/2} = \frac{4}{QR} \]

Step 6: Final conclusion.
Thus, \[ \frac{1}{PS} + \frac{1}{ST}>\frac{4}{QR} \] \[ \boxed{\dfrac{1}{PS} + \dfrac{1}{ST}>\dfrac{4}{QR}} \]