SAT1000 - P842

Since the equation always holds as $l$ rotates, let's examine two convenient slopes for $l$ , namely, $m = 0$ and $m = \frac{1}{2}$ .

By the converse of the angle bisector theorem, the equation $\frac{|QA|}{|QB|} = \frac{|PA|}{|PB|}$ means that $\angle PQA = \angle PQB$ . Therefore, for $m = 0$ , where $B$ , $P$ , and $A$ are evenly spaced out on a horizontal line with $P$ on the $y$ -axis, $Q$ must also lie on the $y$ -axis, so let the coordinates of $Q$ be $Q(0, q)$ .

Then for $m = \frac{1}{2}$ , line $l$ through $P(0, 1)$ would have an equation of $y = \frac{1}{2}x + 1$ , and would intersect the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$ at $B(-2, 0)$ and $A(\frac{2}{3}, \frac{4}{3})$ . Then by the distance formula, $PB = \sqrt{5}$ , $PA = \frac{\sqrt{5}}{3}$ , $QB = \sqrt{4 + q^2}$ , $QA = \frac{1}{3}\sqrt{9q^2 - 24q + 20}$ , and substituting these into $\frac{|QA|}{|QB|} = \frac{|PA|}{|PB|}$ and solving gives $q^2 - 3q + 2 = 0$ which solves to $q = 1$ or $q = 2$ . It can't be $q = 1$ , because then $P$ and $Q$ would be the same point, so $q = 2$ , which means the coordinates of $Q$ are $(0, 2)$ .

Therefore, $x_0 = 0$ , $y_0 = 2$ , and $\lfloor 1000(2y_0 - x_0) \rfloor = \boxed{4000}$ .