Compute the square of the distance between the incenter (center of the inscribed circle) and circumcenter (center of the circumscribed circle) of a 30-60-90 right triangle with hypotenuse of length 2.
1. **Identify the properties of the 30-60-90 triangle:**
   - In a 30-60-90 triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\).
   - Given the hypotenuse is 2, the other sides are:
     \[
     \text{Shorter leg} = 1, \quad \text{Longer leg} = \sqrt{3}
     \]

2. **Calculate the circumradius \(R\):**
   - The circumradius \(R\) of a right triangle is half the hypotenuse.
     \[
     R = \frac{\text{Hypotenuse}}{2} = \frac{2}{2} = 1
     \]

3. **Calculate the inradius \(r\):**
   - The area \(A\) of the triangle is:
     \[
     A = \frac{1}{2} \times \text{Shorter leg} \times \text{Longer leg} = \frac{1}{2} \times 1 \times \sqrt{3} = \frac{\sqrt{3}}{2}
     \]
   - The semiperimeter \(s\) is:
     \[
     s = \frac{1 + \sqrt{3} + 2}{2} = \frac{3 + \sqrt{3}}{2}
     \]
   - Using the formula \(A = r \cdot s\), solve for \(r\):
     \[
     \frac{\sqrt{3}}{2} = r \cdot \frac{3 + \sqrt{3}}{2} \implies r = \frac{\sqrt{3}}{3 + \sqrt{3}}
     \]
   - Simplify \(r\):
     \[
     r = \frac{\sqrt{3}}{3 + \sqrt{3}} \cdot \frac{3 - \sqrt{3}}{3 - \sqrt{3}} = \frac{3\sqrt{3} - 3}{(3 + \sqrt{3})(3 - \sqrt{3})} = \frac{3\sqrt{3} - 3}{9 - 3} = \frac{3\sqrt{3} - 3}{6} = \frac{\sqrt{3} - 1}{2}
     \]

4. **Use the formula \(d^2 = R(R - 2r)\):**
   - Substitute \(R = 1\) and \(r = \frac{\sqrt{3} - 1}{2}\):
     \[
     d^2 = 1 \left(1 - 2 \cdot \frac{\sqrt{3} - 1}{2}\right) = 1 \left(1 - (\sqrt{3} - 1)\right) = 1 \left(1 - \sqrt{3} + 1\right) = 1 \left(2 - \sqrt{3}\right) = 2 - \sqrt{3}
     \]

The final answer is \(\boxed{2 - \sqrt{3}}\).