SAT1000 - P838

The intersection of $x^2 + \frac{y^2}{2} = 1$ and $y = -\sqrt{2}x + 1$ is at $A(\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}, \frac{1}{2} + \frac{\sqrt{3}}{2})$ and $B(\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}, \frac{1}{2} - \frac{\sqrt{3}}{2})$ .

Since $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OP} = 0$ , $A_x + B_x + P_x = 0$ and $A_y + B_y + P_y = 0$ , so $\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} + P_x = 0$ and $\frac{1}{2} + \frac{\sqrt{3}}{2} + \frac{1}{2} - \frac{\sqrt{3}}{2} + P_y = 0$ , which solves to $P(-\frac{\sqrt{2}}{2}, -1)$ . Since $Q$ is symmetrical to $P$ about $O$ , $Q$ has coordinates $Q(\frac{\sqrt{2}}{2}, 1)$ .

The center $C$ of the circle with $A$ , $P$ , $B$ , and $Q$ is on the perpendicular bisector of $PQ$ , which is $y = -\frac{\sqrt{2}}{2}x$ , and the perpendicular bisector of $AB$ , which is $y = \frac{\sqrt{2}}{2}x + \frac{1}{4}$ , and these intersect at $C(-\frac{\sqrt{2}}{8}, \frac{1}{8})$ .

The radius $r$ of the circle can be found by finding the distance $CQ$ , which is $r = \sqrt{(\frac{\sqrt{2}}{2} - -\frac{\sqrt{2}}{8})^2 + (1 - \frac{1}{8})^2} = \frac{3\sqrt{11}}{8}$ .

Therefore, $\lfloor 1000r \rfloor = \boxed{1243}$ .

The radius of the circle is $\dfrac{3\sqrt {11}}{8}\approx 1.2437342963833$