SAT1000 - P839

Let $A$ have coordinates $A(\frac{p^2}{4}, p)$ . Then the equation of line $AK$ is $y = \frac{4p}{p^2 + 4}(x + 1)$ , and the other intersection with parabola $y^2 = 4x$ is $B(\frac{4}{p^2}, \frac{4}{p})$ . As the focus of parabola $y^2 = 4x$ , $F$ has coordinates $F(1, 0)$ , and by symmetry, $D$ has coordinates $D(\frac{p^2}{4}, -p)$ .

Since $\overrightarrow{FA} \cdot \overrightarrow{FB} = \frac{8}{9}$ , $(\frac{p^2}{4} - 1)(\frac{4}{p^2} - 1) = p \cdot \frac{4}{p} = \frac{8}{9}$ , which rearranges to $9p^4 - 184p^2 + 144 = 0$ , or $p = \frac{1}{3}(\pm 8 \pm 2\sqrt{7})$ . Without loss of generality, and to match the given picture, choose $p = \frac{1}{3}(8 - 2\sqrt{7})$ .

Substituting $p = \frac{1}{3}(8 - 2\sqrt{7})$ , the coordinates for $B$ and $D$ are now $B(\frac{1}{9}(23 + 8\sqrt{7}), \frac{2}{3}(4 + \sqrt{7}))$ and $D(\frac{1}{9}(23 - 8\sqrt{7}), \frac{1}{3}(-8 + 2\sqrt{7}))$ .

Using the distance formula, $BD = \frac{64}{9}$ , $BK = \frac{10}{9}\sqrt{23 + 8\sqrt{7}}$ , and $DK = \frac{10}{9}\sqrt{23 - 8\sqrt{7}}$

The center $(a, b)$ of the incircle of $\triangle BDK$ is $a = \frac{BD \cdot K_x + BK \cdot D_x + DK \cdot B_x}{BD + BK + DK}$ and $b = \frac{BD \cdot K_y + BK \cdot D_y + DK \cdot B_y}{BD + BK + DK}$ , which solves to $a = \frac{1}{9}$ and $b = 0$ .

The radius $r$ of the incircle of $\triangle BDK$ is $r = \frac{\sqrt{s(s - BD)(s - BK)(s - DK)}}{s}$ where $s = \frac{1}{2}(BD + BK + DK)$ , which solves to $r = \frac{2}{3}$ .

Therefore, $\lfloor 1000(a + b + r) \rfloor = \boxed{777}$ .

Centre of the circle is at $(\dfrac{1}{9},0)$ , and it's radius is $\dfrac{2}{3}$ .

So the expression is equal to $\lfloor 1000\left (\dfrac{1}{9}+0+\dfrac{2}{3}\right )\rfloor =\boxed {777}$ .