Dynamic Geometry: P14

Let the tangent point of the circle and parabola on the right side be $(p, p^2)$ .

The normal through this point is $y = -\frac{1}{2p}x + p^2 + \frac{1}{2}$ , which means the center of the circle is at $(0, p^2 + \frac{1}{2})$ .

The radius of the circle is the distance between $(p, p^2)$ and $(0, p^2 + \frac{1}{2})$ , which is $r = \sqrt{p^2 + \frac{1}{4}}$ .

Therefore, the green point is at $(0, p^2 + \frac{1}{2} + \sqrt{p^2 + \frac{1}{4}})$ .

The lower right side of the square has a slope of $1$ and passes through the green point, so it follows the equation $y = x + p^2 + \frac{1}{2} + \sqrt{p^2 + \frac{1}{4}}$ .

This equation intersects $y = x^2$ at an $x$ -coordinate of $q = \frac{1}{2}(\sqrt{4p^2 + 2\sqrt{4p^2 + 1} + 3} + 1)$ .

The side of the square is then $s = \sqrt{2}q = \frac{\sqrt{2}}{2}(\sqrt{4p^2 + 2\sqrt{4p^2 + 1} + 3} + 1)$ .

The ratio of the side of the blue square to the radius of the red circle is $\cfrac{r}{s} = \cfrac{\frac{\sqrt{2}}{2}(\sqrt{4p^2 + 2\sqrt{4p^2 + 1} + 3} + 1)}{\sqrt{p^2 + \frac{1}{4}}} = \sqrt{2} \cdot \cfrac{15 + \sqrt{1249}}{17}$ , which solves to $p = \frac{4}{15}$ .

Therefore, the $y$ -coordinate of the green point is $p^2 + \frac{1}{2} + \sqrt{p^2 + \frac{1}{4}} = (\frac{4}{15})^2 + \frac{1}{2} + \sqrt{(\frac{4}{15})^2 + \frac{1}{4}} = \frac{256}{225}$ , so $a = 256$ , $b = 225$ , and $\sqrt{a} - \sqrt{b} = \boxed{1}$ .