SAT1000 - P810

Let $D$ be the intersection of $OP$ and $AB$ , and since $OP$ has an equation of $y = \frac{1}{2}$ , let its coordinates be $D(p, \frac{1}{2}p)$ .

Let the slope of $AB$ be $m$ . Then the equation of the line $AB$ through $D(p, \frac{1}{2}p)$ is $y = mx - mp + \frac{1}{2}p$ , and $x$ -coordinates of $A$ and $B$ at the intersection of $y = mx - mp + \frac{1}{2}p$ and $\frac{x^2}{4} + \frac{y^2}{3} = 1$ are $A_x = \frac{4m^2p - 2mp - \sqrt{(-12m^2 + 12m - 3)p^2 + 48m^2 + 36}}{4m^2 + 3}$ and $B_x = \frac{4m^2p - 2mp + \sqrt{(-12m^2 + 12m - 3)p^2 + 48m^2 + 36}}{4m^2 + 3}$ .

Since $D$ is the midpoint of $A$ and $B$ , $\frac{1}{2}(A_x + B_x) = D_x$ , or $\frac{1}{2}(\frac{4m^2p - 2mp - \sqrt{(-12m^2 + 12m - 3)p^2 + 48m^2 + 36}}{4m^2 + 3} + \frac{4m^2p - 2mp + \sqrt{(-12m^2 + 12m - 3)p^2 + 48m^2 + 36}}{4m^2 + 3}) = p$ , which solves to $m = -\frac{3}{2}$ .

Substituting $m = -\frac{3}{2}$ , the coordinates $A$ and $B$ simplify to $A(p - \frac{1}{3}\sqrt{-3p^2 + 9}, \frac{1}{2}p + \frac{1}{2}\sqrt{-3p^2 + 9})$ and $B(p + \frac{1}{3}\sqrt{-3p^2 + 9}, \frac{1}{2}p - \frac{1}{2}\sqrt{-3p^2 + 9})$ .

The area of $\triangle APB$ is then $A_{\triangle APB} = \frac{1}{2}((A_x - 2)(B_y - 1) - (B_x - 2)(A_y - 1))$ , which after substituting the above values and simplifying solves to $A_{\triangle APB} = \frac{2}{3}(2 - p)\sqrt{-3p^2 + 9}$ .

Since $A''_{\triangle APB} \leq 0$ , the area of the triangle reaches a maximum when the derivative is equal to zero, or $A'_{\triangle APB} = \frac{4p^2 - 4p - 6}{\sqrt{-3p^2 + 9}} = 0$ , which solves to $p = \frac{1}{2}(1 - \sqrt{7})$ for $p < 0$ .

Therefore, after substituting $m = -\frac{3}{2}$ and $p = \frac{1}{2}(1 - \sqrt{7})$ into the equation of the line $AB$ given above as $y = mx - mp + \frac{1}{2}p$ , it solves to $y = -\frac{3}{2}x + 1 - \sqrt{7}$ , so $k = -\frac{3}{2}$ , $b = 1 - \sqrt{7}$ , and $\lfloor 1000(k + b) \rfloor = \boxed{-3146}$ .