A straight line through the vertex 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
(i) `1/(PS)+1/(ST)lt2/(sqrt(QSxxSR))`
(ii) `1/(PS)+1/(ST)gt2/(sqrt(QSxxSR))`
(iii)`1/(PS)+1/(ST)lt4/(QS)`
`1/(PS)+1/(ST)gt4/(QR)`

Options

A. (i) and (iii)
B. (ii) and (iv)
C. (i) and (iv)
D. (ii) and (iii)

Answer

To solve the problem, we need to analyze the given conditions and use properties of triangles and circles. Let's break down the solution step by step.

### Step 1: Understand the Geometry
We have a triangle PQR with a line through vertex P intersecting side QR at point S and the circumcircle of triangle PQR at point T. We know that S is not the center of the circumcircle.

### Step 2: Use the Power of a Point Theorem
According to the Power of a Point theorem, we have:
\[ PS \cdot ST = QS \cdot SR \]
This means that the product of the segments from point S to points P and T is equal to the product of the segments from point Q to S and from S to R.

### Step 3: Apply the Arithmetic Mean-Geometric Mean Inequality
We know that for any two positive numbers \( a \) and \( b \):
\[ \frac{1}{a} + \frac{1}{b} \geq \frac{4}{a + b} \]
This can be applied to \( PS \) and \( ST \):
\[ \frac{1}{PS} + \frac{1}{ST} \geq \frac{4}{PS + ST} \]

### Step 4: Relate PS, ST, QS, and SR
Using the Power of a Point theorem, we can express \( PS \) and \( ST \) in terms of \( QS \) and \( SR \):
\[ PS + ST = QS + SR \]

### Step 5: Establish the Inequalities
From the previous steps, we can derive the inequalities:
1. Since \( PS \cdot ST = QS \cdot SR \), we can write:
\[ \frac{1}{PS} + \frac{1}{ST} < \frac{2}{\sqrt{QS \cdot SR}} \]
This corresponds to option (i).

2. Similarly, we can also show that:
\[ \frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{QS \cdot SR}} \]
This corresponds to option (ii).

3. We can also derive:
\[ \frac{1}{PS} + \frac{1}{ST} < \frac{4}{QS} \]
This corresponds to option (iii).

4. Finally, we can show:
\[ \frac{1}{PS} + \frac{1}{ST} > \frac{4}{QR} \]
This corresponds to option (iv).

### Conclusion
From the analysis:
- Options (i) and (iii) are correct.
- Options (ii) and (iv) are incorrect.

### Final Answer
The correct options are (i) and (iii).