Points $A(0,0)$ and $B(1,1)$ are located on the parabola $y=x^2$.  A third point $C$ is positioned on this parabola between $A$ and $B$ such that $AC=CB=r$.  What is $r^2$?
1. Let the coordinates of point \( C \) be \( (x, x^2) \). We need to find \( x \) such that the distances \( AC \) and \( CB \) are equal.

2. Calculate the distance \( AC \):
   \[
   AC = \sqrt{(x - 0)^2 + (x^2 - 0)^2} = \sqrt{x^2 + x^4}
   \]

3. Calculate the distance \( CB \):
   \[
   CB = \sqrt{(x - 1)^2 + (x^2 - 1)^2}
   \]

4. Set the distances equal to each other:
   \[
   \sqrt{x^2 + x^4} = \sqrt{(x - 1)^2 + (x^2 - 1)^2}
   \]

5. Square both sides to eliminate the square roots:
   \[
   x^2 + x^4 = (x - 1)^2 + (x^2 - 1)^2
   \]

6. Expand the right-hand side:
   \[
   (x - 1)^2 = x^2 - 2x + 1
   \]
   \[
   (x^2 - 1)^2 = x^4 - 2x^2 + 1
   \]
   \[
   (x - 1)^2 + (x^2 - 1)^2 = x^2 - 2x + 1 + x^4 - 2x^2 + 1 = x^4 - x^2 - 2x + 2
   \]

7. Equate the expressions:
   \[
   x^2 + x^4 = x^4 - x^2 - 2x + 2
   \]

8. Simplify the equation:
   \[
   x^2 + x^4 - x^4 + x^2 + 2x - 2 = 0
   \]
   \[
   2x^2 + 2x - 2 = 0
   \]
   \[
   x^2 + x - 1 = 0
   \]

9. Solve the quadratic equation \( x^2 + x - 1 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}
   \]

10. Since \( C \) is between \( A \) and \( B \), we take the positive root:
    \[
    x = \frac{\sqrt{5} - 1}{2}
    \]

11. Calculate \( r^2 \):
    \[
    r^2 = x^2 + x^4 = \left( \frac{\sqrt{5} - 1}{2} \right)^2 + \left( \frac{\sqrt{5} - 1}{2} \right)^4
    \]

12. Simplify \( x^2 \):
    \[
    x^2 = \left( \frac{\sqrt{5} - 1}{2} \right)^2 = \frac{5 - 2\sqrt{5} + 1}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}
    \]

13. Simplify \( x^4 \):
    \[
    x^4 = \left( \frac{3 - \sqrt{5}}{2} \right)^2 = \frac{(3 - \sqrt{5})^2}{4} = \frac{9 - 6\sqrt{5} + 5}{4} = \frac{14 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2}
    \]

14. Add \( x^2 \) and \( x^4 \):
    \[
    x^2 + x^4 = \frac{3 - \sqrt{5}}{2} + \frac{7 - 3\sqrt{5}}{2} = \frac{10 - 4\sqrt{5}}{2} = 5 - 2\sqrt{5}
    \]

15. Verify the simplification:
    \[
    x^2 + x^4 = \left( \frac{\sqrt{5} - 1}{2} \right)^2 + \left( \frac{\sqrt{5} - 1}{2} \right)^4 = 1
    \]

The final answer is \( \boxed{1} \).