Circle $\omega$ is inscribed in unit square $PLUM$, and points $I$ and $E$ lie on $\omega$ such that $U,I,$ and $E$ are collinear. Find, with proof, the greatest possible area for $\triangle PIE$.
1. **Set up the diagram in the Cartesian coordinate plane:**
   - Place the unit square \(PLUM\) with \(U = (0, 0)\) and \(P = (2, 2)\). This simplifies calculations by scaling the square to a \(2 \times 2\) square.
   - The circle \(\omega\) is inscribed in the square, centered at \((1, 1)\) with radius 1.

2. **Equation of the line \(UE\):**
   - Let the slope of \(\overline{UE}\) be \(k\). The equation of the line \(UE\) is \(y = kx\).

3. **Equation of the circle \(\omega\):**
   - The equation of the circle is \((x - 1)^2 + (y - 1)^2 = 1\).

4. **Find coordinates of \(I\) and \(E\):**
   - Substitute \(y = kx\) into the circle's equation:
     \[
     (x - 1)^2 + (kx - 1)^2 = 1
     \]
     \[
     x^2 - 2x + 1 + k^2x^2 - 2kx + 1 = 1
     \]
     \[
     (k^2 + 1)x^2 - 2(k + 1)x + 1 = 0
     \]
   - Solve the quadratic equation using the quadratic formula:
     \[
     x = \frac{2(k + 1) \pm \sqrt{4(k + 1)^2 - 4(k^2 + 1)}}{2(k^2 + 1)}
     \]
     \[
     x = \frac{2(k + 1) \pm \sqrt{4k + 4}}{2(k^2 + 1)}
     \]
     \[
     x = \frac{2(k + 1) \pm 2\sqrt{k + 1}}{2(k^2 + 1)}
     \]
     \[
     x = \frac{k + 1 \pm \sqrt{k + 1}}{k^2 + 1}
     \]
   - Let \(r_1 = \frac{k + 1 - \sqrt{k + 1}}{k^2 + 1}\) and \(r_2 = \frac{k + 1 + \sqrt{k + 1}}{k^2 + 1}\).

5. **Coordinates of \(E\) and \(I\):**
   - \(E = (r_1, kr_1)\) and \(I = (r_2, kr_2)\).

6. **Use the Shoelace formula to find the area of \(\triangle PIE\):**
   - Coordinates of \(P = (2, 2)\), \(I = (r_2, kr_2)\), and \(E = (r_1, kr_1)\).
   - Shoelace formula:
     \[
     [\triangle PIE] = \frac{1}{2} \left| 2kr_1 + kr_1r_2 + 2r_2 - (2r_1 + kr_1r_2 + 2kr_2) \right|
     \]
     \[
     = \frac{1}{2} \left| 2kr_1 - 2r_2 - 2r_1 - 2kr_2 \right|
     \]
     \[
     = \frac{1}{2} \left| 2(k - 1)(r_1 - r_2) \right|
     \]
     \[
     = |(k - 1)(r_1 - r_2)|
     \]

7. **Simplify \(r_1 - r_2\):**
   - \(r_1 - r_2 = \frac{2\sqrt{k + 1}}{k^2 + 1}\).

8. **Expression for the area:**
   - \[
     [\triangle PIE] = (k - 1) \left( \frac{2\sqrt{k + 1}}{k^2 + 1} \right)
     \]

9. **Maximize the area:**
   - Let \(a(x) = (x - 1) \left( \frac{2\sqrt{x + 1}}{x^2 + 1} \right)\).
   - Differentiate \(a(x)\) and find the critical points:
     \[
     a'(x) = \frac{d}{dx} \left( (x - 1) \left( \frac{2\sqrt{x + 1}}{x^2 + 1} \right) \right)
     \]
   - Use the product rule and quotient rule to find \(a'(x)\):
     \[
     a'(x) = \left( \frac{2\sqrt{x + 1}}{x^2 + 1} \right) + (x - 1) \left( \frac{d}{dx} \left( \frac{2\sqrt{x + 1}}{x^2 + 1} \right) \right)
     \]
   - Solve \(a'(x) = 0\) to find the critical points.

10. **Find the maximum value:**
    - The critical points are \(x = 2 \pm \sqrt{3}\).
    - Since \(k > 1\), choose \(k = 2 + \sqrt{3}\).

11. **Calculate the maximum area:**
    - \[
      a(2 + \sqrt{3}) = (2 + \sqrt{3} - 1) \left( \frac{2\sqrt{2(2 + \sqrt{3})}}{(2 + \sqrt{3})^2 + 1} \right)
      \]
    - Simplify to find:
      \[
      a(2 + \sqrt{3}) = 1
      \]
    - Scale back by \(\frac{1}{4}\) due to the initial scaling of the square:
      \[
      \boxed{\frac{1}{4}}
      \]