Let ABC be a triangle with `angleBAC = 2pi//3 and AB = x` such that (AB) (AC) = 1. If x varies, then find the longest possible length of the angle bisector AD

Explanation

To solve the problem step by step, we will analyze the triangle ABC with the given conditions and derive the length of the angle bisector AD.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Angle \( \angle BAC = \frac{2\pi}{3} \)
- Length \( AB = x \)
- The product \( AB \cdot AC = 1 \)

2. **Express \( AC \) in terms of \( x \):**
- Since \( AB \cdot AC = 1 \), we can write:
\[ AC = \frac{1}{AB} = \frac{1}{x} \]

3. **Use the Angle Bisector Theorem:**
- The formula for the length of the angle bisector \( AD \) is given by:
\[ AD = \frac{2 \cdot AB \cdot AC}{AB + AC} \cdot \cos\left(\frac{A}{2}\right) \]
- Here, \( AB = x \) and \( AC = \frac{1}{x} \).

4. **Calculate \( AD \):**
- Substitute \( AB \) and \( AC \) into the formula:
\[ AD = \frac{2 \cdot x \cdot \frac{1}{x}}{x + \frac{1}{x}} \cdot \cos\left(\frac{2\pi}{3} \cdot \frac{1}{2}\right) \]
- Simplifying the expression:
\[ AD = \frac{2}{x + \frac{1}{x}} \cdot \cos\left(\frac{\pi}{3}\right) \]
- Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ AD = \frac{2}{x + \frac{1}{x}} \cdot \frac{1}{2} \]
- Thus:
\[ AD = \frac{1}{x + \frac{1}{x}} \]

5. **Minimize the Denominator:**
- To maximize \( AD \), we need to minimize \( x + \frac{1}{x} \).
- The expression \( x + \frac{1}{x} \) has a minimum value of 2 when \( x = 1 \) (by AM-GM inequality).

6. **Calculate the Maximum Length of \( AD \):**
- Therefore, substituting \( x = 1 \):
\[ AD_{\text{max}} = \frac{1}{2} \]

### Final Answer:
The longest possible length of the angle bisector \( AD \) is \( \frac{1}{2} \).