SAT1000 - P819

If $F(1, 0)$ is the focus of the ellipse, then $b^2 = a^2 - 1$ , which makes the equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{a^2 - 1} = 1$ .

The inequality $|OA|^2 + |OB|^2 < |AB|^2$ always holds when $\angle AOB$ is always an obtuse angle. $\angle AOB$ will be at its smallest when $A$ and $B$ are on the vertical line $x = 1$ , and $\angle AOB$ will start to be a right angle (not obtuse) when $y = 1$ . Substituting $x = 1$ and $y = 1$ into the equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{a^2 - 1} = 1$ gives $\frac{1}{a^2} + \frac{1}{a^2 - 1} = 1$ , which simplifies to $a^2 - a - 1$ and solves to $a = \frac{1 + \sqrt{5}}{2} = \phi$ .

Therefore, the inequality $|OA|^2 + |OB|^2 < |AB|^2$ always holds when $a > \phi$ , so $l = \phi$ , and $\lfloor 1000\phi \rfloor = \boxed{1618}$ .

Let the position coordinates of $A$ be $(h, k)$ . Then those of $B$ are

$(\frac{2a^2-h(a^2+1)}{a^2-2h+1},\frac{k(1-a^2)}{a^2-2h+1})$

The given condition yields

$(3a^2-1)h^2-2a^4h+(a^2-a)^2<0$

So the discriminant of the equation must be negative definite :

$a^6<(3a^2-1)(a^2-1)^2\implies 2a^6-7a^4+5a^2-1>0\implies a^2>2.618$

Hence $a>1.618\implies l=1.618\implies \lfloor 1000l\rfloor =\boxed {1618}$