A point P is given on the circumference of a circle of radius r. Chord QR is parallel to the tangent at P. Determine the maximum possible area of the triangle PQR.

Answer

To solve the problem of finding the maximum possible area of triangle PQR, where P is a point on the circumference of a circle of radius r, and chord QR is parallel to the tangent at P, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let the center of the circle be O and the radius be r.
- Point P is on the circumference, and we draw a tangent at P.
- Chord QR is parallel to this tangent.

2. **Identifying Key Points**:
- Let the angle ∠QOP be θ. Since QR is parallel to the tangent at P, the angle ∠QOP is equal to the angle ∠PQR (alternate interior angles).

3. **Finding the Length of Chord QR**:
- The length of chord QR can be expressed using the sine function:
\[ QR = 2R \sin\left(\frac{\theta}{2}\right) \]
- Here, R is the radius of the circle.

4. **Finding the Height of Triangle PQR**:
- The height from point P to line QR can be calculated as:
\[ h = OP \cos\left(\frac{\theta}{2}\right) = r \cos\left(\frac{\theta}{2}\right) \]

5. **Calculating the Area of Triangle PQR**:
- The area A of triangle PQR is given by:
\[ A = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times QR \times h \]
- Substituting the expressions for QR and h:
\[ A = \frac{1}{2} \times (2r \sin\left(\frac{\theta}{2}\right)) \times (r \cos\left(\frac{\theta}{2}\right)) \]
- Simplifying this, we get:
\[ A = r^2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right) \]

6. **Using the Double Angle Identity**:
- We can use the double angle identity:
\[ \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right) = \frac{1}{2} \sin(\theta) \]
- Therefore, the area becomes:
\[ A = \frac{1}{2} r^2 \sin(\theta) \]

7. **Maximizing the Area**:
- The maximum value of \(\sin(\theta)\) is 1, which occurs at \(\theta = \frac{\pi}{2}\).
- Thus, the maximum area is:
\[ A_{\text{max}} = \frac{1}{2} r^2 \cdot 1 = \frac{1}{2} r^2 \]

### Final Result:
The maximum possible area of triangle PQR is:
\[ \boxed{\frac{1}{2} r^2} \]