In `DeltaAEX, T` is the midpoint of XE and P is the midpoint of ET. If `DeltaAPE` is equilateral of side length equal to unity, then the vaue of `(AX)^(2)` is _____

Explanation

To solve the problem, we need to find the value of \( AX^2 \) in triangle \( AEX \) where \( P \) and \( T \) are midpoints of segments \( XE \) and \( ET \) respectively, and triangle \( APE \) is equilateral with a side length of 1.

### Step-by-Step Solution:

1. **Identify the points and their relationships:**
- Let \( A \), \( E \), and \( X \) be the vertices of triangle \( AEX \).
- \( P \) is the midpoint of \( XE \).
- \( T \) is the midpoint of \( ET \).
- Triangle \( APE \) is equilateral with side length 1, thus \( AP = AE = PE = 1 \).

2. **Determine the coordinates of the points:**
- Place point \( A \) at the origin: \( A(0, 0) \).
- Since triangle \( APE \) is equilateral, we can place point \( E \) at \( (1, 0) \) and point \( P \) at \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \) (using the property of equilateral triangles).

3. **Find the coordinates of point \( T \):**
- Since \( T \) is the midpoint of \( EP \):
\[ T = \left( \frac{1 + \frac{1}{2}}{2}, \frac{0 + \frac{\sqrt{3}}{2}}{2} \right) = \left( \frac{3}{4}, \frac{\sqrt{3}}{4} \right) \]

4. **Determine the coordinates of point \( X \):**
- Since \( P \) is the midpoint of \( XE \), we can express \( X \) in terms of its coordinates:
\[ P = \left( \frac{x + 1}{2}, \frac{y + 0}{2} \right) = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \]
- From this, we can set up equations:
\[ \frac{x + 1}{2} = \frac{1}{2} \implies x + 1 = 1 \implies x = 0 \]
\[ \frac{y}{2} = \frac{\sqrt{3}}{2} \implies y = \sqrt{3} \]
- Thus, \( X(0, \sqrt{3}) \).

5. **Calculate \( AX^2 \):**
- The distance \( AX \) can be calculated as:
\[ AX = \sqrt{(0 - 0)^2 + (0 - \sqrt{3})^2} = \sqrt{3} \]
- Therefore, \( AX^2 = (\sqrt{3})^2 = 3 \).

6. **Final Calculation:**
- The value of \( AX^2 \) is \( 3 \).

### Final Answer:
The value of \( AX^2 \) is \( 3 \).