Dynamic Geometry: P15

Let the pink point be $(p, p^3)$ .

The normal through this point is $y = -\frac{1}{3p^2}x + p^3 + \frac{1}{3p}$ , which means the black point is at $(0, p^3 + \frac{1}{3p})$ .

The ratio of the $y$ -coordinate of the black point to the $y$ -coordinate of the pink point is $\cfrac{p^3 + \frac{1}{3p}}{p^3} = \cfrac{244}{243}$ , which solves to $p = 3$ .

Therefore, the pink point is at $(p, p^3) = (3, 27)$ and the black point is at $(0, p^3 + \frac{1}{3p}) = (0, \frac{244}{9})$

The radius of the circle is the distance between $(3, 27)$ and $(0, \frac{244}{9})$ , which is $r = \frac{\sqrt{730}}{9}$ .

The area of the circle is then $A = \pi r^2 = \pi \cdot \frac{730}{81}$ , so $a = 730$ , $b = 81$ , and $a - \sqrt{b} - \lceil \pi \rceil = \boxed{717}$ .